<s>[INST] Proving $\sum_{n=1}^\infty (-1)^n\tan\left(\frac{n+1}{n^2}\right)$ converges according to the Leibniz test I need to prove the following series converges:
$$\sum_{n=1}^\infty (-1)^n\tan\left(\frac{n+1}{n^2}\right)$$
I'm trying to understand a solution that uses Leibniz's test.
I understand why $\tan\left(\frac{n+1}{n^2}\right)\to0$,
but cannot figure out how to formally prove it monotonically decreases and is positive for every $n\geq2$.
Thank you!
 [/INST] HINT
To begin with, notice the $\tan$ function is strictly increasing.
Moreover, its argument is decreasing, as the following reasoning proves:
\begin{align*}
a_{n + 1} - a_{n} & = \frac{(n + 1) + 1}{(n + 1)^{2}} - \frac{n + 1}{n^{2}}\\\\
& = \frac{n^{2}(n + 1) + n^{2} - (n + 1)^{3}}{n^{2}(n + 1)^{2}}\\\\
& = \frac{n^{3} + 2n^{2} - n^{3} - 3n^{2} - 3n - 1}{n^{2}(n + 1)^{2}}\\\\
& = -\left(\frac{n^{2} + 3n + 1}{n^{2}(n + 1)^{2}}\right) < 0
\end{align*}
Consequently, one has that:
\begin{align*}
a_{n + 1} < a_{n} \Rightarrow \tan(a_{n + 1}) < \tan(a_{n})
\end{align*}
whence it can be concluded that $\tan(a_{n})$ is strictly decreasing.
Based on the previous results, can you take it from here?
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