# Find a formula for the general term anan (not the partial sum) of the infinite series. Assume the infinite series begins at n=1. frac{2}{1^2+1}+frac{1}{2^2+1}+frac{2}{3^2+1}+frac{1}{4^2+1}+...

Question
Series
Find a formula for the general term anan (not the partial sum) of the infinite series. Assume the infinite series begins at n=1.
$$\frac{2}{1^2+1}+\frac{1}{2^2+1}+\frac{2}{3^2+1}+\frac{1}{4^2+1}+...$$

2021-02-12
In the given infinite series, find the term of the series $$a_1, a_2 …\text{ and }a_n$$
The terms of the series is given as:
$$a_1=\frac{2}{1^2+1}$$
$$a_2=\frac{1}{2^2+1}$$
$$a_3=\frac{2}{3^2+1}$$
Here we can see that denominator of each term of the series is represented as $$(n^2+1)$$:
The numerator of the series is the repetition of 2,1,2,1…
Let the terms of the series is calculated as:
when n is odd then k=2
when n is even then k'=1
$$\begin{cases}\frac{k}{n^2+1}\quad n=\text{odd}\\\frac{k'}{n^2+1}\quad n=\text{even}\end{cases}$$
$$k=a+b(-1)^{n-1}=2\Rightarrow a+b=2$$
$$k'=a+b(-1)^{n-1}=1\Rightarrow a-b=1$$
$$a+b=2\text{ and }a-b=1\Rightarrow a=\frac32\text{ and }b=\frac12$$
so, the nummerator of the series is represented as:
$$\frac32+\frac12(-1)^{n-1}=\frac{3+(-1)^{n-1}}{2}$$
Hence, the nth term of the series can be written as:
$$a_n=\frac{(\frac{3+(-1)^{n-1}}{2})}{n^2+1}\Rightarrow a_n=(\frac{3+(-1)^{n-1}}{2(n^2+1)})$$
$$a_1=(\frac{3+(-1)^{1-1}}{2((1)^2+1)})=\frac{(3+1)}{2(1^2+1)}=\frac{2}{1^2+1}$$
$$a_2=(\frac{3+(-1)^{2-1}}{2((2)^2+1)})=\frac{(3-1)}{2(2^2+1)}=\frac{2}{2(2^2+1)}=\frac{1}{2^2+1}$$

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