Question

The sum of the series $\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}....$ upto $\infty$ is equal to

• A. ${\log }_{e}2-1$
• B. ${\log }_{e}2$
• C. ${\log }_e\left(\frac{4}{e}\right)$
• D. $2{\log }_{e}2$

Answer 1

The correct option is C ${\log }_e\left(\frac{4}{e}\right)$

Let us write the series compactly as;

$S=\sum _1^{\infty }\frac{{(-1)}^{n+1}}{\left(n\right)(n+1)}$

Now let there be a similar series but instead of $-1$ we introduce a variable $x$, the reason behind this is to introduce calculus and make the sum much easier, at the end, we shall re-substitute $x=-1$ to get the desired result.

So let there be a function $f\left(x\right)$ such that;

$f\left(x\right)=\sum _1^{\infty }\frac{{\left(x\right)}^{n+1}}{\left(n\right)(n+1)}$

As we can see $f(-1)=S$ which is what we need to find.

Now we shall differentiate twice.

$f\left(x\right)=\sum _1^{\infty }\frac{{\left(x\right)}^{n+1}}{\left(n\right)(n+1)}{f}^{\prime }\left(x\right)=\sum _1^{\infty }\frac{(n+1){\left(x\right)}^n}{\left(n\right)(n+1)}=\sum \frac{x^n}{n}{f}^{\prime \prime }\left(x\right)=\sum _1^{\infty }\frac{n{x}^{n-1}}{n}=\sum _1^{\infty }{x}^{n-1}=1+x+x^2+x^3+...\infty$

Applying infinte GP formula;

$f^{\prime \prime }\left(x\right)=\frac{1}{1-x}$

Now let us integrate twice; also imp. to note $f\left(0\right)=f^{\prime }\left(0\right)=0$;

$f^{\prime }\left(x\right)=\log (1-x)+C$

To find $C$ put $x=0$,

$f^{\prime }\left(0\right)=\log \left(1\right)+C$

$\therefore C=0$

Now for the second integration;

$f\left(x\right)=\int \log (1-x)dx=-x-(1-x)\log (1-x)$

Put $x=0$ and you'll get $C=0$,

Now put $x=-1$ to get $S$;

$S=-(-1)-(1-(-1\left)\right)\log (1-(-1\left)\right)S=1-2{\log }_{e}2=-{\log }_e\left(\frac{4}{e}\right)$