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How to find the $\lim_{n \to \infty} \left(\dfrac{(n+1)(n+2)\cdots(n+2n)}{n^{2n}}\right)^{1/n}$? I know how to find it for the indeterminate form of $1^{\infty}$ by converting it into $0/0$ form, but this cannot be converted into any known indeterminate form: $(0/0)^0$ . Can we convert it into an integral and then try to solve at as infinity is involved?. May someone help? Also please don't use any theorems or formula for limits except perhaps L'Hospital rule which I know. If you do please provide its proof as well.

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  • $\begingroup$ see a technique for similar question math.stackexchange.com/a/1913608/72031 $\endgroup$ – Paramanand Singh Sep 9 '16 at 20:08
  • $\begingroup$ I know, I'm not new here. $\endgroup$ – Matt Sep 12 '16 at 17:16
  • $\begingroup$ @RaghavSingal Just assumed you had forgotten -- this does happen. If not, are you waiting for more details or a more complete answer? $\endgroup$ – Clement C. Sep 12 '16 at 17:37
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First "trick:" convert to the exponential form.

$$ \left(\frac{\prod_{k=1}^{2n}(n+k)}{n^{2n}}\right)^{\frac{1}{n}} =\exp\left({\frac{1}{n}\ln\left(\frac{\prod_{i=1}^{2n}(n+k)}{n^{2n}}\right)}\right) $$ Now, let us focus on the exponent: $$ \frac{1}{n}\ln\left(\frac{\prod_{k=1}^{2n}(n+k)}{n^{2n}}\right) = \frac{1}{n}\ln\left(\prod_{k=1}^{2n}\frac{n+k}{n}\right) = \frac{1}{n}\ln\left(\prod_{k=1}^{2n}\left(1+\frac{k}{n}\right)\right) = \frac{1}{n}\sum_{k=1}^{2n}\ln\left(1+\frac{k}{n}\right) $$ At that point, it starts to really look like a Riemann sum, so let us massage it a little bit more: $$ \frac{1}{n}\ln\left(\frac{\prod_{k=1}^{2n}(n+k)}{n^{2n}}\right) = \frac{1}{n}\sum_{k=1}^{2n}\ln\left(1+\frac{k}{n}\right) = \frac{2}{2n}\sum_{k=1}^{2n}\ln\left(1+2\frac{k}{2n}\right) = 2\cdot\frac{1}{2n}\sum_{k=1}^{2n}f\!\left(\frac{k}{2n}\right) $$ for $f\colon [0,1]\to \mathbb{R}$ defined by $f(x)=\ln(1+2x)$.

We have$^{(\dagger)}$ $$\frac{1}{2n}\sum_{k=1}^{2n}f\!\left(\frac{k}{2n}\right)\xrightarrow[n\to\infty]{} \int_0^1 f = \frac{1}{2}(3\ln 3 -2)$$ and by continuity of the exponential your limit will be $$\left(\frac{\prod_{k=1}^{2n}(n+k)}{n^{2n}}\right)^{\frac{1}{n}} \xrightarrow[n\to\infty]{} e^{2\int_0^1 f} = e^{3\ln 3 -2)} = \frac{27}{e^2}.$$


$(\dagger)$ Here, we use the following theorem, which essentially follows from the definition of Riemann integration:

Theorem. (Riemann sums converge to the integral.) Let $f\colon [a,b]\to\mathbb{R}$ be a continuous function. Then $$ \frac{1}{n}\sum_{k=1}^n f\left(a+k\frac{b-a}{n}\right) \xrightarrow[n\to\infty]{} \int_a^b f $$ and $$ \frac{1}{n}\sum_{k=0}^{n-1} f\left(a+k\frac{b-a}{n}\right) \xrightarrow[n\to\infty]{} \int_a^b f. $$

This is a particular case of a slightly more general theorem ($f$ only needs to be Riemann integrable, and here we took a regular subdivision of $[a,b]$ in intervals of the same length $\frac{b-a}{n}$ instead of an arbitrary subdivision.)

In our case, $a=0$ and $b=1$, and we take a subdivision with $m\stackrel{\rm def}{=} 2n$ points: $$ \frac{1}{m}\sum_{k=1}^m f\left(\frac{k}{m}\right) \xrightarrow[m\to\infty]{} \int_0^1 f. $$

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  • $\begingroup$ How $\frac{n+k}{n^{2n}}$ becomes $\frac{n+k}{n}$? $\endgroup$ – Von Neumann Sep 9 '16 at 18:23
  • $\begingroup$ @FourierTransform It did not. $\frac{1}{n^{2n}} \prod_{k=1}^{2n} (n+k)$ became $\prod_{k=1}^{2n} \frac{n+k}{n}$. $\endgroup$ – Clement C. Sep 9 '16 at 18:24
  • $\begingroup$ Oh! Sorry. I read wrong. Great! $\endgroup$ – Von Neumann Sep 9 '16 at 18:25
  • $\begingroup$ I know this comment doesn't really add much, but I'm just wondering how you came up with this whole process. It seems so genius to me that I can only conclude that it took hours. $\endgroup$ – Polygon Sep 9 '16 at 18:28
  • $\begingroup$ Whenever I see something of the form $a_n^{b_n}$, my first instinct is that I don't understand powers, so I rewrite it $e^{b_n \ln a_n}$ (or $2^{b_n \log_2 a_n}$ if it's computer science). Then, after that, I don't understand products: if there are products inside the logarithm, I "convert" them into a sum of logarithms instead. (The rationale being, I know of more results on sum and series than on products). And a sum of the form $\frac{1}{n}\sum_k g(\frac{k}{n})$ begs for being recognized as a Riemann sum -- it does not always work, but it's worth trying for the 90% cases where it does. $\endgroup$ – Clement C. Sep 9 '16 at 18:32
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Maybe you can use the exponential form.

