# Is the series $\sum_{n=1}^{\infty}\frac{1}{1-\sqrt{n^{2}+5n}}$ convergent or divergent?

$$\sum_{n=1}^{\infty}\frac{1}{1-\sqrt{n^{2}+5n}}$$

in order to simplify the expression I rationalised the denominator and got :

$$\sum_{n=1}^{\infty}\frac{1+\sqrt{n^{2}+5n}}{1-{n^{2}-5n}}$$

This is where I stuck...

I can't use any of the convergence tests because the series itself is not positive. How can I determine if the series is convergent or divergent?

Your series converges if and only if the series$$\sum_{n=1}^\infty\frac1{\sqrt{n^2+5n}-1}$$(which is your series times $$-1$$) converges, and this series happens to be a series of positive numbers. Furthermore,$$\lim_{n\to\infty}\frac{\frac1{\sqrt{n^2+5n}-1}}{\frac1n}=1.$$Therefore, your series diverges.
• Sure: I am comparing the series $\sum_{n=1}^\infty\frac1{\sqrt{n^2+5n}-1}$ with the series $\sum_{n=1}^\infty\frac1n$, which diverges. – José Carlos Santos Jun 13 at 9:57
We can use there asymptotic condition which states that if $$\frac{a_n}{b_n} \rightarrow g$$ where $$g \in \mathbb R \setminus \left\{ 0 \right\}$$ then $$\sum a_n$$ diverges if and only if $$\sum b_n$$ diverges. You see that in denominator you have $$1-\sqrt{n^{2}+5n}$$ which is similar to $$n$$, is that right? To be sure you can do asymptotic test: $$\frac{1-\sqrt{n^{2}+5n}}{n} = \frac{1}{n} - \sqrt{\frac{n^2+5n}{n^2}} \rightarrow 0 - 1 = -1$$ so due to $$\sum \frac{1}{n}$$ diverges, your series diverges too.