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
 A: Hint. One may use the root test, what is
$$
\lim_{n \to \infty} \sqrt[n]{a_n}\,?
$$ 
A: 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 $1$.
A: We have $n/(2n-1) \to 1/2.$ Hence for large $n$ we have $n/(2n-1) < 2/3.$ Since $\sum_n (2/3)^n < \infty,$ the given series converges by the comparison test.
A: You have
$$\left(\frac{n}{2n-1}\right)^n\sim \left( \frac{1}{2}\right)^n\frac{1}{\sqrt{e}}\;  (n\to +\infty)$$
$\implies $ the positive series converge.
A: If $a_n = \left(\frac{n}{2n-1}\right)^n$, then
\begin{align} \frac{a_{n+1}}{a_n} &= \left(\frac{n+1}{2(n+1)-1}\right)^{n+1}/\left(\frac{n}{2n-1}\right)^n \\
&= \left(\frac{n+1}{2n+1}\right)\frac{\left(\frac{n+1}{2n+1}\right)^n}{\left(\frac{n}{2n-1}\right)^n} \\
&= \left(\frac{n+1}{2n+1}\right)\left(\frac{(n+1)(2n-1)}{n(2n+1)}\right)^n \\
&=\left(\frac{n+1}{2n+1}\right)\left(\frac{2n^2+n-1}{2n^2+n}\right)^n.
\end{align}
Now, $\frac{2n^2+n-1}{2n^2+n}<1$, so it follows that $\lim\limits_{n\rightarrow\infty}{\left(\frac{2n^2+n-1}{2n^2+n}\right)^n}\le 1$. (It is not too hard to show, in fact, that the limit is exactly $1$.) Hence
$$\lim\limits_{n\rightarrow\infty}{\left|\frac{a_{n+1}}{a_n}\right|} = \left(\lim\limits_{n\rightarrow\infty}{\frac{n+1}{2n+1}}\right)\left(\lim\limits_{n\rightarrow\infty}{\left(\frac{2n^2+n-1}{2n^2+n}\right)^n}\right)\le\frac{1}{2}\cdot 1 = \frac{1}{2} $$
i.e. $\lim\limits_{n\rightarrow\infty}{\left|\frac{a_{n+1}}{a_n}\right|} < 1$.
