Show $\sum_{n=1}^\infty \left(\frac{n}{2n-1}\right)^n$ converges I'm trying to show that $\sum_{n=1}^\infty \left(\frac{n}{2n-1}\right)^n$ converges. Using the Limit Ratio Test for Series, we want to show that $\lim_{n\to \infty} \left\lvert \frac{a_{n+1}}{a_n}\right\lvert<1$. However, I'm having trouble finding said limit (I know that it is equal to $\frac{1}{2}$, but I don't know how show it). Thanks in advance

Note that we can write
$$\begin{align}
\left(\frac{n}{2n-1}\right)^n&=\frac{1}{2^n}\left(\frac{1}{1-\frac{1}{2n}}\right)^n\tag1\\\\
&\le \frac{1}{2^{n-1}}\tag2
\end{align}$$
where in going from $(1)$ to $(2)$ we invoked Bernoulli's Inequality. 

NOTE:
The OP was pursuing a way forward that relied on the ratio test.  Proceeding, we have
$$\begin{align}
\lim_{n\to \infty}\frac{a_{n+1}}{a_n}&=\lim_{n\to \infty}\frac{\left(\frac{n+1}{2n+1}\right)^{n+1}}{\left(\frac{n}{2n-1}\right)^{n}}\\\\
&=\lim_{n\to \infty}\left(\left(\frac{n+1}{2n+1}\right)\,\left(1-\frac{1}{n(2n+1)}\right)^n\right)\\\\
&=\frac12
\end{align}$$
since from Bernoulli's Inequality we have 
$$1\ge  \left(1-\frac{1}{n(2n+1)}\right)^n\ge 1-\frac{1}{2n+1}$$
whence application of the squeeze theorem reveals that the limit is $1$.
