Convergence of a Series $\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{n^2}$- Which Test? I tried root and ratio tests but it didn't work. Also, i can't use integral tests (and other "uncommon" ones) in this homework. 
(Prove that the series is convergent)
$$\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{n^2} = \frac{1}{2} + \left(\frac{2}{3}\right)^4 + \left(\frac{3}{4}\right)^9 + ...$$
 A: The Bernoulli Inequality says that for $n\ge1$,
$$
\left(1+\frac1n\right)^n\ge1+\frac{n}{n}=2
$$
Thus,
$$
\left(\frac{n}{n+1}\right)^{\large n^2}\le\left(\frac12\right)^n
$$
Thus, we can compare this to a geometric series.
A: Even if you don't know many limits, note that by expanding $(1+{1\over n})^n$ and dropping all terms except the two lowest order terms, one has $(1+{1\over n})^n\ge 1+n\cdot{1\over n}=2$. Thus 
$$0\le\Bigl({n\over n+1}\Bigr)^{n^2}={1\over\bigl[ \,(1+{1\over n})^n\,\bigr]^n}\le {1\over 2^n}.$$
 The convergence of your series now follows by comparison to the convergent geometric series $\sum\limits_{n=1}^\infty (1/2)^n$.
(Using the "well-known" limit $\lim\limits_{n\rightarrow\infty}(1+{1\over n})^n=e$, you can conclude that the terms  $(1+{1\over n})^n$ eventually exceed $2$; this will suffice for  comparing your series with  $\sum\limits_{n=1}^\infty (1/2)^n$. Of course, if you do make use of this limit,  using the Root Test is somewhat easier.)
A: The root test should work.
Notice the following equivalencies:
$$\left(\frac{n}{n+1}\right)^{n^2}=\left[\left(\frac{n}{n+1}\right)^n\right]^n= \frac{1}{\left[\left( 1+{1\over n}\right)^n\right]^n}$$
The right-hand side should look very familiar. You can certainly use the limit comparison test to prove convergence of your series.
A: We can interpret the series as power series evaluated at 1. Using the Hadamard formula we obtain $1/R = \lim_{n \rightarrow \infty} ((\frac{n}{n+1})^{n^2})^\frac{1}{n} = \lim_{n \rightarrow \infty} (1+\frac{-1}{n+1})^{n+1} \frac{n+1}{n} = e^{-1}$ so $R = e > 1$ and the series converges.
