Find the limit:
$
\lim\limits_{n\to\infty}\sqrt[n]{\left(\dfrac{1 + n}{n^2} \right)\left(\dfrac{4 + 2n}{n^2} \right)…\left(\dfrac{n^2 + n^2}{n^2} \right)}
$
I tried simplifying this limit and the one I get to is:
$
\lim\limits_{n\to\infty}\dfrac{1}{n^2}\bigg(\big(2n\big)!\bigg)^{\dfrac{1}{n}}
$
I have an instruction to write the limit as a definite integral and then calculate its value. I think that there should be a way to represent the last limit as a Riemann sum and then calculate it with the integral. But I'm not sure how to get to the Riemann sum.
Looking forward for any ideas!
Best Answer
The logarithm of the initial expression is $$ \frac 1n \sum_{k=1}^n \ln\left(\frac{k^2+kn}{n^2}\right) = \frac 1n \sum_{k=1}^n \ln\left(\left(\frac kn \right)^2 + \frac kn \right) $$ which is a Riemann sum for $$ \int_0^1 \ln(x^2 + x) \, dx = \int_0^2 \ln x \, dx = 2 \ln 2 - 2 \, . $$