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Sum of series $\displaystyle \frac{15}{16}+\frac{15}{16}\cdot\frac{21}{24}+\frac{15}{16}\cdot\frac{21}{24}\cdot \frac{27}{32}+\cdots\cdots $

what i try

$\displaystyle S =\frac{15}{16}+\frac{15\cdot 21}{16\cdot 24}+\frac{15\cdot 21\cdot 27}{16\cdot 24\cdot 32}+\cdots \cdots $

i am trying to convert numerator and denomiantor terms into arithmetic progression

$\displaystyle \frac{9S}{8}=\frac{9\cdot 15}{8\cdot 16}+\frac{9\cdot 15\cdot 21}{8\cdot 16\cdot 24}+\cdots \cdots $

$\displaystyle \frac{9S}{8}+\frac{9}{8}+1=1+\frac{9}{8}+\frac{9\cdot 15}{8\cdot 16}+\frac{9\cdot 15\cdot 21}{8\cdot 16\cdot 24}+\cdots \cdots $

but it is divergent series

i did not know how i solve that infinite series

Help me how to solve

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marked as duplicate by user, Abcd, Lord Shark the Unknown, Shailesh, Cesareo Jan 13 at 9:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Based on the hint of lab bhattacharjee

$$S=\frac{5}{2}\left(\frac{3}{8}\right)+\frac{5\cdot7}{2\cdot3}\left(\frac{3}{8}\right)^2+\frac{5\cdot7\cdot9}{2\cdot3\cdot4}\left(\frac{3}{8}\right)^3+ \cdots\\ =\frac{4}{3}\cdot\frac{2}{3}\sum_{n=2}^\infty \frac{-\frac{3}{2}\left(-\frac{3}{2}-1\right)\cdots\left(-\frac{3}{2}-n+1\right)}{n!}\left(-\frac{3}{4}\right)^n\\ =\frac{8}{9}\left[\left(1-\frac{3}{4}\right)^{-\frac{3}{2}}-1-\frac{-\frac{3}{2}}{1!}\left(-\frac{3}{4}\right)\right]\\=\frac{8}{9}\left[8-1-\frac{9}{8}\right]=\frac{47}{9}. $$


Generally for arbitrary $a\in\mathbb{R}_+$, $k\in\mathbb{Z}_+$, $|x|<1$: $$ \sum_{N=1}^\infty \prod_{n=1}^N\left(1+\frac{a-1}{k+n}\right)x= \frac{k!}{a(a+1)\cdots(a+k-1)x^k}\sum_{N=k+1}^\infty\frac{a(a+1)\cdots(a+N-1)}{N!}x^N\\ =\left[\binom{a+k-1}{k}x^k\right]^{-1}\left[\left(1-x\right)^{-a}-\sum_{N=0}^k\binom{a+N-1}{N}x^N\right]. $$

For your problem: $a=\frac{3}{2}$, $k=1$, $x=\frac{3}{4}$.

Note: $$ \binom{a}{k}:=\frac{1}{k!}\prod_{i=0}^{k-1} (a-i);\quad \binom{a+k-1}{k}x^k\equiv \binom{-a}{k}(-x)^k. $$

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We have $S=\frac{3}{4}\sum_{n=2}^{\infty}a_n$ with

$a_n=\frac{5 \cdot 7 \cdot ... \cdot (2n+1)}{4 \cdot 6 \cdot ... \cdot (2n)}$. It is easy to see that $a_n \ge 1$, hence $\sum_{n=2}^{\infty}a_n$ is divergent, thus $S= \infty$.

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  • $\begingroup$ answer is $47/9$ $\endgroup$ – jacky Jan 7 at 8:47
  • $\begingroup$ Think of the harmonic series $$1+ \frac 12 + \frac 13 + \frac 14 + \cdots$$ The terms are strictly decreasing and tend to $0$, yet the series still diverges. Here, you have a series where not only do the terms dont tend to $0$, but are in fact strictly increasing (when put into the form as suggested in this answer)! There is thus no way that this series is ever going to converge. $\endgroup$ – glowstonetrees Jan 7 at 9:19

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