How to prove the following ?
$$\sum_{n=1}^{\infty}\frac{n(3n−2)}{n!}=4e$$
I do know the series expansion for $e^x$.You may use it...
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Related : http://math.stackexchange.com/questions/576976/evaluate-the-series-lim-limits-n-to-infty-sum-limits-i-1n-fracn22 – lab bhattacharjee Jul 16 '16 at 08:50
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One may rewrite $$ \sum_{n=1}^{\infty}\frac{n(3n−2)}{n!}=\sum_{n=1}^{\infty}\frac{3n(n-1)+n}{n!}=3\sum_{n=2}^{\infty}\frac{1}{(n-2)!}+\sum_{n=1}^{\infty}\frac{1}{(n-1)!}=3e+e, $$ where we have used changes of index in the last step.
Olivier Oloa
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How does the that become n=2 in the third equality....?It was n=1 right initially.I know the denominator becomes undefined....but I have a little confusion in that part... – Jul 16 '16 at 08:38
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@ZOZ Observe that the term for $n=1$ vanishes... so the summation really starts at $n=2$. Isn't it? – Olivier Oloa Jul 16 '16 at 08:40
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@ZOZ This term $\frac{3n(n-1)}{n!}$. It is equal to $0$ for $n=1$. That's why we start the sum from $n=2$. – Olivier Oloa Jul 16 '16 at 09:41
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Hint
Consider $$A=\sum_{n=1}^{\infty}\frac{n(3n−2)}{n!}x^n=\sum_{n=1}^{\infty}\frac{3n^2−2n}{n!}x^n=\sum_{n=1}^{\infty}\frac{3n(n-1)+n}{n!}x^n$$ So $$A=3\sum_{n=1}^{\infty}\frac{n(n-1)}{n!}x^n+\sum_{n=1}^{\infty}\frac{n}{n!}x^n=3x^2\sum_{n=1}^{\infty}\frac{n(n-1)}{n!}x^{n-2}+x\sum_{n=1}^{\infty}\frac{n}{n!}x^{n-1}$$ Now recognize that the first sum is the second derivative of $e^x$ and that the second sum is the first derivative of $e^x$. When you finish, set $x=1$.
Claude Leibovici
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