I am interested to find $$\sum_{k=1}^n\frac{1}{k(k+1)}$$ without using telescoping series method. I tried very hard but still could not think of a way to find it without using telescoping series method. Can someone perhaps give me a hint?

  • 1
    $\begingroup$ What's wrong with telescoping? But you could always guess the answer and prove it by induction... $\endgroup$ Apr 17, 2015 at 10:05
  • 2
    $\begingroup$ @HansLundmark: Nothing wrong with telescoping series. I am just curious if I can find this sum without using telescoping series method. That is all. $\endgroup$
    – mathema
    Apr 17, 2015 at 10:06
  • $\begingroup$ And I want to avoid induction too. $\endgroup$
    – mathema
    Apr 17, 2015 at 10:07
  • $\begingroup$ Be sure to put on your hair shirt first :) $\endgroup$ Apr 17, 2015 at 10:34

3 Answers 3


First, let us recall following integral representation:

$$\frac{1}{k(k+1)} = \int_0^1 \int_0^y x^{k-1} dx dy$$

We have $$ \sum_{k=1}^n \frac{1}{k(k+1)} = \int_0^1 \int_0^y \left( \sum_{k=1}^n x^{k-1} \right) dx dy = \int_0^1 \int_0^y \left( \frac{1-x^n}{1-x} \right) dx dy\\ = \int_0^1 \int_x^1 \left( \frac{1-x^n}{1-x} \right) dy dx = \int_0^1 (1-x^n) dx = 1 - \frac{1}{n+1} $$

  • $\begingroup$ very ingenious. love it! $\endgroup$ Apr 17, 2015 at 12:32
  • $\begingroup$ That's brilliant! $\endgroup$ Apr 17, 2015 at 15:55

One neat way you can do this is via generating functions but I think all it really does is obfuscate the fact that there is a telescoping happening. Nevertheless here's how it works. Let $s_n$ be the sum in question. Let $f(x)$ be the generating function of the the sequence in the sum, that is

$$f(x) = \sum_{k=1}^\infty \frac{x^k}{k(k+1)}$$

To get at $s_n$ you need the partial summations of coefficients of $f(x)$. This can be done by multiplying $f(x)$ by $1/(1-x)$, that is

$$\sum_{n=1}^\infty s_n x^n = \frac{f(x)}{1-x}$$

So all we need to do is find $f(x)$ itself. First note that

$$\frac{1}{k(k+1)}x^k = \frac{1}{x}\iint x^{k-1} dxdx$$

Then write $f(x)$ as

$$f(x) = \frac{1}{x}\iint \sum_{k=1}^\infty x^{k-1} dxdx = \frac{1}{x}\iint \frac{1}{1-x} dxdx$$

It's not too difficult to show that the integration leads to

$$f(x) = \frac{1}{x}\left(x + (1-x)\log(1 - x)\right)$$

and finally

$$\frac{f(x)}{1-x} = \frac{1}{1 - x} + \frac{\log(1 - x)}{x}$$

If you expand each of the terms here the coefficient of $x^n$ in the first one is a 1 and in the second one it is $-1/(n+1)$, which yields

$$s_n = 1- \frac{1}{n+1}$$

the hard way.


By partial summation we have $$\sum_{k\leq n}\frac{1}{k\left(k+1\right)}=\frac{H_{n}}{N+1}-\sum_{k\leq n-1}H_{k}\left(\frac{1}{k+2}-\frac{1}{k+1}\right) $$ where $H_{k} $ is the $k-th $ armonic number. These sums have a closed form $$\sum_{k\leq n-1}\frac{H_{k}}{k+1}=\frac{1}{2}\left(H_{n}^{2}-H_{n}^{(2)}\right) $$ $$\sum_{k\leq n-1}\frac{H_{k}}{k+2}=\frac{1}{2}\left(H_{n+1}^{2}-H_{n+1}^{(2)}-\frac{2n}{n+1}\right) $$ where $H_{n}^{(r)} $ is the generalized harmonic number. Then $$\sum_{k\leq n-1}\frac{H_{k}}{k+1}-\sum_{k\leq n-1}\frac{H_{k}}{k+2}=\frac{1}{2}\left(H_{n}^{2}-H_{n}^{(2)}-H_{n+1}^{2}+H_{n+1}^{(2)}+\frac{2n}{n+1}\right)=\frac{n-H_{n}}{n+1} $$ and so $$\sum_{k\leq n}\frac{1}{k\left(k+1\right)}=\frac{n}{n+1}. $$

  • $\begingroup$ Nice! Just to confirm, by partial summation, do you mean Summation By Parts? $\endgroup$
    – MathGod
    Nov 28, 2016 at 10:59
  • $\begingroup$ @IshanSingh Yes, exactly. $\endgroup$ Nov 28, 2016 at 21:41

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