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Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dotsb = \frac{e}{2}$.

This is problem 4 from page 303 of S.L.Loney's 'Plane Trigonometry'.

It seems fairly obvious that the series expansion $e^x$ will be used. However, I am unsure where to start. Should I consider other series?

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4 Answers 4

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$$\sum\limits_{n=0}^{+\infty} \frac{n+1}{(2n+1)!} = \frac{1}{2} \cdot \sum\limits_{n=0}^{+\infty} \frac{(2n+1)+1}{(2n+1)!} = \frac{1}{2} \cdot \sum\limits_{n=0}^{+\infty} \bigg(\frac{1}{(2n)!}+\frac{1}{(2n+1)!}\bigg) = \frac{1}{2} \sum\limits_{k=0}^{+\infty} \frac{1}{k!} = \frac{e}{2}$$

Maybe the details are not explained well enough. If there is anything unclear just ask.

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    $\begingroup$ Ingenious. I think you ought to have some parentheses around the terms in the penultimate sum, though. $\endgroup$ Commented Aug 26, 2018 at 12:41
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    $\begingroup$ Well, of course I mean that the sum should be made for both terms (otherwise you would $n$ be?). Anyway, I am editing it. $\endgroup$
    – LucaMac
    Commented Aug 26, 2018 at 12:45
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Another approach is $$\sinh x=x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\cdots$$ then $$(x\sinh x)'\Big|_{x=1}=2\left(1+\dfrac{2}{3!}+\dfrac{3}{5!}+\dfrac{4}{7!}+\cdots\right)$$ which gives the result.

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  • $\begingroup$ +1 very nice. you deserved it. $\endgroup$
    – Khosrotash
    Commented May 10, 2020 at 5:19
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$$e=(1+1)+\left(\frac {1}{2!}+\frac {1}{3!}\right) +\left(\frac {1}{4!}+\frac {1}{5!}\right) +\left(\frac {1}{6!}+\frac {1}{7!}\right) +\cdots=2\left(1+\frac {2}{3!}+\frac {3}{5!}+\frac {4}{7!}+\cdots\right) $$

Q. E. D

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Your series is $$\sum_{n=1}^{\infty}\frac{n}{(2n-1)!}.$$ Let $a_n=\frac{n}{(2n-1)!}$, then $$a_n=\frac{1}{2}\frac{2n}{(2n-1)!}=\frac{1}{2}\frac{2n-1+1}{(2n-1)!} =\frac{1}{2}\left(\frac{1}{(2n-2)!}+\frac{1}{(2n-1)!}\right).$$

So $$\sum_{n=1}^{\infty}\frac{n}{(2n-1)!}=\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{1}{(2n-2)!}+\frac{1}{(2n-1)!}\right)$$ $$=\frac{1}{2}\sum_{n=0}\frac{1}{n!}=\frac{1}{2}e.$$

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  • $\begingroup$ You are 4 minutes late ... all the credit goes to @lucamac $\endgroup$
    – rtybase
    Commented Aug 26, 2018 at 12:43
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    $\begingroup$ Yes, it does not matter! $\endgroup$
    – Riemann
    Commented Aug 26, 2018 at 12:44

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