Forgive me for what I'm going to write, I know it's bad formalism but it may be effective:

$$\left(\frac{0}{0}\right)^0 = e^{0\cdot \ln\left(\frac{0}{0}\right)}$$

Then you may use hospital for the limit inside the logarithm and proceed...


Answer:-

$$ \text{let } y = \left(\dfrac{(n+1)(n+2)\cdots(n+2n)}{n^{2n}}\right)^{1/n}$$

$$\implies\log_e y = {1\over n}\log_e \left(\dfrac{(3n)!}{n! \times n^{2n}}\right)$$

$$\implies e^{{1\over n}\log_e \left(\dfrac{(3n)!}{n! \times n^{2n}}\right)} = y$$


$$\text{let } z = \log_e \left(\dfrac{(3n)!}{n! \times n^{2n}}\right)$$

$$\implies z = \log_e (3n)!- \log_e n! - 2n\log_e n$$

$$\color{red}{\implies z = 3n\log_e (3n) - 3n - (n\log_e n - n) - 2n\log_e n}$$

$$\implies z = 3n\log_e (n) + 3n\log_e 3 - 2n - n\log_e n - 2n\log_e n$$

$$\implies z = n(3\log_e 3 - 2) $$


$$\text{substituting the value of z in y}$$

$$y= e^{{1\over n}z}$$

$$\implies y= e^{3\log_e 3 - 2}$$

$$\text{Finally finding the limit}$$ $$\implies \lim_{n \to \infty} y = \lim_{n \to \infty} e^{3\log_e 3 - 2} = e^{3\log_e 3 - 2} = {e^{3\log_e 3}\over e^2} = {27\over e^2}$$

$$\color{red}{RED} \leftarrow \text{Stirling approximation}$$

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  • $\begingroup$ I am getting $$e^{{1\over n}\log_e \left(\dfrac{(3n)!}{n! \times n^{2n}}\right)} = y$$, any idea how should i proceed ? $\endgroup$ – A---B Sep 9 '16 at 18:50
  • $\begingroup$ @A---B Stirling approximation for example! $\endgroup$ – Von Neumann Sep 9 '16 at 18:51
  • $\begingroup$ Then you should complete you answer. $\endgroup$ – A---B Sep 9 '16 at 18:55
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    $\begingroup$ @A---B Nope. I just gave a Hint, since the question was about the indeterminate form. $\endgroup$ – Von Neumann Sep 9 '16 at 18:56
  • $\begingroup$ hey, I got the answer. i will more than happy to edit it inside your answer. maybe that downvote will turn into upvote. $\endgroup$ – A---B Sep 9 '16 at 19:36
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HINT:

Let $A=\left(\dfrac{(n+1)(n+2)\cdots(n+2n)}{n^{2n}}\right)^{1/n}$

$$\implies\ln A=\dfrac1n\sum_{r=1}^{2n}\ln\left(1+\dfrac rn\right)$$

Now like Find $\lim\limits_{n \to \infty} \frac{1}{n}\sum\limits^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}}$, let $2n=m$

$$\implies\ln A=2\cdot\dfrac1m\sum_{r=1}^m\ln\left(1+2\cdot\dfrac rm\right)$$

$$\text{As }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

$$\implies\ln A=2\int_0^1\ln(1+2x)\ dx$$

Now integrate by parts,

$$\int\ln(1+2x)\ dx=\ln(1+2x)\int dx-\int\left(\dfrac{d\{\ln(1+2x)\}}{dx}\cdot\int dx\right)dx$$

$$=x\ln(1+2x)-\int\dfrac x{1+2x}dx$$

Now for $\int\dfrac x{1+2x}dx,$ set $1+2x=y$ to ultimately find that

$$\ln A=2\left(\dfrac32\ln(1+2)-1\right)=3\ln 3-2=\ln\dfrac{3^3}{e^2}\text{ as }\ln(e)=1$$

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  • $\begingroup$ should it not be ..$ln\left(1+\dfrac{r}{2n}\right)$? $\endgroup$ – G Cab Sep 9 '16 at 18:51
  • $\begingroup$ @GCab, No, Please find the updated answer. $\endgroup$ – lab bhattacharjee Sep 10 '16 at 1:58
  • $\begingroup$ @right, so .. one mis-attention per each .. $\endgroup$ – G Cab Sep 10 '16 at 16:30

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