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


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.

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    $\begingroup$ 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. $\endgroup$ – 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.


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