I would like to show that $\sum_{k=1}^{\infty}\frac{1}{k(k+1)}=1$

I was wondering if I could get a hint to get me started.

I tried trying to rewrite it in some sort of familiar sum like a Taylor Series for a function I know like the exponential but no luck. I also tried writing it as the sum of two terms but I'm having trouble knowing how to break it down (if this is even the right way to attack this problem?). Any advice would be appreciated. Thanks.

  • 2
    $\begingroup$ Have you heard of partial fraction decompositions? That gives you a nice telescoping series here. $\endgroup$ Jan 18 '16 at 20:22



  • $\begingroup$ Thanks for the hint. I'll try using this fact now. Out of curiosity, is there a way you knew that this particular decomposition would be useful, or is this just from past experience? $\endgroup$
    – Chio
    Jan 18 '16 at 20:23
  • 1
    $\begingroup$ @Chio The decomposition comes from Partial Fraction Decomposition. I knew that it works from experience: whenever when the denominator is the product of terms in arithmetic progression, the (partial fraction) decomposition makes the sum into a telescopic one. $\endgroup$
    – N. S.
    Jan 18 '16 at 20:25

Note that $$ \sum_{k=1}^{\infty}\frac{1}{k(k+1)} = \frac{1}{2} + \sum_{k=2}^{\infty}\frac{1}{k(k+1)} = \frac{1}{2} + \sum_{j=1}^{\infty}\frac{1}{(j+1)(j+2)}, \ \text{with} \ j = k - 1 $$ But $$ \frac{1}{(j+1)(j+2)} = \frac{1}{j+1} - \frac{1}{j+2} = \int_{0}^{1}x^jdx - \int_{0}^{1}x^{j+1}dx = \int_{0}^{1}x^j(1 - x)dx\\ \text{and} \ \sum_{j=1}^{\infty}x^jdx = \frac{x}{1 - x}, \quad |x| < 1 $$ Hence, $$ \sum_{k=1}^{\infty}\frac{1}{k(k+1)} = \frac{1}{2} + \sum_{k=1}^{\infty}\int_{0}^{1}x^j(1 - x)dx = \frac{1}{2}+\int_{0}^{1}(1 - x)\sum_{j=1}^{\infty}x^jdx\\ = \frac{1}{2} + \int_{0}^{1}(1 - x)\frac{x}{1 - x}dx = \frac{1}{2} + \int_{0}^{1}xdx = \frac{1}{2} + \frac{1}{2} = 1 $$


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