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Find Sum of Series $$2+\frac5{2!\cdot3}+\frac{5\cdot7}{3!\cdot3^2}+\frac{5\cdot7\cdot9}{4!\cdot3^3}+ \cdots\cdots $$

I tried breaking the terms into differences or finding a generalised term but did not get it right. Can someone please help me to proceed with this?

Rohit Singh
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Aditya Singh
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1 Answers1

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Assuming the first term is $1$ (and not $2$ as written), the general term of the series is, for $n\geq 1$, $$ a_n \stackrel{\rm def}{=} \frac{\prod_{k=2}^{n}(2k+1)}{n!3^{n-1}} = \frac{\prod_{k=1}^{n}(2k+1)}{n!3^{n}} = \frac{(2n+1)!}{n!3^{n}\prod_{k=1}^n(2k)} = \frac{(2n+1)!}{n!3^{n}2^nn!} = \frac{(2n+1)!}{(n!)^26^{n}} $$ or, equivalently, $a_n= \binom{2n}{n}\left(\frac{1}{6}\right)^n$.

Now, either you work towards finding the general form for $$f(x) = \sum_{n=1}^\infty (2n+1) \binom{2n}{n}x^n$$ (a power series with radius of convergence $1/4$), which you can find by relating it to both $ g(x) = \sum_{n=1}^\infty n\binom{2n}{n}x^{n-1} $ (recognize a derivative) and $ h(x) = \sum_{n=1}^\infty \binom{2n}{n}x^{n} $, since $$f(x) = 2xg(x)+h(x)\,;$$ or, by other means (there may be?) you establish that $f(1/6) = 3\sqrt{3}$, leading to $$ \sum_{n=1}^\infty a_n = 3\sqrt{3}. $$

Clement C.
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  • If $n=1$, the product is empty, not 5 since it goes from 2 to $n$. – marty cohen Nov 04 '17 at 22:43
  • @martycohen for $n=1$, you get 1, the purported first term. $n=2$ gives 5/(6.3) – Clement C. Nov 04 '17 at 22:45
  • Sorry, the last part above should of course read 5/6, which matches the 5/(2! 3) from the question. I blame typing on my phone. – Clement C. Nov 04 '17 at 23:26
  • @martycohen but the question says first term=2 and can you please help me with the third step on how you managed the question or can you do it in simple terms without any complex terms (radius of convergence). I would be very grateful to you :) – Aditya Singh Nov 05 '17 at 07:08
  • @AdityaSingh take the result, remove 1 (my first term), add 2 (your first term). – Clement C. Nov 05 '17 at 15:06
  • For the second question (how to derive the result once you know the form of the series, have a look at this, which solves the problem of computing $\sum_n \binom{2n}{n}x^n$ (and stems from wanting to compute $\sum_n (2n+1)\binom{2n}{n}x^n$ as in your case, but for $x=1/8$ instead of $1/6$) – Clement C. Nov 05 '17 at 16:20
  • @AdityaSingh And besides this (most relevant) question, see also this one and that one. – Clement C. Nov 05 '17 at 16:22
  • @AdityaSingh Let me know if something is still unclear. – Clement C. Nov 06 '17 at 16:03