# Evaluate $\lim_{n \rightarrow \infty } \frac {[(n+1)(n+2)\cdots(n+n)]^{1/n}}{n}$ [duplicate]

Evaluate $$\lim_{n \rightarrow \infty~} \dfrac {[(n+1)(n+2)\cdots(n+n)]^{\dfrac {1}{n}}}{n}$$ using Cesáro-Stolz theorem.

I know there are many question like this, but i want to solve it using Cesáro-Stolz method and no others.

I took log and applied Cesáro-Stolz, I get $$\log{2}+n\log\cfrac{n}{n+1}$$

Which gives me answer as $$\frac{2}{e}$$ . But answer is $$\frac{4}{e}$$. Could someone help?.

Edit: On taking log, $$\lim_{n \to \infty} \frac{-n\log n + \sum\limits_{k=1}^{n} \log \left(k+n\right)}{n} \\= \lim_{n \to \infty} \left(-(n+1)\log (n+1) + \sum\limits_{k=1}^{n+1} \log \left(k+n\right)\right) - \left(-n\log n + \sum\limits_{k=1}^{n} \log \left(k+n\right)\right) \\ = \lim_{n \to \infty} \log \frac{2n+1}{n+1} - n\log \left(1+\frac{1}{n}\right) = \log 2 - 1$$ Which gives $$2/e$$

## marked as duplicate by Henning Makholm, Cesareo, Yanior Weg, José Carlos Santos sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 1 at 21:52

• I don't know where i miss '2'. And its been 4 hours , still didn't find it. :( – Cloud JR Apr 30 at 18:23
• Well, $4/e$ is correct. We might need more details on how you applies Cesaro Stolz. [Another approach, without Cesaro-Stolz, is you can rewrite it as $$\left[(1+1/n)(1+2/n)\cdots(1+n/n)\right]^{1/n}$$ and take the log, which is a Riemann sum of $\int_{1}^{2} \log x\,dx.$ The anti-derivative of $\log x$ is $x\log x - x$ and you get $4/e.$ ] – Thomas Andrews Apr 30 at 18:33
• @Thomas Andrews the second method works and i got 4/e. I want to clarify where it went wrong in first method. I don't want to give it up. I will add more details asap – Cloud JR Apr 30 at 18:40
• The expression can be rewritten as $\left(\frac{(2n!)}{n!n^n}\right)^{1/n}$ Taking logarithm gives you $\frac{\log(2n!) - log(n!) - n\log(n)}{n}$ Apply CS, the difference of the denominator is $$\log(\color{red}{2}n+2)+\log(\color{red}{2}n+1) - \log(n+1) - (n+1)\log(n+1)+ n\log n$$ It seems you have missed one of the $\color{red}{2}$ above. – achille hui Apr 30 at 18:52
• @achille hui. Thanks, writing it as factorial, helps a lot in simplify the notation. – Cloud JR Apr 30 at 18:56

Let $$a_n=\sum_{i=1}^{n} \log\left(1+\frac{i}n\right)$$ be the numerator of the logarithm, with denominator $$b_n=n.$$

The key is that the first term $$\log(1+1/n)$$ of $$a_n$$ doesn't cancel with any of the terms $$\log(1+i/(n+1)).$$ It alone is subtracted, so, while there are $$n$$ occurrences of $$-\log(1+1/n)$$, there are still two more terms for $$a_{n+1}.$$ So you get:

$$a_{n+1} - a_n = \left(\frac{2n+1}{n+1}\right)+\log\left(\frac{2n+2}{n+1}\right)-n\log(1+1/n)$$

Basically, the "cancellation" happens when we have $$\log\left(1+\frac{i}{n+1}\right)-\log\left(1+\frac{i+1}{n}\right)= \log\left(\frac{n}{n+1}\right)$$

While we can assume there is a $$i=0$$ in $$a_{n+1},$$ since it adds $$0=\log 1$$ to $$a_{n+1},$$ that means there are $$n+2$$ values of $$i$$ in $$\sum_{i=0}^{n+1},$$ and hence there is not cancellation of the $$i=n$$ and $$i=n+1$$ terms from $$a_{n+1}.$$

After took log, we have $$\frac{\sum _{i=1}^n \log (i+n)-n \log (n)}{n}$$ Applied cesaro stolez, we have \begin{align} & \left(\sum _{i=1}^{n+1} \log (i+n+1)-(n+1) \log (n+1)\right)-\left(\sum _{i=1}^n \log (i+n)-n \log (n)\right) \\&= \left(\sum _{i=2}^{n+2} \log (i+n)-(n+1) \log (n+1)\right)-\left(\sum _{i=1}^n \log (i+n)-n \log (n)\right)\\&= (\log (2 n+1)+\log (2 n+2)-(n+1) \log (n+1))-(\log (n+1)-n \log (n)) \\ &= n \log \left(\frac{n}{n+1}\right)+\log \left(\frac{(2 n+1) (2 n+2)}{(n+1)^2}\right) \\ &\rightarrow -1+\log (4) \end{align}

• Thanks man !, i got it. – Cloud JR Apr 30 at 18:54