Finding sum to infinity [duplicate]
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This question already has an answer here:
What's the value of $sumlimits_{k=1}^{infty}frac{k^2}{k!}$?
5 answers
I am trying to find what this value will converge to
$$sum_{n = 1}^{ infty}frac{n^2}{n!}$$
I tried using the Taylor series for $e^x$ but couldn’t figure out how to manipulate it to get the above expression, can someone help me out.
Edit: I have seen the solution, the manipulation required didn’t come to me, is there any resource that you guys can tell me about where I can find/practice more questions like this?
calculus sequences-and-series taylor-expansion
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marked as duplicate by lab bhattacharjee
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27 secs ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
What's the value of $sumlimits_{k=1}^{infty}frac{k^2}{k!}$?
5 answers
I am trying to find what this value will converge to
$$sum_{n = 1}^{ infty}frac{n^2}{n!}$$
I tried using the Taylor series for $e^x$ but couldn’t figure out how to manipulate it to get the above expression, can someone help me out.
Edit: I have seen the solution, the manipulation required didn’t come to me, is there any resource that you guys can tell me about where I can find/practice more questions like this?
calculus sequences-and-series taylor-expansion
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marked as duplicate by lab bhattacharjee
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27 secs ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
In general, $$ sum_{n=0}^{infty} frac{n^k}{n!} = B_k e, $$ where $B_k$ is the $k$-th Bell-number. This is called the Dobinski's formula. This can be computed by expanding $n^k$ into the sum of falling factorials as in Olivier Oloa's answer.
$endgroup$
– Sangchul Lee
1 hour ago
$begingroup$
If $R>0$ and a power series $sum_{j=0}^{infty}a_jx^j$ converges to $f(x)$ for each $xin (-R,R)$ then $f$ is differentiable on $(-R,R)$ and $sum_{j=1}^{infty}ja_jx^{j-1} =f'(x)$ for all $(-R,R).$ Notice also that the previous sentence now applies with $f$ replaced by $f'$. For example, for $|x|<2$ let $f(x) =1/(1-x)=sum_{j=0}^{infty}x^j .$ Then $sum_{j=1}^{infty}jx^{j-1}=f'(x)=1/(1-x)^2,$ and $sum_{j=2}^{infty}j(j-1)x^{j-2}=f''(x)=2/(1-x)^3.$
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– DanielWainfleet
32 mins ago
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See : math.stackexchange.com/questions/1711318/…
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– lab bhattacharjee
6 secs ago
add a comment |
$begingroup$
This question already has an answer here:
What's the value of $sumlimits_{k=1}^{infty}frac{k^2}{k!}$?
5 answers
I am trying to find what this value will converge to
$$sum_{n = 1}^{ infty}frac{n^2}{n!}$$
I tried using the Taylor series for $e^x$ but couldn’t figure out how to manipulate it to get the above expression, can someone help me out.
Edit: I have seen the solution, the manipulation required didn’t come to me, is there any resource that you guys can tell me about where I can find/practice more questions like this?
calculus sequences-and-series taylor-expansion
$endgroup$
This question already has an answer here:
What's the value of $sumlimits_{k=1}^{infty}frac{k^2}{k!}$?
5 answers
I am trying to find what this value will converge to
$$sum_{n = 1}^{ infty}frac{n^2}{n!}$$
I tried using the Taylor series for $e^x$ but couldn’t figure out how to manipulate it to get the above expression, can someone help me out.
Edit: I have seen the solution, the manipulation required didn’t come to me, is there any resource that you guys can tell me about where I can find/practice more questions like this?
This question already has an answer here:
What's the value of $sumlimits_{k=1}^{infty}frac{k^2}{k!}$?
5 answers
calculus sequences-and-series taylor-expansion
calculus sequences-and-series taylor-expansion
edited 1 hour ago
user601297
asked 1 hour ago
user601297user601297
38019
38019
marked as duplicate by lab bhattacharjee
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27 secs ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by lab bhattacharjee
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27 secs ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
In general, $$ sum_{n=0}^{infty} frac{n^k}{n!} = B_k e, $$ where $B_k$ is the $k$-th Bell-number. This is called the Dobinski's formula. This can be computed by expanding $n^k$ into the sum of falling factorials as in Olivier Oloa's answer.
