# 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

Hint. One may use the root test, what is $$\lim_{n \to \infty} \sqrt[n]{a_n}\,?$$

• @Oliver Oloa Thank you for your answer. I haven't covered the Root Test, but I looked it up and I see how to get the answer. Is there another way to show this? It's because my professor will probably mark me down for using something not covered in lectures. Dec 2, 2016 at 22:41

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 $1$.

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.

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. $\lim\limits_{n\rightarrow\infty}{\left|\frac{a_{n+1}}{a_n}\right|} < 1$.

• While you're correct that showing that the limit is $1$, that is a key to this development. See my solution for an elementary approach. Dec 2, 2016 at 23:16
• @Dr.MV can you explain what you mean by "that is a key to this development"? Dec 3, 2016 at 0:04
• Without showing that $\lim_{n\to \infty}\left(\frac{2n^2+n-1}{2n^2+n}\right)^n=1$, then how do you propose to continue to arrive at the limit of interest? Dec 3, 2016 at 0:16
• @Dr.MV I showed that it is at most $1$; hence the final ratio is at most $1/2$. That's all we need to apply the ratio test, right? Dec 3, 2016 at 1:35
• Yes, it's fine. Dec 3, 2016 at 2:00

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.

• Actually, no: $[(n/(2n-1)]^n \sim (1/\sqrt e)\cdot (1/2^n).$
– zhw.
Dec 3, 2016 at 0:06
• @zhw. Yes .Thanks. Dec 3, 2016 at 3:45