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:

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

\(a_1=(\frac{3+(-1)^{1-1}}{2((1)^2+1)})=\frac{(3+1)}{2(1^2+1)}=\frac{2}{1^2+1}\)

\(a_2=(\frac{3+(-1)^{2-1}}{2((2)^2+1)})=\frac{(3-1)}{2(2^2+1)}=\frac{2}{2(2^2+1)}=\frac{1}{2^2+1}\)

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:

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

\(a_1=(\frac{3+(-1)^{1-1}}{2((1)^2+1)})=\frac{(3+1)}{2(1^2+1)}=\frac{2}{1^2+1}\)

\(a_2=(\frac{3+(-1)^{2-1}}{2((2)^2+1)})=\frac{(3-1)}{2(2^2+1)}=\frac{2}{2(2^2+1)}=\frac{1}{2^2+1}\)