MHB Point of convergence of a series

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The discussion focuses on calculating the sum of the series $\sum_{k=1}^\infty \frac{k^2}{k!}$. The user confirms the series converges using the ratio test and provides a detailed breakdown of the calculation, ultimately concluding that the sum equals $2e$. The steps include transforming the original series and utilizing known sums of exponential series. The conversation emphasizes understanding series manipulation and convergence in the context of GRE preparation.
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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