$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$

Question

How to find the sum of the following series:

$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$

Any hints.

Answer 1

The sum may be written as

$$\sum_{k=1}^{\infty} (-1)^{k+1} \frac{1\cdot 4 \cdots (3k-2)}{5^k k!} $$

Now, based on a suggestion from @Robert Israel:

$$(1+t)^{-1/3} = 1 -\frac1{1!}\frac13 t + \frac{1}{2!}\left ( -\frac13 \right ) \left ( -\frac{4}{3} \right )t^2 - \cdots$$

so that

$$1-(1+t)^{-1/3} = \frac1{1!}\frac13 t - \frac{1}{2!}\left ( -\frac13 \right ) \left ( -\frac{4}{3} \right )t^2 + \cdots $$

The series on the RHS reproduces the series in question when $t=3/5$, so that the sum is

$$1-\left ( \frac{8}{5} \right )^{-1/3} = 1-\frac12 5^{1/3}$$

Answer 2

According to Maple it's $1 - 5^{1/3}/2$. It looks like this comes from the Maclaurin series for $(1-t)^{-1/3}$.

Answer 3

I am applying Robert Israel's suggestion - I have learned that this is generally a good thing to do.

The general term in $\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot$ seems to be

$\begin{array}\\ a_n &=(-1)^n\dfrac{\prod_{k=1}^n (3k-2)}{5^n n!}\\ &=\dfrac{\prod_{k=1}^n (2-3k)}{5^n n!}\\ &=\dfrac{3^n\prod_{k=1}^n (\frac23-k)}{5^n n!}\\ &=\dfrac{3^n\prod_{k=0}^{n-1} (\frac23-(k+1))}{5^n n!}\\ &=\dfrac{3^n\prod_{k=0}^{n-1} (-\frac13-k)}{5^n n!}\\ &=\left(\dfrac{3}{5}\right)^n\dfrac{\prod_{k=0}^{n-1} (-\frac13-k)}{ n!}\\ \end{array} $

The expansion of $(1+t)^r$ is

$\begin{array}\\ (1+t)^r &=\sum_{n=0}^{\infty} t^n \dfrac{\prod_{k=0}^{n-1} (r-k)}{n!}\\ \end{array} $

If $t=\frac35$ and $r=-\frac13$, this gives $(1+\frac35)^{-1/3} =\sum_{n=0}^{\infty} (\frac35)^n \dfrac{\prod_{k=0}^{n-1} (-\frac13-k)}{n!} =1-\sum_{n=1}^{\infty} a_n $.

Therefore $\sum_{n=1}^{\infty} a_n =1-(1+\frac35)^{-1/3} =1-(\frac85)^{-1/3} =1-(\frac58)^{1/3} =1-\frac{5^{1/3}}{2} $.

Cowabunga!!!!!

Generalizations:

If $a, b, c, d$ are non-zero integers such that $gcd(a, b) =gcd(c, d) =1 $,

$\begin{array}\\ (1+\frac{a}{b})^{c/d} &=\sum_{n=0}^{\infty} \left(\frac{a}{b}\right)^n \dfrac{\prod_{k=0}^{n-1} (\frac{c}{d}-k)}{n!}\\ &=\sum_{n=0}^{\infty} \left(\frac{a}{b}\right)^n \dfrac{\prod_{k=0}^{n-1} (c-kd)}{d^n n!}\\ &=\sum_{n=0}^{\infty} \dfrac{a^n\prod_{k=0}^{n-1} (c-kd)}{(bd)^n n!}\\ \end{array} $

Putting $-c/d$ for $c/d$,

$\begin{array}\\ (1+\frac{a}{b})^{-c/d} &=\sum_{n=0}^{\infty} \left(\frac{a}{b}\right)^n \dfrac{\prod_{k=0}^{n-1} (-\frac{c}{d}-k)}{n!}\\ &=\sum_{n=0}^{\infty} (-1)^n\left(\frac{a}{b}\right)^n \dfrac{\prod_{k=0}^{n-1} (c+kd)}{d^n n!}\\ &=\sum_{n=0}^{\infty} (-1)^n\dfrac{a^n\prod_{k=0}^{n-1} (c+kd)}{(bd)^n n!}\\ \end{array} $