<s>[INST] proving $\sum_{k=1}^{\infty}\frac{1}{k(k+1)}=1$ 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.
 [/INST] 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,
$$
\sum_{k=1}^{\infty}\frac{1}{k(k+1)} = \frac{1}{2} + \sum_{k=1}^{\infty}\int_{0}^{1}x^j(1 - x)dx = \frac{1}{2}+\int_{0}^{1}(1 - x)\sum_{j=1}^{\infty}x^jdx\\
= \frac{1}{2} + \int_{0}^{1}(1 - x)\frac{x}{1 - x}dx = \frac{1}{2} + \int_{0}^{1}xdx = \frac{1}{2} + \frac{1}{2} = 1
$$
</s>