Prove that this series converges: ${(\frac{n}{n+1}})^{n^2}$ How to prove that this series converges: $\sum_{n = 1}^{\infty}{({\frac{n}{n+1}})^{n^2}}$ ?
I tried root test, ratio test, comparison test, Cauchy condensation test, but none works. Any hints?
 A: By using  Root test we get
$$\\ \lim _{ n\rightarrow \infty  }{ \sqrt [ n ]{ { \left( \frac { n }{ n+1 }  \right)  }^{ { n }^{ 2 } } }  } =\lim _{ n\rightarrow \infty  }{ { \left( \frac { n }{ n+1 }  \right)  }^{ n } } =\lim _{ n\rightarrow \infty  }{ { \left( 1+\frac { -1 }{ n+1 }  \right)  }^{ -\left( n+1 \right) \frac { -n }{ n+1 }  } } =\lim _{ n\rightarrow \infty  }{ { e }^{ \frac { -n }{ n+1 }  } } ={ e }^{ -1 }<1$$
A: 
I thought it might be instructive to present a way forward that uses the comparison test.  To that end, we proceed.


It is straightforward to show that $a_n=\left(\frac{n}{n+1}\right)^n$ is monotonically decreasing (See the note at the end of this post) and bounded below trivially by $0$.  
Thus, $a_n$ converges.  Furthermore, for $n\ge 1$, $\left(\frac{n}{n+1}\right)^n\le \frac12$.
Obviously then, can write
$$\left(\frac{n}{n+1}\right)^{n^2}\le \left(\frac{1}{2}\right)^n$$
where $\sum_{n=1}^\infty \left(\frac{1}{2}\right)^n =1<\infty$.

Therefore, the series $\sum_{n=1}^\infty \left(\frac{n}{n+1}\right)^{n^2}$ converges by comparison to the geometric series $\sum_{n=1}^\infty \left(\frac{1}{2}\right)^n$.


NOTE:
To show that $a_n=\left(\frac{n}{n+1}\right)^n$ is monotone decreasing, we form the ratio $\frac{a_n}{a_{n+1}}$ and show that this ratio is greater than or equal to $1$.  We write
$$\begin{align}
\frac{a_n}{a_{n+1}}&=\frac{\left(\frac{n}{n+1}\right)^n}{\left(\frac{n+1}{n+2}\right)^{n+1}}\\\\
&=\left(\frac{n(n+1)}{(n+1)^2}\right)^{n+1}\,\left(\frac{n+1}{n}\right) \\\\
&=\left(1-\frac{1}{(n+1)^2}\right)^{n+1}\,\left(\frac{n+1}{n}\right) \tag1\\\\
&\ge \left(1-\frac{1}{(n+1)}\right)\,\left(\frac{n+1}{n}\right) \tag 2\\\\
&=1
\end{align}$$
as was to be shown.  In going from $(1)$ to $(2)$, we applied Bernoulli's Inequality.
