The sum of series `x/(1-x^2)+(x^2)/(1-x^4)+(x^4)/(1-x^8)+` to infinite terms, if `|x|<1,` is `x/(1-x)` b. `1/(1-x)` c. `(1+x)/(1-x)` d. `1`

Options

A. `x/(1-x)`
B. `(1)/(1-x)`
C. `(1+x)/(1-x)`
D. 1

Answer

A

Explanation

The general term of the given series is
`t_(n)=(x^(2n-1))/(1-x^(2n))=(1+x^(2n-1)-1)/((1+x^(2n-1))(1-x^(2n-1)))`
`=1/(1-x^(2n-1))-1/(1-x^(2n))`
Now, `S_(n)=sum_(n=1)^(n)t_(n)`
`=[{:({1/(1-x)-1/(1-x^(2))}+{1/(1-x^(2))-1/(1-x^(4))}),(" "+...+{1/(1-x^(2n-1))-1/(1-x^(2n))}):}]`
`=1/(1-x)-1/(1-x^(2n))`
Therefore, the sum to infinite terms is
`lim_(ntooo)S_(n)=1/(1-x)-1`
`=x/(1-x) [because lim_(ntooo)x^(2n)=0,` as `absxlt1]`