Question

If $f\left(x\right)=\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}+\frac{x}{(1+2x)(1+3x)}+....\infty$, then

• A. $f\left(x\right)$ is continuous for all $x$
• B. $f\left(x\right)$ is discontinuous for finite number of points
• C. $f\left(x\right)$ is discontinuous for positive values of x
• D. None of the above

Answer 1

The correct option is B $f\left(x\right)$ is discontinuous for finite number of points

For $x\ne 0$

$f\left(x\right)=(1-\frac{1}{1+x})+(\frac{1}{1+x}-\frac{1}{1+2x})+(\frac{1}{1+2x}-\frac{1}{1+3x})+....+(\frac{1}{1+(n-1)x}-\frac{1}{1+nx})$

$=1-\frac{1}{1+nx}$
$\Rightarrow f\left(x\right)=\underset{n\to \infty }{lim}(1-\frac{1}{1+nx})=1-0=1$
and for $x=0,f\left(0\right)=0$

$\Rightarrow f\left(x\right)=\{\begin{array}{cc}1,& x\ne 0\\ 0,& x=0\end{array}$

Clearly, $f\left(x\right)$ is discontinuous at $x=0$.