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
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5 Answers
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Hint. One may use the root test, what is $$ \lim_{n \to \infty} \sqrt[n]{a_n}\,? $$
answered Dec 2, 2016 at 22:31
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$\begingroup$ @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. $\endgroup$– aL_eXDec 2, 2016 at 22:41
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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 $1$.
answered Dec 2, 2016 at 22:44
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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.
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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. $\lim\limits_{n\rightarrow\infty}{\left|\frac{a_{n+1}}{a_n}\right|} < 1$.
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$\begingroup$ 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. $\endgroup$ Dec 2, 2016 at 23:16
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$\begingroup$ @Dr.MV can you explain what you mean by "that is a key to this development"? $\endgroup$– Joey ZouDec 3, 2016 at 0:04
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$\begingroup$ 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? $\endgroup$ Dec 3, 2016 at 0:16
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$\begingroup$ @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? $\endgroup$– Joey ZouDec 3, 2016 at 1:35
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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.
answered Dec 2, 2016 at 22:37
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$\begingroup$ Actually, no: $[(n/(2n-1)]^n \sim (1/\sqrt e)\cdot (1/2^n).$ $\endgroup$– zhw.Dec 3, 2016 at 0:06
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