$endgroup$
– Sangchul Lee
1 hour ago
$begingroup$
If $R>0$ and a power series $sum_{j=0}^{infty}a_jx^j$ converges to $f(x)$ for each $xin (-R,R)$ then $f$ is differentiable on $(-R,R)$ and $sum_{j=1}^{infty}ja_jx^{j-1} =f'(x)$ for all $(-R,R).$ Notice also that the previous sentence now applies with $f$ replaced by $f'$. For example, for $|x|<2$ let $f(x) =1/(1-x)=sum_{j=0}^{infty}x^j .$ Then $sum_{j=1}^{infty}jx^{j-1}=f'(x)=1/(1-x)^2,$ and $sum_{j=2}^{infty}j(j-1)x^{j-2}=f''(x)=2/(1-x)^3.$
$endgroup$
– DanielWainfleet
32 mins ago
$begingroup$
See : math.stackexchange.com/questions/1711318/…
$endgroup$
– lab bhattacharjee
6 secs ago
add a comment |
$begingroup$
In general, $$ sum_{n=0}^{infty} frac{n^k}{n!} = B_k e, $$ where $B_k$ is the $k$-th Bell-number. This is called the Dobinski's formula. This can be computed by expanding $n^k$ into the sum of falling factorials as in Olivier Oloa's answer.
$endgroup$
– Sangchul Lee
1 hour ago
$begingroup$
If $R>0$ and a power series $sum_{j=0}^{infty}a_jx^j$ converges to $f(x)$ for each $xin (-R,R)$ then $f$ is differentiable on $(-R,R)$ and $sum_{j=1}^{infty}ja_jx^{j-1} =f'(x)$ for all $(-R,R).$ Notice also that the previous sentence now applies with $f$ replaced by $f'$. For example, for $|x|<2$ let $f(x) =1/(1-x)=sum_{j=0}^{infty}x^j .$ Then $sum_{j=1}^{infty}jx^{j-1}=f'(x)=1/(1-x)^2,$ and $sum_{j=2}^{infty}j(j-1)x^{j-2}=f''(x)=2/(1-x)^3.$
$endgroup$
– DanielWainfleet
32 mins ago
$begingroup$
See : math.stackexchange.com/questions/1711318/…
$endgroup$
– lab bhattacharjee
6 secs ago
$begingroup$
In general, $$ sum_{n=0}^{infty} frac{n^k}{n!} = B_k e, $$ where $B_k$ is the $k$-th Bell-number. This is called the Dobinski's formula. This can be computed by expanding $n^k$ into the sum of falling factorials as in Olivier Oloa's answer.
$endgroup$
– Sangchul Lee
1 hour ago
$begingroup$
In general, $$ sum_{n=0}^{infty} frac{n^k}{n!} = B_k e, $$ where $B_k$ is the $k$-th Bell-number. This is called the Dobinski's formula. This can be computed by expanding $n^k$ into the sum of falling factorials as in Olivier Oloa's answer.
