I have been asked to find the limit of the following series by expressing it as definite integral:
If $na=1$ always and $n$ tends to infinity, find the limiting value
$$\lim_{n \to \infty}\left(\prod_{r=1}^{n}[1+(ra)^2]^{1/r}\right)$$
The answer given is $\exp\left(\dfrac{\pi^2}{24}\right)$.
I took the log of both sides and set up my integral as $$\int_{0}^{1}\log(1+x^2)dx$$
But my answer came out wrong.
Thanks in advance
Best Answer
Since $ na = 1$ we can write the product as $\displaystyle \mathcal{P} := \prod_{r=1}^\infty\left(1+\frac{r^2}{n^2}\right)^{1/r}.$
Then taking logs and using the continuity of logs
\begin{align} \log(\mathcal P) & = \log\left(\lim_{n\to\infty}\prod_{r=1}^n\left(1+\frac{r^2}{n^2}\right)^{1/r}\right) \\&=\lim_{n\to\infty}\log\left(\prod_{r=1}^n\left(1+\frac {r^2}{ n^2}\right)^{1/r}\right)\\ &=\lim_{n\to\infty}\sum_{r=1}^n\frac1r\log\left(1+ \frac {r^2}{ n^2}\right)\\ &=\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^n\frac{n}{r} \log\left(1+ \frac {r^2}{ n^2}\right)\\ &=\int_0^1\frac1x\log(1+x^2)\,\mathrm{d}x\\ &=\int_0^1\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{k+1}\,\mathrm{d}x\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)(2k+2)} \\& = \frac{1}{2}\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)^2}\\& = \frac{1}{4}\sum_{k=0}^\infty\frac{1}{(k+1)^2} \\&=\frac{\pi^2}{24}. \end{align}
Therefore $\displaystyle \mathcal P = \exp\left(\frac{\pi^2}{24}\right).$