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Prove that $ \sum \limits_{n=1}^{\infty} \frac{n}{(n+1)^{(n+2)} (n+2)!}(-1)^{(n+1)} = \frac{23}{24} - \frac 2 3 \sqrt 2 $

This question was asked in Math Tripos. and is taken from the classic Hall and Knight Question Number - 131 of Miscellaneous Examples.

My try:

So, I tried using the expansion of $e^x = \sum \limits_{i=0}^\infty \frac{x^i}{i!}$ to get factorial part in the denominator and no idea how to get alternate +, - in the series... My try was based on finding functions which can do that and multiply them to get suitable denominator, but I don't seem to find any.

Any help is appreciated!

Wiktoria
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  • Not sure what I'm missing, but when I try to evaluate both sides, the left hand side is about $0.0198$ and the right hand side is about $0.0155$. Since I don't have a copy of Hall and Knight, could you check the statement of the problem? – WhatsUp Mar 18 '20 at 18:27
  • @whatsup Here's a copy : https://archive.org/details/higheralgebraseq00hall – Wiktoria Mar 18 '20 at 19:57
  • I didn't find Question Number -131 ... And there are just so many things called "Miscellaneous" in that book. Could you locate the exact page? – WhatsUp Mar 18 '20 at 20:28
  • On page 502 of the link, problem 131 has the series: $$\frac{1}{2^33} - \frac{1\cdot 3}{2^44} + \frac{1\cdot 3\cdot 5}{2^55} - \cdots.$$

    Hopefully I am interpreting this correctly.

     – ljeabmreosn Mar 18 '20 at 20:31
  • @ljeabmreosn I see that equation now. Thanks. And it seems that this is not the same as in the statement of this question... – WhatsUp Mar 18 '20 at 20:33
  • @ljeabmreosn And obviously that "L" shape notation should mean something different - perhaps continued fractions? The series you typed doesn't converge... – WhatsUp Mar 18 '20 at 20:36
  • @WhatsUp I suspected that meant something, though I'm not sure what those bars mean. – ljeabmreosn Mar 18 '20 at 20:40
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    @ljeabmreosn From previous sections in the book, I think I figured it out: the "L" shape thing means "factorial". So the real formula should be $$\frac1{2^3 3!} - \frac{1\cdot 3}{2^4 4!} + \frac{1\cdot 3 \cdot 5}{2^5 5!} - \dotsc$$. – WhatsUp Mar 18 '20 at 20:40

1 Answers1

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Gathering various informations from the book, I finally figured out the real formula to prove: $$\sum_{n = 1}^\infty \frac{(-1)^{n + 1} (2n - 1)!!}{2^{n + 2}(n + 2)!} = \frac{23}{24} - \frac23\sqrt2,$$ where the $!!$ symbol means double factorial.

For a proof, first rewrite $$(2n - 1)!! = \frac{(2n)!}{(2n)!!} = \frac{(2n)!}{2^nn!}$$ and hence $$\frac{(2n - 1)!!}{2^{n + 2}(n + 2)!} = \frac{(2n)!}{2^{2n + 2}(n + 2)!n!} = \frac{\binom{2n}n}{2^{2n + 2}(n + 1)(n + 2)}.$$

Thus the sum on the left hand side can be written as $$\sum_{n = 1}^\infty (-\frac14)^{n + 1}\frac{\binom{2n}n}{(n + 1)(n + 2)}.$$

The next input is the following famous formula: $$\sum_{n = 0}^\infty \binom{2n}nx^n = (1 - 4x)^{-1/2}.$$ This formula has been discuss many times on this site, see e.g. here.

Integrating the above formula with respect to $x$, we get: $$\sum_{n = 0}^\infty \binom{2n}n\frac{x^{n + 1}}{n + 1} = -\frac 12 ((1 - 4x)^{1/2} - 1).$$

Integrating once more, we get:$$\sum_{n = 0}^\infty \binom{2n}n\frac{x^{n + 2}}{(n + 1)(n + 2)} = \frac1{12}((1 - 4x)^{3/2} - 1) + \frac 12 x.$$

Setting $x = -\frac14$ in the above formula gives the willing result (note the extra term for $n = 0$ on the left).

WhatsUp
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