Question

${\sum }_{n=1}^{\infty }\frac{1}{2n-1}.x^{2n}$ is equal to

• A. $\frac{x}{2}\log {\left(\frac{1+x}{1-x}\right)}^x$
• B. $\frac{1}{2}\log \left(\frac{1+x^2}{1-x^2}\right)$
• C. $x\log \frac{1+x}{1-x}$
• D. none of these

Answer 1

The correct option is D none of these

$S\left(x\right)=x^2+\frac{x^4}{3}+\frac{x^6}{5}...\infty$

Now

Consider

$\frac{1}{1-x^2}=1+x^2+x^4+x^6...\infty$

$\int \frac{1}{1-x^2}.dx=x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+...\infty$

$\frac{1}{2}ln\left(\frac{1+x}{1-x}\right)+C=x+\frac{x^3}{3}+\frac{x^5}{5}+..\infty$

At $x=0$, we get both $LHS=RHS=0$.

Hence $C=0$.

Therefore

$\frac{1}{2}ln\left(\frac{1+x}{1-x}\right)=x+\frac{x^3}{3}+\frac{x^5}{5}+...\infty$

Multiplying by $x$ on both sides, we get

$\frac{x}{2}ln\left(\frac{1+x}{1-x}\right)=x^2+\frac{x^4}{4}+\frac{x^6}{6}+..\infty$

Thus

$\frac{x}{2}ln\left(\frac{1+x}{1-x}\right)=S\left(x\right)$