$endgroup$
– Sangchul Lee
1 hour ago
$begingroup$
If $R>0$ and a power series $sum_{j=0}^{infty}a_jx^j$ converges to $f(x)$ for each $xin (-R,R)$ then $f$ is differentiable on $(-R,R)$ and $sum_{j=1}^{infty}ja_jx^{j-1} =f'(x)$ for all $(-R,R).$ Notice also that the previous sentence now applies with $f$ replaced by $f'$. For example, for $|x|<2$ let $f(x) =1/(1-x)=sum_{j=0}^{infty}x^j .$ Then $sum_{j=1}^{infty}jx^{j-1}=f'(x)=1/(1-x)^2,$ and $sum_{j=2}^{infty}j(j-1)x^{j-2}=f''(x)=2/(1-x)^3.$
$endgroup$
– DanielWainfleet
32 mins ago
$begingroup$
If $R>0$ and a power series $sum_{j=0}^{infty}a_jx^j$ converges to $f(x)$ for each $xin (-R,R)$ then $f$ is differentiable on $(-R,R)$ and $sum_{j=1}^{infty}ja_jx^{j-1} =f'(x)$ for all $(-R,R).$ Notice also that the previous sentence now applies with $f$ replaced by $f'$. For example, for $|x|<2$ let $f(x) =1/(1-x)=sum_{j=0}^{infty}x^j .$ Then $sum_{j=1}^{infty}jx^{j-1}=f'(x)=1/(1-x)^2,$ and $sum_{j=2}^{infty}j(j-1)x^{j-2}=f''(x)=2/(1-x)^3.$
$endgroup$
– DanielWainfleet
32 mins ago
$begingroup$
See : math.stackexchange.com/questions/1711318/…
$endgroup$
– lab bhattacharjee
6 secs ago
$begingroup$
See : math.stackexchange.com/questions/1711318/…
$endgroup$
– lab bhattacharjee
6 secs ago
add a comment |
3 Answers
3
active
oldest
votes
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One may write
begin{align}
sum_{n = 1}^{ infty}frac{n^2}{n!}&=sum_{n = 1}^{ infty}frac{n(n-1)+n}{n!}
\\&=sum_{n = 1}^{ infty}frac{n(n-1)}{n!}+sum_{n = 1}^{ infty}frac{n}{n!}
\\&=sum_{n = 2}^{ infty}frac{1}{(n-2)!}+sum_{n = 1}^{ infty}frac{1}{(n-1)!}
end{align}Can you take it from here?
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2
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Further hints: Since $e^1 = sum_{n=0}^{infty} frac{1}{n!}$, we have $$ sum_{n=2}^{infty} frac{1}{(n-2)!}= 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ and $$ sum_{n=1}^{infty} frac{1}{(n-1)!} = 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ So, ans: $boxed{ 2 mathrm{e} } $
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– Jimmy Sabater
1 hour ago
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Ok both expressions sum to $e$, i get it, thanks a lot
$endgroup$
– user601297
1 hour ago
add a comment |
$begingroup$
$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Taking $d/dx$ on both sides,
$$e^x=sum_{ngeq1}frac{n}{n!}x^{n-1}$$
Multiplying both sides by $x$,
$$xe^x=sum_{ngeq1}frac{n}{n!}x^n$$
Then taking $d/dx$ on both sides again,
$$(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^{n-1}$$
Then plug in $x=1$:
$$sum_{ngeq1}frac{n^2}{n!}=2e$$
Edit
This is a really neat trick that is widely used. Whenever you see an $n^k$ in the numerator, think applying the $xfrac{d}{dx}$ operator $k$ times. Example:
$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Apply $xfrac{d}{dx}$:
$$xe^x=sum_{ngeq1}frac{x}{n!}x^n$$
Apply $xfrac{d}{dx}$:
$$x(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^n$$
The pattern continues:
$$left(xfrac{d}{dx}right)^k[e^x-1]=sum_{ngeq1}frac{n^k}{n!}x^n$$
A similar thing can be done with integration. Example:
Evaluate $$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}$$
Start by recalling that (use geometric series)
$$frac1{1+t^2}=sum_{ngeq0}(-1)^nt^{2n}$$
Then integrate both sides from $0$ to $x$ to get
$$arctan x=sum_{ngeq0}frac{(-1)^n}{2n+1}x^{2n+1}$$
integrate both sides from $0$ to $1$ now to produce
$$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}=fracpi4-frac12log2$$
$endgroup$
1
$begingroup$
Amazing, this is exactly what I was looking for
$endgroup$
– user601297
1 hour ago
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@user601297 you are very welcome :)
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– clathratus
1 hour ago
add a comment |
$begingroup$
Just to give a slightly different approach,
$$sum_{n=1}^infty{n^2over n!}=sum_{n=1}^infty{nover(n-1)!}=sum_{m=0}^infty{m+1over m!}=sum_{m=0}^infty{mover m!}+e=sum_{m=1}^infty{mover m!}+e=sum_{m=1}^infty{1over(m-1)!}+e=sum_{k=0}^infty{1over k!}+e=e+e$$
The trick in reading this is to note what changes across each equal sign as you proceed from left to right and understand what justifies the equality for each change.
