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Find the sum of the infinite series $$\sum^{\infty}_{k=1} \frac{k-1}{k!}.$$

Using the ratio test, for example, one can easily show that the series is convergent.

But how does one go about finding the sum?

Perhaps the power series will be useful, such as $$e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}.$$

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Sketch:

$$ \sum_{k=1}^\infty \frac{k-1}{k!}=\sum_{k=1}^\infty\frac{k}{k!}-\sum_{k=1}^\infty\frac{1}{k!}=\sum_{k=1}^\infty\frac{1}{(k-1)!}-\sum_{k=1}^\infty\frac{1}{k!}=\sum_{k=1}^\infty\left(\frac{1}{(k-1)!}-\frac{1}{k!}\right) $$

This is telescoping. Make sure to understand and be able to explain what happens in each equality. Can you take it from here?

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  • $\begingroup$ Clearly the best answer since a student need not know anything about the series representation of $e$. (+1) $\endgroup$ – Mark Viola May 13 '18 at 17:36
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$$ \sum^{\infty}_{k=1}\frac{k-1}{k!} = \sum^{\infty}_{k=1}\frac{k}{k!} - \sum^{\infty}_{k=1} \frac1{k!} = \sum^{\infty}_{k=1}\frac{1}{(k-1)!} - \sum^{\infty}_{k=1} \frac1{k!} = \cdots $$

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Note that$$\sum_{k=1}^\infty\frac1{k!}=e-1$$and that$$\sum_{k=1}^\infty\frac k{k!}=\sum_{k=1}^\infty\frac1{(k-1)!}=e.$$

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HINT

$$\sum^{\infty}_{k=1} \frac{k-1}{k!}=\sum^{\infty}_{k=1} \frac{k}{k!}-\sum^{\infty}_{k=1} \frac{1}{k!}$$

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$$e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}.$$

$$e=1+1+ \frac {1}{2!}+ \frac {1}{3!}+....$$

$$\sum^{\infty}_{k=1} \frac{k-1}{k!}= \sum^{\infty}_{k=1} \frac{k}{k!} + \sum^{\infty}_{k=1} \frac{-1}{k!} =$$

$$\sum^{\infty}_{k=1} \frac{1}{(k-1)!} -\sum^{\infty}_{k=1} \frac{1}{k!} =$$

$$(1+1+\frac {1}{2!}+...)-(1+\frac {1}{2!}+...)=e-(e-1)=1$$

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