# $\lim_{n \rightarrow \infty} n \left( \frac{1}{(n+1)^2} + \frac{1}{(n+2)^2} + \cdots + \frac{1}{(2n)^2} \right)$ as Riemann sum?

I am trying to evaluate the limit $$\lim_{n \rightarrow \infty} n \left( \frac{1}{(n+1)^2} + \frac{1}{(n+2)^2} + \cdots + \frac{1}{(2n)^2} \right).$$

I have been trying to convert it to the Riemann sum of some integral, but have been unable to recongize what the integral should be. How should I go about solving this problem?

• Does it matter? It is not part of the index of summation, after all, so you can just distribute it. – user153224 Jan 30 '17 at 3:53
• – lab bhattacharjee Jan 30 '17 at 4:11

$$\lim_{n \rightarrow \infty} n \left(\sum_{k =1}^{n} \frac{1}{(n+k)^2}\right) =\lim_{n \rightarrow \infty} n \left(\sum_{k =1}^{n} \frac{1}{n^2(1+\dfrac kn)^2}\right) \\\lim_{n \rightarrow \infty} \left(\sum_{k =1}^{n} \frac{n}{n^2(1+\dfrac kn)^2}\right)=\\\lim_{n \rightarrow \infty} \left(\sum_{k =1}^{n} \frac{1}{(1+\dfrac kn)^2}\dfrac 1n\right)=\\\int_{0}^{1}\dfrac{1}{(1+x)^2}dx=\\ \int_{0}^{1}{(1+x)^{-2}}dx=\dfrac{(1+x)^{-1}}{-1} \space [0,1]\\\dfrac{-1}{1+x}\space [0,1] =\dfrac {-1}{1+1}-( \dfrac {-1}{1+0})=\\\dfrac12$$

• Darn, beat me to it ! – Antonios-Alexandros Robotis Jan 30 '17 at 3:56
• Can you go on ,and find the answer of integral ? – Khosrotash Jan 30 '17 at 3:59
• @Khosrotash can you explain how you went from the second last line to the final line? – IntegrateThis Jan 30 '17 at 4:01
• I think you did this $$\int_{0}^{1}\dfrac{1}{1+x^2}dx=\dfrac{\pi}{4}$$ – Khosrotash Jan 30 '17 at 4:01
• @RealMath you evaluated the integral incorrectly, use a substitution. – IntegrateThis Jan 30 '17 at 4:03

The polygamma functions are great to solve this kind of problems.

$$\sum_{k=1}^N \frac{1}{(x+k)^2}=\psi^{(1)}(x+1)-\psi^{(1)}(N+x+1)$$

$\psi^{(1)}(z)$ is the trigamma function, i.e.: the polygamma[1,z] function.

With $z=n$ and $N=n$ :

$$n\sum_{k=1}^n \frac{1}{(n+k)^2}=n\left(\psi^{(1)}(n+1)-\psi^{(1)}(2n+1) \right)$$

The asymptotic expansion of the trigamma function is : $\psi^{(1)}(z+1)=\frac{1}{z}-\frac{2}{z^2}+O\left(\frac{1}{z^3}\right)$

$$n\sum_{k=1}^n \frac{1}{(n+k)^2}=n\left(\frac{1}{n}-\frac{2}{n^2}-\frac{1}{2n}+\frac{2}{4n^2}+O\left(\frac{1}{n^3}\right) \right) = \frac{1}{2}+\frac{3}{2n}+O\left(\frac{1}{n^2}\right)$$

$$\lim_{n \rightarrow \infty} n\sum_{k=1}^n \frac{1}{(n+k)^2}=\frac{1}{2}$$