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One may write
begin{align}
sum_{n = 1}^{ infty}frac{n^2}{n!}&=sum_{n = 1}^{ infty}frac{n(n-1)+n}{n!}
\\&=sum_{n = 1}^{ infty}frac{n(n-1)}{n!}+sum_{n = 1}^{ infty}frac{n}{n!}
\\&=sum_{n = 2}^{ infty}frac{1}{(n-2)!}+sum_{n = 1}^{ infty}frac{1}{(n-1)!}
end{align}Can you take it from here?
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2
$begingroup$
Further hints: Since $e^1 = sum_{n=0}^{infty} frac{1}{n!}$, we have $$ sum_{n=2}^{infty} frac{1}{(n-2)!}= 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ and $$ sum_{n=1}^{infty} frac{1}{(n-1)!} = 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ So, ans: $boxed{ 2 mathrm{e} } $
$endgroup$
– Jimmy Sabater
1 hour ago
$begingroup$
Ok both expressions sum to $e$, i get it, thanks a lot
$endgroup$
– user601297
1 hour ago
add a comment |
$begingroup$
One may write
begin{align}
sum_{n = 1}^{ infty}frac{n^2}{n!}&=sum_{n = 1}^{ infty}frac{n(n-1)+n}{n!}
\\&=sum_{n = 1}^{ infty}frac{n(n-1)}{n!}+sum_{n = 1}^{ infty}frac{n}{n!}
\\&=sum_{n = 2}^{ infty}frac{1}{(n-2)!}+sum_{n = 1}^{ infty}frac{1}{(n-1)!}
end{align}Can you take it from here?
$endgroup$
2
$begingroup$
Further hints: Since $e^1 = sum_{n=0}^{infty} frac{1}{n!}$, we have $$ sum_{n=2}^{infty} frac{1}{(n-2)!}= 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ and $$ sum_{n=1}^{infty} frac{1}{(n-1)!} = 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ So, ans: $boxed{ 2 mathrm{e} } $
$endgroup$
– Jimmy Sabater
1 hour ago
$begingroup$
Ok both expressions sum to $e$, i get it, thanks a lot
$endgroup$
– user601297
1 hour ago
add a comment |
$begingroup$
One may write
begin{align}
sum_{n = 1}^{ infty}frac{n^2}{n!}&=sum_{n = 1}^{ infty}frac{n(n-1)+n}{n!}
\\&=sum_{n = 1}^{ infty}frac{n(n-1)}{n!}+sum_{n = 1}^{ infty}frac{n}{n!}
\\&=sum_{n = 2}^{ infty}frac{1}{(n-2)!}+sum_{n = 1}^{ infty}frac{1}{(n-1)!}
end{align}Can you take it from here?
$endgroup$
One may write
begin{align}
sum_{n = 1}^{ infty}frac{n^2}{n!}&=sum_{n = 1}^{ infty}frac{n(n-1)+n}{n!}
\\&=sum_{n = 1}^{ infty}frac{n(n-1)}{n!}+sum_{n = 1}^{ infty}frac{n}{n!}
\\&=sum_{n = 2}^{ infty}frac{1}{(n-2)!}+sum_{n = 1}^{ infty}frac{1}{(n-1)!}
end{align}Can you take it from here?
