Finding $\lim_{n\to \infty}\left((1+\frac{1}{n^2})(1+\frac{2^2}{n^2})^2\cdots(1+\frac{n^2}{n^2})^n\right)^{\frac{1}{n}}$

Question

Calculate the following limit: $$\lim_{n\to \infty}\left((1+\frac{1}{n^2})(1+\frac{2^2}{n^2})^2\cdots(1+\frac{n^2}{n^2})^n\right)^{\frac{1}{n}}$$

My attempt:

Let $$y=\lim_{n\to \infty}\left((1+\frac{1}{n^2})(1+\frac{2^2}{n^2})^2\cdots(1+\frac{n^2}{n^2})^n\right)^{\frac{1}{n}}=\lim_{n\to \infty}\left((1+\frac{1}{n^2})^{\frac{1}{n}}(1+\frac{2^2}{n^2})^{\frac{2}{n}}\cdots(1+\frac{n^2}{n^2})^{\frac{n}{n}}\right)$$ Now, taking logarithm on both sides to get: $$\log y=\lim_{n \to \infty}\sum_{k=1}^n\frac{k}{n}\log(1+(\frac{k}{n})^2)$$ But I am unable to convert it into integral form (riemann sum to integral ) as the expression $\frac{k}{n}$ is present instead of $\frac{1}{n}$. Can someone please help me in solving this question. Thanks in advance.

Answer 1

By your work $$\lim_{n\to +\infty}\left(\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right)^2\cdots\left(1+\frac{n^2}{n^2}\right)^n\right)^{\frac{1}{n^2}}=$$ $$=e^{\int\limits_0^1x\ln(1+x^2)dx}=e^{\frac{1}{2}\int\limits_0^1\ln(1+x)dx}=e^{\frac{1}{2}\left((1+x)\ln(1+x)-x\right)|_0^1}=e^{\ln2-\frac{1}{2}}=\frac{2}{\sqrt e}.$$ Can you end it now?

I used your work and $$\lim_{n \to \infty}\sum_{k=1}^n\frac{k}{n}\ln\left(1+\left(\frac{k}{n}\right)^2\right)\frac{1}{n}=\int\limits_0^1x\ln(1+x^2)dx.$$ Since $$\lim_{n\rightarrow+\infty}\left(\frac{2}{\sqrt{e}}\right)^n=+\infty,$$ we see that starting limit does not exist.

Answer 2

Considert the factors

$$\left (1+\frac{k^2}{n^2}\right )^k,\,\, n/2< k \le n.$$

Each of these is greater than

$$\left (1+\frac{(n/2)^2}{n^2}\right )^{n/2} = \left (\frac{5}{4}\right)^{n/2}.$$

There are more than $n/2$ such factors. Thus the product of these factors is at least $(5/4)^{n^2/4}.$ Now take the $n$th root to get $(5/4)^{n/4}\to \infty.$ Thus the limit of the original expression is $\infty.$