# Convergence of series $\sum\limits_{n=0}^\infty\frac{3n}{n!}$

Investigate the series for convergence and if possible, determine its limit: $\sum\limits_{n=0}^\infty\frac{3n}{n!}$

My solution

$$\sum\limits_{n=0}^\infty\frac{3n}{n!} = \sum\limits_{n=0}^\infty\frac{3n}{(n-1)!n} = \sum\limits_{n=0}^\infty\frac{3}{(n-1)!}$$

But since $n \to \infty$, the sum is the same as

$$\sum\limits_{n=0}^\infty\frac{3}{n!} = 3\sum\limits_{n=0}^\infty\frac{1}{n!} = 3\sum\limits_{n=0}^\infty\frac{1}{n!}1^n$$

which is the definition of $e$, so the series is convergent and it equals to $3e^1$.

Is this proof correct? How could I improve it by making it more formal?

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You have to be careful when $n=0$. Also your assertion that "the sum is the same since $n\to\infty$" is incorrect - it does converge similarly, but not to the same value... –  gt6989b Nov 29 '12 at 18:03
@gt6989b Could you elaborate in an answer how I could alleviate those problems? –  Flavius Nov 29 '12 at 18:11
@Flavius The first term in the given sum is $0$. So you may omit it and save yourself some $0/0$ grief by noting that your sum equals $\sum_{n=1}^{\infty}\frac{3n}{n!}$. Then reindex by replacing all "n"s by "n+1": $\sum_{n+1=1}^{n+1=\infty}\frac{3}{(n+1-1)!}$. –  alex.jordan Nov 29 '12 at 18:18

This should be a comment, but it is a little long.

Note that the term corresponding to $n=0$ is $0$. This is because $0!$, by definition, is equal to $1$. So, as pointed out by Phira, our sum is equal to $\sum_{n=1}^\infty \frac{3n}{n!}$. As in your answer, this simplifies to $\sum_{n=1}^\infty \frac{3}{n-1}!$. Writing $k$ for $n-1$, we see that our sum is $\sum_{k=0}^\infty \frac{3}{k!}$, which we recognize.

Conceivably, this is not quite the desired answer! Perhaps the problem setter expects us to investigate the series for convergence while pretending we don't know the sum, and as a second part, to find the sum. If so, the Ratio Test quickly tells us that the series converges.

Even though you end up with the right number, the first line of your answer is problematic. For in the case $n=0$, you are cancelling $0$'s (in general forbidden). Then your first term is $\frac{3}{(-1)!}$, whatever that may mean. Then for no clear reason the $(n-1)!$ gets replaced by $n!$.

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On the contrary, this is exactly the type of answer I was looking for, insightful and with an eye on mathematical formalism (which unfortunately I lack). –  Flavius Nov 29 '12 at 18:28

If you replace $$\sum_{n=0}^\infty \frac{3n}{n!}=\sum_{n=0}^\infty\frac{3n}{(n−1)!n}$$

by $$\sum_{n=0}^\infty \frac{3n}{n!}=\sum_{n=1}^\infty \frac{3n}{n!}=\sum_{n=1}^\infty\frac{3n}{(n−1)!n},$$

you can just replace the vague $n \to \infty$ line by an equal sign.

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Thanks, +1. Your answer is really nice, but I had to accept the one who matched best my way of thinking. –  Flavius Nov 29 '12 at 18:31

Beside the point that @Phira noted: $$\sum_{n=0}^\infty\frac{3n}{(n−1)!n}\to\sum_{n=1}^\infty\frac{3n}{(n−1)!n}$$ I think, since the partial sums of the latter series is the same patial sums for $\exp(1)$, then your claim looks right.

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Thanks, +1. Your answer is really nice, but I had to accept the one who matched best my way of thinking. –  Flavius Nov 29 '12 at 18:29
@Flavius: Of course you had. :) I, personally like the points Andre pointed. Those points means a lot to me. +1 for him. –  Babak S. Nov 29 '12 at 18:32

$\sum_{n=0}^\infty \frac{3n}{n!} < \sum_{n=0}^\infty \frac{3^n}{n!}=e^3$

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Thanks, +1. Your answer is really nice, but I had to accept the one who matched best my way of thinking. –  Flavius Nov 29 '12 at 18:30