answered 1 hour ago
Olivier OloaOlivier Oloa
108k17176293
108k17176293
2
$begingroup$
Further hints: Since $e^1 = sum_{n=0}^{infty} frac{1}{n!}$, we have $$ sum_{n=2}^{infty} frac{1}{(n-2)!}= 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ and $$ sum_{n=1}^{infty} frac{1}{(n-1)!} = 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ So, ans: $boxed{ 2 mathrm{e} } $
$endgroup$
– Jimmy Sabater
1 hour ago
$begingroup$
Ok both expressions sum to $e$, i get it, thanks a lot
$endgroup$
– user601297
1 hour ago
add a comment |
2
$begingroup$
Further hints: Since $e^1 = sum_{n=0}^{infty} frac{1}{n!}$, we have $$ sum_{n=2}^{infty} frac{1}{(n-2)!}= 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ and $$ sum_{n=1}^{infty} frac{1}{(n-1)!} = 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ So, ans: $boxed{ 2 mathrm{e} } $
$endgroup$
– Jimmy Sabater
1 hour ago
$begingroup$
Ok both expressions sum to $e$, i get it, thanks a lot
$endgroup$
– user601297
1 hour ago
2
2
$begingroup$
Further hints: Since $e^1 = sum_{n=0}^{infty} frac{1}{n!}$, we have $$ sum_{n=2}^{infty} frac{1}{(n-2)!}= 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ and $$ sum_{n=1}^{infty} frac{1}{(n-1)!} = 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ So, ans: $boxed{ 2 mathrm{e} } $
$endgroup$
– Jimmy Sabater
1 hour ago
$begingroup$
Further hints: Since $e^1 = sum_{n=0}^{infty} frac{1}{n!}$, we have $$ sum_{n=2}^{infty} frac{1}{(n-2)!}= 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ and $$ sum_{n=1}^{infty} frac{1}{(n-1)!} = 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ So, ans: $boxed{ 2 mathrm{e} } $
$endgroup$
– Jimmy Sabater
1 hour ago
$begingroup$
Ok both expressions sum to $e$, i get it, thanks a lot
$endgroup$
– user601297
1 hour ago
$begingroup$
Ok both expressions sum to $e$, i get it, thanks a lot
$endgroup$
– user601297
1 hour ago
add a comment |
$begingroup$
$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Taking $d/dx$ on both sides,
$$e^x=sum_{ngeq1}frac{n}{n!}x^{n-1}$$
Multiplying both sides by $x$,
$$xe^x=sum_{ngeq1}frac{n}{n!}x^n$$
Then taking $d/dx$ on both sides again,
$$(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^{n-1}$$
Then plug in $x=1$:
$$sum_{ngeq1}frac{n^2}{n!}=2e$$
Edit
This is a really neat trick that is widely used. Whenever you see an $n^k$ in the numerator, think applying the $xfrac{d}{dx}$ operator $k$ times. Example:
$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Apply $xfrac{d}{dx}$:
$$xe^x=sum_{ngeq1}frac{x}{n!}x^n$$
Apply $xfrac{d}{dx}$:
$$x(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^n$$
The pattern continues:
$$left(xfrac{d}{dx}right)^k[e^x-1]=sum_{ngeq1}frac{n^k}{n!}x^n$$
A similar thing can be done with integration. Example:
Evaluate $$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}$$
Start by recalling that (use geometric series)
$$frac1{1+t^2}=sum_{ngeq0}(-1)^nt^{2n}$$
Then integrate both sides from $0$ to $x$ to get
$$arctan x=sum_{ngeq0}frac{(-1)^n}{2n+1}x^{2n+1}$$
integrate both sides from $0$ to $1$ now to produce
$$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}=fracpi4-frac12log2$$
$endgroup$
1
$begingroup$
Amazing, this is exactly what I was looking for
$endgroup$
– user601297
1 hour ago
$begingroup$
@user601297 you are very welcome :)
$endgroup$
– clathratus
1 hour ago
add a comment |
$begingroup$
$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Taking $d/dx$ on both sides,
$$e^x=sum_{ngeq1}frac{n}{n!}x^{n-1}$$
Multiplying both sides by $x$,
$$xe^x=sum_{ngeq1}frac{n}{n!}x^n$$
Then taking $d/dx$ on both sides again,
$$(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^{n-1}$$
Then plug in $x=1$:
$$sum_{ngeq1}frac{n^2}{n!}=2e$$
Edit
This is a really neat trick that is widely used. Whenever you see an $n^k$ in the numerator, think applying the $xfrac{d}{dx}$ operator $k$ times. Example:
$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Apply $xfrac{d}{dx}$:
$$xe^x=sum_{ngeq1}frac{x}{n!}x^n$$
Apply $xfrac{d}{dx}$:
$$x(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^n$$
The pattern continues:
$$left(xfrac{d}{dx}right)^k[e^x-1]=sum_{ngeq1}frac{n^k}{n!}x^n$$
A similar thing can be done with integration. Example:
Evaluate $$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}$$
Start by recalling that (use geometric series)
$$frac1{1+t^2}=sum_{ngeq0}(-1)^nt^{2n}$$
Then integrate both sides from $0$ to $x$ to get
$$arctan x=sum_{ngeq0}frac{(-1)^n}{2n+1}x^{2n+1}$$
integrate both sides from $0$ to $1$ now to produce
