3

$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+v)}=\frac{1}{vv!}$$ I am struggling to find a solution for this but no luck yet. How can I analyze it to get to second part?

GorillaApe
  • 1,071
  • 1
    $$\frac{1}{n(n+1)\cdots(n+a)} = \frac{1}{a}\frac{(n+a)-n}{n(n+1)\cdots(n+a)} =,, ???$$ – achille hui Nov 10 '14 at 09:56
  • I have tried this but cant get any further. – GorillaApe Nov 10 '14 at 09:59
  • 2
    It's a telescopic sum. Follow achille's advice, split into two fractions, write two consecutive terms and you will see. – orion Nov 10 '14 at 10:05
  • Here is an idea (which might be a wrong idea): $\forall x \in [-1,1[, , -\ln(1-x) = \displaystyle \sum_{n=1}^{+\infty} \frac{x^{n}}{n}$. Taking the antiderivative leads to : $\displaystyle \int -\ln(1-x) , dx = \sum_{n=1}^{+\infty} \frac{1}{n(n+1)} x^{n+1}$. Therefore, $x - (x-1)\ln(1-x) = \displaystyle \sum_{n=1}^{+\infty} \frac{1}{n(n+1)}x^{n+1}$. Taking the limit as $x \to 1$ gives : $\displaystyle \sum_{n=1}^{+\infty} \frac{1}{n(n+1)} = 1$. Taking the antiderivative again and, then, the limit as $x \to 1$ gives : $\displaystyle \sum_{n=1}^{+\infty} \frac{1}{n(n+1)(n+2)} = \frac{1}{4}$... – pitchounet Nov 10 '14 at 10:09
  • 1
    @jibounet. Interesting idea ! Cheers – Claude Leibovici Nov 10 '14 at 10:28
  • http://math.stackexchange.com/questions/1013030/how-to-prove-that-sum-n-1-infty-frac1nn1n2-nk-frac1k – lab bhattacharjee Nov 10 '14 at 10:38
  • its a dupe ,.,, – GorillaApe Nov 10 '14 at 10:55

2 Answers2

2

\begin{eqnarray} \frac{1}{n(n+1)...(n+v)}&=&\frac{1}{v}\frac{(n+v)-n}{n(n+1)...(n+v)}\\ &=&\frac{1}{v}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}] \end{eqnarray} $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)...(n+v)} =\frac{1}{v}\sum_{n=1}^{\infty}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}]$$ $$\frac{1}{v}\sum_{n=1}^{k}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}] =\frac{1}{v}[\frac{1}{v!}-\frac{1}{(k+1)...(k+v)}]$$ Let $k\rightarrow\infty$, then $\frac{1}{v}\sum_{n=1}^{\infty}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}] =\frac{1}{v}\frac{1}{v!}$

Alfred Chern
  • 2,658
1

Let $n^{\overline{r}}=\underbrace{n(n+1)(n+2)\cdots (n+r-1)}_{r\text{ terms}}$.

Hence $$\begin{align}\require{cancel} &\sum_{n=1}^{\infty}\frac1{n(n+1)(n+2)\cdots (n+v)}\\ &=\sum_{n=1}^{\infty}\frac1{n(n+1)^{\overline{v}}}=\sum_{n=1}^{\infty}\frac1{n^\overline{v}(n+v)}\\ &=\frac1v\sum_{n=1}^{\infty}\frac1{(n)^\overline{v}}-\frac1{(n+1)^{\overline{v}}}\\ &=\frac1v\left[\left(\frac1{1^\overline{v}}-\cancel{\frac1{2^\overline{v}}}\right)+\left(\cancel{\frac1{2^\overline{v}}}-\bcancel{\frac1{3^\overline{v}}}\right)+\left(\bcancel{\frac1{3^\overline{v}}}-\cancel{\frac1{4^\overline{v}}}\right)+\cdots\right]\\ &=\frac1v \left[\frac1{1^\overline{v}}\right]\\ &=\frac1{vv!}\qquad\blacksquare\end{align}$$

Hypergeometricx
  • 22,657