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Does the following converge or diverge? I should use the ratio test.

$$\sum_{x=1}^\infty \frac{x}{(x-1)!}$$

Don Shrinkle
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  • You should definitely use it. –  Apr 09 '16 at 05:25
  • Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote to reopen this. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – Martin Sleziak Apr 09 '16 at 06:58
  • You could also have a look at some questions with similar sums, such as http://math.stackexchange.com/questions/521861/proof-by-induction-that-sum-i-1n-fracii1-1-frac1n1 or http://math.stackexchange.com/questions/764065/summation-of-a-series-help-sum-fracn-1n – Martin Sleziak Apr 09 '16 at 07:02

2 Answers2

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$$f(x)=\dfrac x{(x-1)!}=\dfrac{x-1+1}{(x-1)!}=\dfrac1{(x-2)!}+\dfrac1{(x-1)!}$$

Now use $e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$

$f(1)=\dfrac1{0!}$ as $\dfrac1{n!}=0$ for integer $n<0$

lab bhattacharjee
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The answer by lab bhattacharjee is clever and even allows you to compute the limit to which the series converges, but of course you can also use the ratio test if you just want to know whether the series converges or not. Let $$a_n = \frac{n}{(n-1)!}$$ Then $$\begin{aligned} \left| \frac{a_{n+1}}{a_n} \right| &= \left| \frac{n+1}{n} \cdot \frac{(n-1)!}{n!} \right| \\ &= \left| \frac{n+1}{n^2} \right| \\ &\to 0 \quad \text{as }n \to \infty \\ \end{aligned}$$ and consequently the series $\sum a_n$ converges absolutely.