# 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

## 5 Answers

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 $1$.

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. – aL_eX Dec 2 '16 at 22:41

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. – Mark Viola Dec 2 '16 at 23:16
• @Dr.MV can you explain what you mean by "that is a key to this development"? – Joey Zou Dec 3 '16 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? – Mark Viola Dec 3 '16 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? – Joey Zou Dec 3 '16 at 1:35
• Yes, it's fine. – Mark Viola Dec 3 '16 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 '16 at 0:06
• @zhw. Yes .Thanks. – hamam_Abdallah Dec 3 '16 at 3:45