Limit of a function $p_n$ when $n \rightarrow +\infty$

Question

We have the following function : $$p_n = \frac{n}{(2n+1)^2} \frac{1}{\sum_{d=1}^n \dfrac{1}{(2n+d)^2} }$$

And we would like to know the limit of $p_n$ when $n \rightarrow +\infty$.

I don't know at all how to do it... I just know that $\dfrac{n}{(2n+1)^2} \rightarrow_{n \rightarrow +\infty} 0$. But how to study the second member ?

Answer 1

Put first $$v_n=\sum_{d=1}^n \frac{1}{(2n+d)^2}=\frac{1}{4n}u_n$$ with $$u_n=\frac{1}{n}\sum_{d=1}^n\frac{1}{(1+\frac{d}{2n})^2}$$

The sum $u_n$ is a Riemann sum, so $u_n$ is convergent to $L=\displaystyle \int_0^1 f(t)dt$, with $f(t)=\frac{1}{(1+t/2)^2}$.

Now your expression is $\frac{4n^2}{(2n+1)^2u_n}$, so it is easy to see that the limit is $\displaystyle \frac{1}{L}$. It remain to compute $L$.

Answer 2

at a first step you must calculate the sum $$ \sum_{d=1}^n\frac{1}{(2n+d)^2}$$

Answer 3

Write $\ p_n$ as :$$\ p_n = (\frac{n^2}{(2n+1)^2})(\frac{1}{\frac{1}{n}\sum_{d=1}^n\frac{1}{(\frac{d}{n}+2)^2}})$$Now , You can use "limit of a sum" for second part and limit of first can be calculated by dividing numerator - denominator by $\ n^2$.. Final limit will be product of limits of both parts.