# Thread: Find the sum of the following series

1. ## Find the sum of the following series

Find the sum of the following series as $\displaystyle n \to \infty$

$\displaystyle \frac{n}{n^2+1^2} + \frac{n}{n^2+2^2} + \frac{n}{n^2+3^2}+...+\frac{n}{n^2+n^2}$

2. The sum of the first n terms is...

$\displaystyle S_{n}= \frac{n}{n^{2}+1^{2}} + \frac{n}{n^{2}+2^{2}} + \dots + \frac{n}{n^{2}+(n-1)^{2}} + \frac{n}{n^{2}+n^{2}} =$

$\displaystyle = \frac{1}{n}\cdot \{\frac{1}{1+(\frac{1}{n})^{2}} + \frac{1}{1+(\frac{2}{n})^{2}} + \dots + \frac{1}{1+(\frac{n-1}{n})^{2}} + \frac{1}{2}\}$

But is...

$\displaystyle \frac{1}{1+(\frac{1}{n})^{2}} = 1 - (\frac{1}{n})^{2} + (\frac{1}{n})^{4} - \dots + (-1)^{i} (\frac{1}{n})^{2i} + \dots$

$\displaystyle \frac{1}{1+(\frac{2}{n})^{2}} = 1 - (\frac{2}{n})^{2} + (\frac{2}{n})^{4} - \dots + (-1)^{i} (\frac{2}{n})^{2i} + \dots$

$\displaystyle \dots$

$\displaystyle \frac{1}{1+(\frac{n-1}{n})^{2}} = 1 - (\frac{n-1}{n})^{2} + (\frac{n-1}{n})^{4} - \dots + (-1)^{i} (\frac{n-1}{n})^{2i} + \dots$

... so that after some easy steps we obtain...

$\displaystyle S_{n}= 1 - \frac{1}{n^{3}}\cdot \sum_{k=1}^{n} k^{2} + \frac{1}{n^{5}}\cdot \sum_{k=1}^{n} k^{4} -\dots \frac{(-1)^{i}}{n^{2i+1}}\cdot \sum_{k=1}^{n} k^{2i} + \dots$

Now if we remember the formula...

$\displaystyle \sum_{k=1}^{n} k^{2i} = \frac{B_{2i+1}(n+1) - B_{2i+1}(0)}{2i+1}$

... from that we derive...

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

... so that we conclude that...

$\displaystyle \lim_{n \rightarrow \infty} S_{n}= 1 - \frac{1}{3} + \frac{1}{5} - \dots + \frac{(-1)^{i}}{2i+1} + \dots = \frac{\pi}{4}$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. we can think this also as a Riemann sum, 'cause it's $\displaystyle \sum_{k=1}^n\frac n{n^2+k^2}=\frac1n\sum_{k=1}^n\frac1{1+\left(\frac kn\right)^2}\to\int_0^1\frac{dx}{1+x^2}$ as $\displaystyle n\to\infty.$