# How can I sum the infinite series $\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad$

How can I find the sum of the infinite series $$\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad ?$$

My attempt at a solution - I saw that I could rewrite it as $$\frac{1}{5}\left(1 - \frac{4}{10} \left( 1 - \frac{7}{15} \left(\cdots \left(1 - \frac{3k - 2}{5k}\left( \cdots \right)\right)\right.\right.\right.$$ and that $\frac{3k - 2}{5k} \to \frac{3}{5}$ as $k$ grows larger. Using this I thought it might converge to $\frac{1}{8}$, but I was wrong, the initial terms deviate significantly from $\frac{3}{5}$.

According to Wolfram Alpha it converges to $1-\frac{\sqrt{5}}{2}$. How can I get that ?

\begin{align*}(-1)^{n-1}\frac{1\cdot 4 \dots (3n-2)}{5\cdot 10 \dots 5n}&= (-1)^{n-1}\frac{3^n}{5^n}(-1)^n\frac{(-\frac{1}{3})\cdot (-\frac{4}{3}) \dots (-\frac{3n-2}{3})}{1\cdot 2 \dots n}\\ &= -(3/5)^n\binom{-1/3}{n}\end{align*}

Therefore, you can obtain

$$\sum_{n=1}^{\infty} -(3/5)^n\binom{-1/3}{n} = 1 - \sum_{n=0}^{\infty} (3/5)^n\binom{-1/3}{n} = 1- (1+\frac35)^{-1/3} = 1- \sqrt{5/8} =1-\frac{\sqrt5}{2}$$

Here is a way that uses the Gamma function, and that was strangely the first thing to come to mind. (Warning!: It is perhaps more complicated than necessary.)

Let $I$ denote your infinite series. We can write this as $$I=\sum_{n=0}^{\infty}(-1)^{n}\prod_{m=0}^{n}\frac{3m+1}{5m+5}$$ which becomes

$$I=\sum_{n=0}^{\infty}\left((-1)^{n}\frac{3^{n+1}}{5^{n+1}(n+1)!}\prod_{m=0}^{n}\left(m+\frac{1}{3}\right)\right).$$
Now since $$\Gamma\left(\frac{1}{3}\right)\prod_{m=0}^{n}\left(m+\frac{1}{3}\right)=\Gamma\left(n+\frac{4}{3}\right)$$ we have $$I=\frac{1}{\Gamma\left(1/3\right)}\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{n+1}}{5^{n+1}(n+1)!}\Gamma\left(n+\frac{4}{3}\right).$$ By the definition of $\Gamma(s)$ we have $$\Gamma\left(n+\frac{4}{3}\right)=\int_{0}^{\infty}t^{n+\frac{1}{3}}e^{-t}dt,$$ which implies that $$I=\frac{1}{\Gamma\left(1/3\right)}\int_{0}^{\infty}t^{-\frac{2}{3}}e^{-t}\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{n+1}t^{n+1}}{5^{n+1}(n+1)!}dt$$ by switching the order of summation and integration. As $$1-\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{n+1}t^{n+1}}{5^{n+1}(n+1)!}=\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{n}t^{n}}{5^{n}n!}=e^{-\frac{3t}{5}}$$ it follows that $$I=\frac{1}{\Gamma\left(1/3\right)}\int_{0}^{\infty}t^{-\frac{2}{3}}e^{-t}\left(1-e^{-\frac{3t}{5}}\right)dt=1-\frac{1}{\Gamma\left(1/3\right)}\int_{0}^{\infty}t^{-\frac{2}{3}}e^{-\frac{8}{5}t}dt$$ Setting $u=\frac{8}{5}t$ we get $$\frac{1}{\Gamma\left(1/3\right)}\int_{0}^{\infty}t^{-\frac{2}{3}}e^{-\frac{8}{5}t}dt=\frac{1}{\Gamma\left(1/3\right)}\left(\frac{5}{8}\right)^{\frac{1}{3}}\int_{0}^{\infty}u^{-\frac{2}{3}}e^{-u}du=\frac{\left(5\right)^{\frac{1}{3}}}{2}$$and hence $$I=1-\frac{\left(5\right)^{\frac{1}{3}}}{2}$$ as desired.

I hope that helps,

• This is so complicated. – user9413 Apr 24 '11 at 2:30
• @Chandru: There is already a simpler solution. Eric's example is still quite interesting and there's a lot to learn from it. – Jacob Apr 24 '11 at 2:47
• Of course no problem is permitted with a complicated solution! – GEdgar Jun 17 '11 at 17:52

You use this formula $$(1+x)^{-p/q} = 1 - \frac{\displaystyle\frac{p}{q}}{1!}\cdot x + \frac{ \displaystyle\frac{p}{q} \cdot \Bigl(\frac{p}{q}+1\Bigr)}{2!} x^{2} - \cdots$$

So your series can be rewritten as $$\frac{1}{3} \times \frac{3}{5} - \frac{\displaystyle\frac{1}{3}\cdot\Bigl(\frac{1}{3} + 1\Bigr)}{2!} \times \Bigl(\frac{3}{5}\Bigr)^{2} + \frac{\displaystyle\frac{1}{3} \cdot \Bigl(\frac{1}{3}+1\Bigr) \cdot \Bigl(\frac{1}{3}+2\Bigr)}{3!} \times \Bigl(\frac{3}{5}\Bigr)^{3} - \cdots$$

So this is nothing but $$\Bigl(1+\frac{3}{5}\Bigr)^{-1/3} -1 = \frac{1}{3} \times \frac{3}{5} - \frac{\displaystyle\frac{1}{3}\cdot\Bigl(\frac{1}{3} + 1\Bigr)}{2!} \times \Bigl(\frac{3}{5}\Bigr)^{2} + \cdots$$

So our answer will be $$1 - \Bigl(\frac{8}{5}\Bigr)^{-1/3} = 1 -\frac{5^{1/3}}{2}$$