$$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}=fracpi4-frac12log2$$
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1
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Amazing, this is exactly what I was looking for
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– user601297
1 hour ago
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@user601297 you are very welcome :)
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– clathratus
1 hour ago
add a comment |
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$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Taking $d/dx$ on both sides,
$$e^x=sum_{ngeq1}frac{n}{n!}x^{n-1}$$
Multiplying both sides by $x$,
$$xe^x=sum_{ngeq1}frac{n}{n!}x^n$$
Then taking $d/dx$ on both sides again,
$$(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^{n-1}$$
Then plug in $x=1$:
$$sum_{ngeq1}frac{n^2}{n!}=2e$$
Edit
This is a really neat trick that is widely used. Whenever you see an $n^k$ in the numerator, think applying the $xfrac{d}{dx}$ operator $k$ times. Example:
$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Apply $xfrac{d}{dx}$:
$$xe^x=sum_{ngeq1}frac{x}{n!}x^n$$
Apply $xfrac{d}{dx}$:
$$x(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^n$$
The pattern continues:
$$left(xfrac{d}{dx}right)^k[e^x-1]=sum_{ngeq1}frac{n^k}{n!}x^n$$
A similar thing can be done with integration. Example:
Evaluate $$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}$$
Start by recalling that (use geometric series)
$$frac1{1+t^2}=sum_{ngeq0}(-1)^nt^{2n}$$
Then integrate both sides from $0$ to $x$ to get
$$arctan x=sum_{ngeq0}frac{(-1)^n}{2n+1}x^{2n+1}$$
integrate both sides from $0$ to $1$ now to produce
$$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}=fracpi4-frac12log2$$
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$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Taking $d/dx$ on both sides,
$$e^x=sum_{ngeq1}frac{n}{n!}x^{n-1}$$
Multiplying both sides by $x$,
$$xe^x=sum_{ngeq1}frac{n}{n!}x^n$$
Then taking $d/dx$ on both sides again,
$$(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^{n-1}$$
Then plug in $x=1$:
$$sum_{ngeq1}frac{n^2}{n!}=2e$$
Edit
This is a really neat trick that is widely used. Whenever you see an $n^k$ in the numerator, think applying the $xfrac{d}{dx}$ operator $k$ times. Example:
$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Apply $xfrac{d}{dx}$:
$$xe^x=sum_{ngeq1}frac{x}{n!}x^n$$
Apply $xfrac{d}{dx}$:
$$x(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^n$$
The pattern continues:
$$left(xfrac{d}{dx}right)^k[e^x-1]=sum_{ngeq1}frac{n^k}{n!}x^n$$
A similar thing can be done with integration. Example:
Evaluate $$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}$$
Start by recalling that (use geometric series)
$$frac1{1+t^2}=sum_{ngeq0}(-1)^nt^{2n}$$
Then integrate both sides from $0$ to $x$ to get
$$arctan x=sum_{ngeq0}frac{(-1)^n}{2n+1}x^{2n+1}$$
integrate both sides from $0$ to $1$ now to produce
$$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}=fracpi4-frac12log2$$
edited 46 mins ago
answered 1 hour ago
clathratusclathratus
3,651332
3,651332
1
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Amazing, this is exactly what I was looking for
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– user601297
1 hour ago
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@user601297 you are very welcome :)
$endgroup$
– clathratus
1 hour ago
add a comment |
1
$begingroup$
Amazing, this is exactly what I was looking for
$endgroup$
– user601297
1 hour ago
$begingroup$
@user601297 you are very welcome :)
$endgroup$
– clathratus
1 hour ago
1
1
$begingroup$
Amazing, this is exactly what I was looking for
$endgroup$
– user601297
1 hour ago
$begingroup$
Amazing, this is exactly what I was looking for
$endgroup$
– user601297
1 hour ago
$begingroup$
@user601297 you are very welcome :)
$endgroup$
– clathratus
1 hour ago
$begingroup$
@user601297 you are very welcome :)
$endgroup$
– clathratus
1 hour ago
add a comment |
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Just to give a slightly different approach,
$$sum_{n=1}^infty{n^2over n!}=sum_{n=1}^infty{nover(n-1)!}=sum_{m=0}^infty{m+1over m!}=sum_{m=0}^infty{mover m!}+e=sum_{m=1}^infty{mover m!}+e=sum_{m=1}^infty{1over(m-1)!}+e=sum_{k=0}^infty{1over k!}+e=e+e$$
The trick in reading this is to note what changes across each equal sign as you proceed from left to right and understand what justifies the equality for each change.
