<s>[INST] Convergence of infinite $ \sum (\frac{n}{n+1})^{n^2} $? I have being trying to solve this convergence but with no success. Using the ratio test I have reached here:
$$ 
a_n = \left( \frac{n}{n+1} \right)^{n^2}
$$
And
$$ 
\frac{1}{a_n} = \left( \frac{n+1}{n} \right)^{n^2}
$$
Also
$$ 
a_{n+1} = \left(\frac{n+1}{n+2}\right)^{{(n+1)}^2} = \left(\frac{n+1}{n+2}\right)^{n^2+2n+1} = \\ a_{n+1} = \left(\frac{n+1}{n+2}\right)^{n^2} \cdot \left(\frac{n+1}{n+2}\right)^{2n} \cdot \left(\frac{n+1}{n+2}\right)
$$
Therefore
$$
\lim \frac{a_{n+1}}{a_n} = \lim \frac{1}{a_n} \cdot a_{n+1}  = \\
\lim \left[ \left( \frac{n+1}{n} \right)^{n^2} \cdot \left(\frac{n+1}{n+2}\right)^{n^2} \cdot \left(\frac{n+1}{n+2}\right)^{2n} \cdot \left(\frac{n+1}{n+2}\right) \right]
$$
And I dont know how to go any further, I have tried many possibilities none of them reaching any result.
 [/INST] The Bernoulli inequality proves that the sequence $\{a_n\}_{n\in\mathbb{N}^+}$ defined by:
$$ a_n = \left(1+\frac{a}{n}\right)^n $$
is an increasing sequence converging towards $e^a$. By taking $a=-1$, this gives that your series is positive and bounded by:
$$\sum_{n\geq 1}e^{1-n} = \frac{e^2}{e-1}.$$
</s>