MHB Point of convergence of a series

caffeinemachine
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Hello MHB.

I have been preparing for my subject GRE and I need help on the following problem.

Find $\displaystyle\sum_{k=1}^\infty \frac{k^2}{k!}$.

Using the ratio test we know that the series converges but how to we find what it converges to?
 
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caffeinemachine said:
Hello MHB.

I have been preparing for my subject GRE and I need help on the following problem.

Find $\displaystyle\sum_{k=1}^\infty \frac{k^2}{k!}$.

Using the ratio test we know that the series converges but how to we find what it converges to?

Is...

$\displaystyle \sum_{k=1}^{\infty} \frac{k^{2}}{k!} = \sum_{k=1}^{\infty} \frac{k}{(k-1)!} = \sum_{k=0}^{\infty} \frac{k+1}{k!} = \sum_{k=0}^{\infty} \frac{k}{k!} + \sum_{k=0}^{\infty} \frac{1}{k!} = 2\ e$

Kind regards

$\chi$ $\sigma$
 
chisigma said:
Is...

$\displaystyle \sum_{k=1}^{\infty} \frac{k^{2}}{k!} = \sum_{k=1}^{\infty} \frac{k}{(k-1)!} = \sum_{k=0}^{\infty} \frac{k+1}{k!} = \sum_{k=0}^{\infty} \frac{k}{k!} + \sum_{k=0}^{\infty} \frac{1}{k!} = 2\ e$

Kind regards

$\chi$ $\sigma$
Taught me a lot. Thanks. :)
 
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