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add a comment |
$begingroup$
Just to give a slightly different approach,
$$sum_{n=1}^infty{n^2over n!}=sum_{n=1}^infty{nover(n-1)!}=sum_{m=0}^infty{m+1over m!}=sum_{m=0}^infty{mover m!}+e=sum_{m=1}^infty{mover m!}+e=sum_{m=1}^infty{1over(m-1)!}+e=sum_{k=0}^infty{1over k!}+e=e+e$$
The trick in reading this is to note what changes across each equal sign as you proceed from left to right and understand what justifies the equality for each change.
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add a comment |
$begingroup$
Just to give a slightly different approach,
$$sum_{n=1}^infty{n^2over n!}=sum_{n=1}^infty{nover(n-1)!}=sum_{m=0}^infty{m+1over m!}=sum_{m=0}^infty{mover m!}+e=sum_{m=1}^infty{mover m!}+e=sum_{m=1}^infty{1over(m-1)!}+e=sum_{k=0}^infty{1over k!}+e=e+e$$
The trick in reading this is to note what changes across each equal sign as you proceed from left to right and understand what justifies the equality for each change.
$endgroup$
Just to give a slightly different approach,
$$sum_{n=1}^infty{n^2over n!}=sum_{n=1}^infty{nover(n-1)!}=sum_{m=0}^infty{m+1over m!}=sum_{m=0}^infty{mover m!}+e=sum_{m=1}^infty{mover m!}+e=sum_{m=1}^infty{1over(m-1)!}+e=sum_{k=0}^infty{1over k!}+e=e+e$$
The trick in reading this is to note what changes across each equal sign as you proceed from left to right and understand what justifies the equality for each change.
answered 1 hour ago
Barry CipraBarry Cipra
59.4k653125
59.4k653125
add a comment |
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In general, $$ sum_{n=0}^{infty} frac{n^k}{n!} = B_k e, $$ where $B_k$ is the $k$-th Bell-number. This is called the Dobinski's formula. This can be computed by expanding $n^k$ into the sum of falling factorials as in Olivier Oloa's answer.
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– Sangchul Lee
1 hour ago
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If $R>0$ and a power series $sum_{j=0}^{infty}a_jx^j$ converges to $f(x)$ for each $xin (-R,R)$ then $f$ is differentiable on $(-R,R)$ and $sum_{j=1}^{infty}ja_jx^{j-1} =f'(x)$ for all $(-R,R).$ Notice also that the previous sentence now applies with $f$ replaced by $f'$. For example, for $|x|<2$ let $f(x) =1/(1-x)=sum_{j=0}^{infty}x^j .$ Then $sum_{j=1}^{infty}jx^{j-1}=f'(x)=1/(1-x)^2,$ and $sum_{j=2}^{infty}j(j-1)x^{j-2}=f''(x)=2/(1-x)^3.$
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– DanielWainfleet
32 mins ago
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See : math.stackexchange.com/questions/1711318/…
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– lab bhattacharjee
6 secs ago