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How to find the sum of the following series

$$\frac{1}{5} - \frac{1\cdot 4}{5\cdot 10} + \frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15} - \dots\,.?$$

I have no idea. I have written the general term and tested its convergence by Gauss' test for convergence, but they are neither the question nor the answer.

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Could you reveal what your general term is? It is not clear what the pattern in "1, 4, 7, ..." is supposed to be. –  Henning Makholm Jan 12 at 2:39
The general term is $$\frac{(-1)^{n-1}}{5^n}. \frac{1.4.7. \dots (3n-2)}{n!}$$. $1, 4, 7,\dots$ is in an AP. What to do after it? –  Dutta Jan 12 at 2:45
See binomial series. –  Lucian Jan 12 at 3:02

3 Answers 3

up vote 4 down vote accepted

From the Generalized Binomial Theorem (for $|x|<1$), $$\left(1+x\right)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots$$

Method $1:$

$\displaystyle -\frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15}=\frac{-\frac13\left(-\frac13-1\right)\left(-\frac13-2\right)3^3}{3!\cdot 5^3}=\frac{-\frac13\left(-\frac13-1\right)\left(-\frac13-2\right)}{3!}\left(\frac35\right)^3 $

and $\displaystyle \frac{1\cdot 4}{5\cdot 10}=\frac{-\frac13\left(-\frac13-1\right)3^2}{2!\cdot 5^2}=\frac{-\frac13\left(-\frac13-1\right)}{2!}\left(\frac35\right)^2$

and $\displaystyle-\frac15=\left(-\frac13\right)\left(\frac35\right)$

So, the given series $\displaystyle=1-\left(1+\frac35\right)^{-\frac13}$

Method $2:$

If $\displaystyle nx=-\frac15 \ \ \ \ (1)$

and $\displaystyle\frac{n(n-1)}{2!}x^2=-\frac{1\cdot4}{5\cdot10}=\frac2{25} \ \ \ \ (2)$

Divide $(2)$ with the square of $(1)$ to get $n$ and then $x$ using $(1)$

and check that $x,n$ satisfies $\displaystyle\frac{n(n-1)(n-2)}{3!}x^3=-\frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15}$

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@HopelessFool, how about this method? –  lab bhattacharjee Jan 12 at 5:39
Higher Secondary mathematics. Half forgotten. My fault no doubt. I do not understand why these are asked in NBHM! Thank you. –  Dutta Jan 12 at 6:08

For any $\gamma \in \mathbb{R}$ and $k \in \mathbb{N}$, let $(\gamma)_k = \gamma(\gamma+1)\cdots(\gamma+k-1)$ be the rising Pochhammer symbol. Using following expansion $$\frac{1}{(1-z)^\gamma} = \sum_{k=0}^\infty \frac{(\gamma)_k}{k!} z^k$$ We can evaluate the sum as

$$\sum_{k=1}^\infty (-1)^{k-1}\frac{\prod_{j=0}^{k-1}(3j+1)}{5^kk!} = - \sum_{k=1}^\infty \frac{(\frac13)_k}{k!}\left(-\frac{3}{5}\right)^k = 1 - \frac{1}{\sqrt[3]{1+\frac{3}{5}}} = 1 - \frac{\sqrt[3]{5}}{2} $$

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According to Maple, $$ \sum_{m=0}^\infty (-1)^m \dfrac{\prod_{j=0}^m (1+3 j)}{\prod_{j=0}^m (5 + 5 j)} = 1 - \dfrac{5^{1/3}}{2}$$

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Dear Sir, I do not know Maple. The left hand side is the general term written differently. How to get the the left hand side by manual computation? –  Dutta Jan 12 at 2:56

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