Mistake in converting Riemann Sum

Question

EDIT: I know how to integrate the last part. I'm just try to find mistake in converting Sum to integral

Question:

$$a_n=\left(\left(1+\left(\frac1n\right)^2\right)\left(1+\left(\frac2n\right)^2\right)\cdots\left(1+\left(\frac{n}n\right)^2\right)\right)^n$$ find
$$\lim_{n\to\infty}a_n^{-1/n^2}$$

My Approach: Let $$y=\lim_{n\to\infty}a_n^{-1/n^2}$$ I Converted this to $$\ln y=\lim_{n\to\infty}{-1\over n}\sum^n_{k=1}\left(\ln\left(1+{k^2\over n^2}\right) \right)$$

and then to (here's where i think the mistake is in conversion, not integration):

$$\ln y=-\int_0^1\ln(1+x^2)dx$$

However, the answer is $\ln y=1/2-\ln2$

Please the help me find mistake

Answer 1

You're right, your book is wrong.

Answer 2

You are correct but..

By parts,

$$\int_0^1\ln (1+x^2)dx=$$ $$[x\ln (1+x^2)]_0^1-2\int_0^1\frac {x^2+1-1}{1+x^2}dx =$$

$$\ln (2)-2+2\arctan (1) =$$ $$\ln (2)-2+\frac {\pi}{2} =-\ln y$$