Determine the value of \sum_{k=1}^{\infty}\frac{k^2}{k!}.
Solution :
- e
- 2e
- (e+1)(e-1)
- e^2
- \infty
Solution :
Note that \sum_{k=1}^{\infty}\frac{k^{2}}{k!}=\sum_{k=1}^{\infty}\frac{k}{(k-1)!} =\sum_{k=1}^{\infty}\frac{k-1}{(k-1)!}+\sum_{k=1}^{\infty}\frac{1}{(k-1)!}
Now if we adjust the indices a little, we obtain:
\begin{eqnarray*} \sum_{k=1}^{\infty}\frac{k-1}{(k-1)!}+\sum_{k=1}^{\infty}\frac{1}{(k-1)!} & = & \sum_{k=0}^{\infty}\frac{k}{k!}+\sum_{k=0}^{\infty}\frac{1}{k!} \\ & = & \sum_{k=1}^{\infty}\frac{k}{k!}+\sum_{k=0}^{\infty}\frac{1}{k!} \\ & = & \sum_{k=1}^{\infty}\frac{1}{(k-1)!}+\sum_{k=0}^{\infty}\frac{1}{k!} \\ & = & \sum_{k=0}^{\infty}\frac{1}{k!}+\sum_{k=0}^{\infty}\frac{1}{k!} \end{eqnarray*}
What we have left is just two series definition of e added to each other, hence the answer is 2e.
Now if we adjust the indices a little, we obtain:
\begin{eqnarray*} \sum_{k=1}^{\infty}\frac{k-1}{(k-1)!}+\sum_{k=1}^{\infty}\frac{1}{(k-1)!} & = & \sum_{k=0}^{\infty}\frac{k}{k!}+\sum_{k=0}^{\infty}\frac{1}{k!} \\ & = & \sum_{k=1}^{\infty}\frac{k}{k!}+\sum_{k=0}^{\infty}\frac{1}{k!} \\ & = & \sum_{k=1}^{\infty}\frac{1}{(k-1)!}+\sum_{k=0}^{\infty}\frac{1}{k!} \\ & = & \sum_{k=0}^{\infty}\frac{1}{k!}+\sum_{k=0}^{\infty}\frac{1}{k!} \end{eqnarray*}
What we have left is just two series definition of e added to each other, hence the answer is 2e.
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I'd actually cancel the k's first and then re-index immediately; then instead of the substitution k=(k-1) + 1 coming out of nowhere you see a k+1 in the numerator, which is more intuitive to split up.
Also, here's another way to do it:
First cancel the k's and re-index the sum to start at 0, obtaining \sum_{k=0}^\infty \frac{k+1}{k!}.
Now note that this is the same as the series \sum_{k=0}^\infty \frac{k+1}{k!} x^k evaluated at x=1.
Observe now that the above sum is the derivative of another sum: \frac{\mathrm{d}}{\mathrm{d}x}\sum_{k=0}^\infty \frac{x^{k+1}}{k!}.
Factor out an x and you're left with \frac{\mathrm{d}}{\mathrm{d}x} x \sum_{k=0}^\infty \frac{x^{k}}{k!}. Apply the series definition of e^x and you get a very workable formula: \frac{\mathrm{d}}{\mathrm{d}x} (xe^x). This is just e^x + xe^x.
Thus for the original series you have \sum_{x=1}^\infty \frac{k^2}{k!} = \left(e^x + xe^x\right|_{x=1}, from which evaluation at 1 gives you e+e=2e.
Very clever! I think your technique is more useful than my solution since many of these summation identities are about recognizing the correct power series representation.
Thanks! I'll add a page below the header covering basics of MathJax tonight.
Thanks again! There is now a message above the comment box indicating the correct control characters.
Testing reply system.
it's LaTeX with a capital T and you can write it with a backslash in the LaTeX non-math environment. Not sure if it works in math mode. Let's see. \LaTeX
Thanks again! The typo is corrected.
Nice.
I'd actually cancel the k's first and then re-index immediately; then instead of the substitution k=(k-1) + 1 coming out of nowhere you see a k+1 in the numerator, which is more intuitive to split up.
Also, here's another way to do it:
First cancel the k's and re-index the sum to start at 0, obtaining \sum_{k=0}^\infty \frac{k+1}{k!}.
Now note that this is the same as the series \sum_{k=0}^\infty \frac{k+1}{k!} x^k evaluated at x=1.
Observe now that the above sum is the derivative of another sum: \frac{\mathrm{d}}{\mathrm{d}x}\sum_{k=0}^\infty \frac{x^{k+1}}{k!}.
Factor out an x and you're left with \frac{\mathrm{d}}{\mathrm{d}x} x \sum_{k=0}^\infty \frac{x^{k}}{k!}. Apply the series definition of e^x and you get a very workable formula: \frac{\mathrm{d}}{\mathrm{d}x} (xe^x). This is just e^x + xe^x.
Thus for the original series you have \sum_{x=1}^\infty \frac{k^2}{k!} = \left(e^x + xe^x\right|_{x=1}, from which evaluation at 1 gives you e+e=2e.
@Dave:
Very clever! I think your technique is more useful than my solution since many of these summation identities are about recognizing the correct power series representation.
A comment about MathJax: you should mention somewhere (sidebar?) how MathJax syntax works. I had to poke around the MathJax site to figure it out.
@Dave:
Thanks! I'll add a page below the header covering basics of MathJax tonight.
I think you can get away with just saying what the control characters to enter the math environment are. The rest is just LaTeX, for which there exist lots of resources and even more sample LaTeX code.
@Dave:
Thanks again! There is now a message above the comment box indicating the correct control characters.
Testing reply system.
didn't go through, posting again:
it's LaTeX with a capital T and you can write it with a backslash in the LaTeX non-math environment. Not sure if it works in math mode. Let's see. \LaTeX
Thanks again! The typo is corrected.
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