3
$\begingroup$

I was trying to find a formula for the series

$$\sum_{k=1}^n \frac{1}{k(k+1)(k+2)} =? $$

I tried to break this into partial fractions...To see if I could telescope this series.. The partial fraction went like this $$\frac{1}{k(k+1)(k+2)}=\frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2k+4}$$

But the terms are so random that they hardly cancel.... I also tried partial sums but couldn't make much headway....Any help to solve this would be appreciated

$\endgroup$
  • $\begingroup$ With problems like this (an AP in the denominator, constant in numerator) you want to multiple by $$\frac{\text{last term} - \text{first term}}{\text{difference between them}}=1$$ to get the right telescope. $\endgroup$ – Ian Miller Jan 22 '17 at 12:01
  • $\begingroup$ @IanMiller Lol, that's perfect. $\endgroup$ – Simply Beautiful Art Jan 22 '17 at 12:03
  • $\begingroup$ See also : math.stackexchange.com/questions/2100299/… $\endgroup$ – lab bhattacharjee Jan 22 '17 at 12:04
  • $\begingroup$ "But the terms are so random that they hardly cancel" The terms very much cancel out. Did you try a few sums to see that they do? $\endgroup$ – Did Jan 22 '17 at 12:08
  • $\begingroup$ Thanks...I figured it out...Maybe , now I can generalize for 4 or 5 terms in denominator $\endgroup$ – user35508 Jan 22 '17 at 12:09
8
$\begingroup$

$$\dfrac2{k(k+1)(k+2)}=\dfrac{k+2-k}{k(k+1)(k+2)}=\dfrac1{k(k+1)}-\dfrac1{(k+1)(k+2)}=f(k)-f(k+1)$$

where $f(m)=\dfrac1{m(m+1)}$

$\endgroup$
6
$\begingroup$

HINT$$\sum_{k=1}^n \frac{1}{k(k+1)(k+2)} = \sum_{k=1}^n \frac{1}{2} \left(\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)} \right)$$ Can you take it from here?

$\endgroup$
3
$\begingroup$

Generalization:

Indeed, I will show you an approach to evaluate the series:

$$\displaystyle{\sum_{k=1}^{\infty}\frac{1}{k(k+1)\cdots(k+l)}}$$

You asked for the evaluation of a series (not finite summation, from $1$ to $n$). However, in a similar manner and without the Dominated Convergence Theorem, you can evaluate the above sum when it is finite.

First we give some definitions.
We have the definition of the beta function:

$$\displaystyle{B(x,y)=\int_{0}^{1}{(1-t)}^{x-1}t^{y-1}dt,\ x,y>0}$$

We also have the definition of the Euler $\Gamma$ function:

$$\displaystyle{\Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}dt,\ x>0}$$

For the latter function, with repeated integration by parts we get that:

$$\Gamma(n)=(n-1)!,\ \forall n \in \mathbb N$$

Furthermore we need the equality below, that relates the two functions.

$$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)},\ x,y>0$$

Now we will evaluate the series, that I mentioned in the beginning of the post.

$$\displaystyle{\sum_{k=1}^{\infty}\frac{1}{k(k+1)\cdots(k+l)}=\frac{1}{l!}\sum_{k=1}^{\infty}\frac{l!(k-1)!}{(k+l)!}=\frac{1}{l!}\sum_{k=1}^{\infty}\frac{\Gamma(l+1)\Gamma(k)}{\Gamma(l+k+1)}=\frac{1}{l!}\sum_{k=1}^{\infty}B(l+1,k)=}$$ $$\displaystyle{\frac{1}{l!}\sum_{k=1}^{\infty}\int_{0}^{1}x^l{(1-x)}^{k-1}dx=\frac{1}{l!}\int_{0}^{1}x^l\sum_{k=1}^{\infty}{(1-x)}^{k-1}dx}\ (1)$$

In the last step, where we interchanged the order of summation and integration, we applied the Theorem of Dominated Convergence. We also pulled the $x^l$ out of the sigma notation. By the geometric series we have that: $$\displaystyle{\sum_{k=1}^{\infty}{(1-x)}^{k-1}=\sum_{k=0}^{\infty}{(1-x)}^{k}=\frac{1}{1-(1-x)}=\frac{1}{x}}$$

We substitute into $(1)$ and we get that: $$\displaystyle{\sum_{k=1}^{\infty}\frac{1}{k(k+1)\cdots(k+l)}=\frac{1}{l!}\int_{0}^{1}x^{l-1}dx=\frac{1}{l\cdot l!}}$$

Finally:$$\displaystyle{\sum_{k=1}^{\infty}\frac{1}{k(k+1)\cdots(k+l)}=\frac{1}{l\cdot l!}}$$

$\endgroup$
  • $\begingroup$ Wonderfully done ...... $\endgroup$ – user35508 Jan 23 '17 at 6:31
  • $\begingroup$ @user35508 Thanks! $\endgroup$ – Stelios Sachpazis Jan 23 '17 at 7:48
0
$\begingroup$

\begin{align*} S&=\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}=\sum_{k=1}^n\left(\frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2k+4}\right)\\ &=\frac12\sum_{k=1}^n\frac1k-\sum_{k=1}^n\frac1{k+1}+\frac12\sum_{k=1}^n\frac1{k+2}\\ &=\frac12H_n-\sum_{k=2}^{n+1}\frac1{k}+\frac12\sum_{k=3}^{n+2}\frac1{k}\\ &=\frac12H_n-\left(-1+\sum_{k=1}^{n+1}\frac1{k}\right)+\frac12\left(-1-\frac12+\sum_{k=1}^{n+2}\frac1{k}\right)\\ &=\frac14+\frac12H_n-H_{n+1}+\frac12H_{n+2}\\ &=\frac{n(n+3)}{4(n+1)(n+2)} \end{align*} where in the second last line we used $$H_{n+1}=H_n+\frac1{n+1}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.