# Sum of infinite series $\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)…(i+n)}$

Lets define the following series $$S_n =\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}.$$ I know that $S_0$ does not converge so let's suppose $n \in N$ and define $S_n$ for some $n$. We have $S_1=1$ , $S_2=\frac1 4$ , $S_3=\frac 1 {18}$ , $S_4=\frac 1{96}$ , $S_5=\frac 1{600}$ etc..

the numerator in all results is 1
the pattern in denominator is $[1,4,1,96,600...]$ and can be found here that is equal to $n*n!$ Finally I want to prove the general equality:

$\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}=\frac 1 {n*n!}$

Hint: Telescopic sum and $$\frac{n}{i(i+1)(i+2)\cdots(i+n)}=\frac{1}{i(i+1)\cdots(i+n-1)}-\frac{1}{(i+1)(i+2)\cdots(i+n)}.$$

• I don't think there is n in the numerator – Rohan Shinde Jan 5 '18 at 4:00
• Of course, You made the point with separating the fraction for telescopic sum,that is what i have missed – Ramin Rahimzada Jan 5 '18 at 4:07
• @Manthanein n is a constant, the summation is over i, so... – Landon Carter Jan 5 '18 at 4:15
• @Landon Carter I totally agree with that but Please check my answer for what I meant to say – Rohan Shinde Jan 5 '18 at 4:16

If you failed to see the telescoping nature of the series, here is an alternative approach that makes use of the Beta function.

\begin{align*} S_n &= \sum_{i = 1}^\infty \frac{1}{i (i + 1) (i + 2) \cdots (i + n)} \quad n \in \mathbb{N}\\ &= \sum_{i = 1}^\infty \frac{1 \cdot 2 \cdots (i - 1)}{1 \cdot 2 \cdots (i - 1)i(i + 1) \cdots (i + n)}\\ &= \sum_{i = 1}^\infty \frac{(i - 1)!}{(i + n)!}\\ &= \sum_{i = 1}^\infty \frac{\Gamma (i)}{\Gamma (i + n + 1)}\\ &= \frac{1}{\Gamma (n + 1)} \sum_{i = 1}^\infty \frac{\Gamma (i) \Gamma (n + 1)}{\Gamma (i + n + 1)}\\ &=\frac{1}{\Gamma (n + 1)} \sum_{i = 1}^\infty \text{B}(i, n + 1)\\ &=\frac{1}{\Gamma (n + 1)} \sum_{i = 1}^\infty \int_0^1 t^{i - 1} ( 1 - t)^n \, dt\\ &=\frac{1}{\Gamma (n + 1)} \int_0^1 (1 - t)^n \left [\sum_{i = 1}^\infty t^{i - 1} \right ] \, dt\\ &= \frac{1}{\Gamma (n + 1)} \int^1_0 (1 - t)^n \cdot \frac{1}{1 - t} \, dt\\ &= \frac{1}{\Gamma (n + 1)} \int^1_0 (1 - t)^{n - 1} \, dt\\ &= \frac{1}{\Gamma (n + 1)} \left [-\frac{1}{n} (1 - t)^n \right ]^1_0\\ &= \frac{1}{\Gamma (n + 1) n}\\ &= \frac{1}{n \cdot n!}, \end{align*} as expected.

• I really like this approach (+1). – clathratus Nov 27 '18 at 4:11

$$\frac{1}{i(i+1)(i+2)\cdots(i+n)}$$ $$=\frac{1}{n}\left(\frac{1}{i(i+1)\cdots(i+n-1)}-\frac{1}{(i+1)(i+2)\cdots(i+n)}.\right)$$

The first term of this series becomes $$=\frac{1}{n}\left(\frac{1}{n!}-\frac{1}{(n+1)!}\right)$$

The second term becomes $$=\frac{1}{n}\left(\frac{1}{(n+1)!}-\frac{2}{(n+2)!}\right)$$

The third term becomes $$=\frac{1}{n}\left(\frac{2}{(n+2)!}-\frac{6}{(n+3)!}\right)$$

Hence we see that the sum telescopes to $\frac {1}{n. n!}$

• Isn't this exactly like Math Lover's answer? Perhaps you can include how you arrived at the final step... – Crescendo Jan 5 '18 at 4:04

I will add another answer with a different approach (less elementary but also interesting). From finite calculus theory your series can also be stated as $$\sum_{k=1}^\infty\frac1{k^\overline n}=\frac1{1^\overline n}+\sum_{k=2}^\infty\frac1{k^\overline n}=\frac1{n!}+\sum\nolimits_2^\infty (k-1)^\underline{-n}\delta k$$ where $$a^\overline c$$ is a rising factorial and $$a^\underline c$$ is a falling factorial. Hence

\begin{align}\sum_{k=1}^\infty\frac1{k^\overline n}&=\frac1{n!}+\frac{(k-1)^\underline{1-n}}{1-n}\bigg|_{k=2}^{k\to\infty}\\&=\frac1{n!}+0-\frac{1^\underline{1-n}}{1-n}\\&=\frac1{n!}+\frac1{(n-1)2^\overline{n-1}}\\&=\frac1{n!}+\frac1{(n-1)n!}\\ &=\frac1{n!\, n}\end{align}

because $$\frac{a^\underline m}{m}$$ is a primitive (in the finite calculus sense) of $$a^\underline {m-1}$$, in the domain where both expressions are well-defined.