Question

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

Answer 1

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:
\(a_n=\frac{(\frac{3+(-1)^{n-1}}{2})}{n^2+1}\Rightarrow a_n=(\frac{3+(-1)^{n-1}}{2(n^2+1)})\)
\(a_1=(\frac{3+(-1)^{1-1}}{2((1)^2+1)})=\frac{(3+1)}{2(1^2+1)}=\frac{2}{1^2+1}\)
\(a_2=(\frac{3+(-1)^{2-1}}{2((2)^2+1)})=\frac{(3-1)}{2(2^2+1)}=\frac{2}{2(2^2+1)}=\frac{1}{2^2+1}\)