Question

The sum of the infinite series,
$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$is

• A. $\pi$
• B. infinity
• C. 4
• D. $\frac{{\pi }^2}{4}$

Answer 1

The correct option is B infinity

Let $S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$
we know that,
$log(1-x)=-[x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...]$
$-log(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+...$
$log\left(\frac{1}{1-x}\right)=x+\frac{x^2}{2}+\frac{x^3}{3}+...$
put $x=1$
$1+\frac{1}{2}+\frac{1}{3}+...=log\frac{1}{0}$
$=log\infty$
$1+\frac{1}{2}+\frac{1}{3}+...=\infty$