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I need to find the value of $$L=\lim_{n \rightarrow \infty}\displaystyle\prod_{r=1}^{n} \left(1+\dfrac{r^2}{n^2}\right)^{1/n}$$

Is doing this OK?--

$$\begin{align} L &=\lim_{n \rightarrow \infty}\left(1+\dfrac{1^2}{n^2}\right)^{1/n}\left(1+\dfrac{2^2}{n^2}\right)^{1/n} \ ... \ \left(1+\dfrac{n^2}{n^2}\right)^{1/n} \\ &=\lim_{n \rightarrow \infty} e^{1^2/n^3} e^{2^2/n^3} \ ... \ e^{n^2/n^3} \\ &= \lim_{n \rightarrow \infty} e^{n (n+1)(2n+1)/6n^3}\\ &=e^{1/3} \end{align}$$

I get $L=e^{1/3}$ but the answer is $2e^{\pi/2-2}$.

Where have I gone wrong? How to arrive at the correct answer $2e^{\pi/2-2}$?

Thank you.

user1001001
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  • Careful, the limit of a product isn't (generally) the product of the limit (and it is what you're doing here). Please take the logarithm of the product and play with the sum. – user37238 Sep 12 '14 at 07:47
  • @user37238 Can't we take the limit on the terms individually if they are multiplied? That's what I have been taught.. – user1001001 Sep 12 '14 at 07:51
  • You can if the number of terms that are multiplied is fixed. Here the number of terms is increasing with $n$. – user37238 Sep 12 '14 at 08:03

1 Answers1

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$$L=\lim_{n \rightarrow \infty}\displaystyle\prod_{r=1}^{n} \left(1+\dfrac{r^2}{n^2}\right)^{1/n}$$

$$\ln L=\lim_{n \rightarrow \infty}\frac1n\sum_{r=1}^n\ln\left(1+\dfrac{r^2}{n^2}\right)=\int_0^1\ln(1+x^2)dx$$

$$\int\ln(1+x^2)dx=\ln(1+x^2)\int dx-\int\left(\frac{d[\ln(1+x^2)]}{dx}\int dx\right)dx$$

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

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

$$=x\ln(1+x^2)-2\int dx+2\int\frac1{1+x^2}dx$$

lab bhattacharjee
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  • Sir could you please explain how you made the transition from the sum to the integral (the 3rd step)? – user1001001 Sep 12 '14 at 07:54
  • @Pkwssis, See http://math.stackexchange.com/questions/465075/find-lim-limits-n-to-infty-frac1n-sum-limits2n-r-1-fracr-sq/465081#465081 and http://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2?lq=1 – lab bhattacharjee Sep 12 '14 at 07:58