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Find sum to infinity terms of the series $$S=\frac{4}{5}+\frac{4.7}{5.8}+\frac{4.7.10}{5.8.11}+\cdots$$

My Try:

we have $$1+S=1+\frac{4}{5}+\frac{4.7}{5.8}+\frac{4.7.10}{5.8.11}+\cdots$$

now $$(1+x)^n=1+nx+\frac{n(n-1)x^2}{2}+\cdots $$ So comparing we get

$$nx=\frac{4}{5}$$ and

$$\frac{n(n-1)x^2}{2}=\frac{7}{10}$$

solving for $x$ and $n$ we get

$$n=\frac{-16}{19}$$ and $$x=\frac{-19}{20}$$

Hence

$$1+S=(1+x)^n=20^{\frac{16}{19}}$$

$$S=20^{\frac{16}{19}}-1$$

Is this alright?

Batominovski
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Ekaveera Gouribhatla
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    Apparently, $S=\infty$ because the series diverges: $$\prod_{k=1}^n,\frac{3k+1}{3k+2}=\frac{2^{2/3},\Gamma(5/6)}{\sqrt{\pi}},\frac{\Gamma(n+4/3)}{\Gamma(n+5/3)} \approx \frac{2^{2/3},\Gamma(5/6)}{\sqrt{\pi},\sqrt[3]{n}}$$ for large positive integers $n$. – Batominovski Dec 02 '17 at 07:26
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    Unless you perhaps miss something like the factor $\frac{1}{n!}$ for the $n$-th term, then you will get a generalized hypergeometric series $$1+S={_1}F_1\left(\frac{4}{3};\frac{5}{3};1\right)\approx 2.28616,.$$ – Batominovski Dec 02 '17 at 07:30
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    The comparison with $(1+x)^n$ is incorrect. Even if the first two terms match some $x$ and $n$, there's no guarantee that the third term will match. – Sarvesh Ravichandran Iyer Dec 02 '17 at 08:10
  • @labbhattacharjee How is this question a duplicate of the link you gave? – Batominovski Dec 02 '17 at 09:44
  • Is it possible to atleast sum to $n$ terms – Ekaveera Gouribhatla Dec 02 '17 at 16:25

1 Answers1

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$$S=\sum_{n\geq 1}\prod_{k=1}^{n}\frac{3k+1}{3k+2}=\sum_{n\geq 1}\frac{2^{2/3} \,\Gamma\left(\tfrac{5}{6}\right)\, \Gamma\left(\tfrac{4}{3}+n\right)}{\sqrt{\pi}\,\Gamma\left(\tfrac{5}{3}+n\right)} $$ is divergent by Gautschi's inequality and the p-test: the main term of such series behaves like $\frac{C}{n^{1/3}}$.

Jack D'Aurizio
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