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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 $2/e$

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marked as duplicate by Henning Makholm, Cesareo, Yanior Weg, José Carlos Santos sequences-and-series May 1 at 21:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I don't know where i miss '2'. And its been 4 hours , still didn't find it. :( $\endgroup$ – Cloud JR Apr 30 at 18:23
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    $\begingroup$ 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 $4/e.$ ] $\endgroup$ – Thomas Andrews Apr 30 at 18:33
  • $\begingroup$ @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 $\endgroup$ – Cloud JR Apr 30 at 18:40
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    $\begingroup$ 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. $\endgroup$ – achille hui Apr 30 at 18:52
  • $\begingroup$ @achille hui. Thanks, writing it as factorial, helps a lot in simplify the notation. $\endgroup$ – Cloud JR Apr 30 at 18:56
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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 $a_{n+1}.$

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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}

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

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