# Does $\sum_{n=1}^{\infty} (\frac{1}{n}-\frac{1}{\sqrt{n(n+1)}})$ converge

$$\sum_{n=1}^{\infty} \left ( \frac{1}{n}-\frac{1}{\sqrt{n(n+1)}}\right )=\sum_{n=1}^{\infty} \left ( \frac{\sqrt{n(n+1)}-n}{n\sqrt{n(n+1)}}\right )$$

By ratio test I get $$\lim_{n\to \infty} \frac{\frac{\sqrt{n(n+1)}-n}{n\sqrt{n(n+1)}}}{\frac{1}{n}}=\frac{0}{0}$$

So i can't conclude anything. I tried comparison tests, but i get $\le$ something divergent.

• Surely you could do a better attempt at the limit? Flip those fractions and L'hospital's rule like pancakes on a stove. – Simply Beautiful Art Sep 2 '16 at 12:05
• Look in your book for the material on "indeterminate forms". – GEdgar Sep 2 '16 at 12:11

Use Taylor's expansion at order $1$: \begin{align*}\frac1n-\frac1{\sqrt{n(n+1)}}&=\frac1n\Biggl(1-\frac1{\sqrt{1+\frac1n}}\Biggr)=\frac1n\biggl(1-1+\frac1{2n}+o\Bigl(\frac1n\Bigr)\biggr)\\ &=\frac1{2n^2}+o\biggl(\frac1{n^2}\biggr)\sim_\infty\frac1{2n^2} \end{align*} and the latter converges.
Compare to $$\sum_{n=1}^\infty\left(\frac1n-\frac1{\sqrt{n\cdot n}}\right)$$ on one side and $$\sum_{n=1}^\infty\left(\frac1n-\frac1{\sqrt{(n+1)(n+1)}}\right)$$on the other.
$$\frac{1}{n}-\frac{1}{\sqrt{n(n+1)}} = \frac{1}{2n^2} + o\left(\frac{1}{n^2}\right)$$