Question

The value of $\underset{t\to 1^{-}}{\lim }\left(1-t\right)\sum _{r=1}^{\in fty}\frac{t^r}{1+t^r}$ is

• A. $ln2$
• B. $ln3$
• C. $ln\left(\frac{2e}{1+e}\right)$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

$\underset{t\to 1^{-}}{\lim }\left(1-t\right)\sum \left\{\underset{G.P.}{\underbrace{\underbrace{t^r-t^{2r}+t^{3r}-t^{4r}+\dots upto\infty }}}\right\}$
$=\underset{t\to 1^{-}}{\lim }\left(1-t\right)\left[\underset{G.P.}{\underbrace{\sum t^r}}-\underset{G.P.}{\underbrace{\sum t^{2r}}}+\underset{G.P.}{\underbrace{\sum t^{3r}}}-\underset{G.P.}{\underbrace{\sum t^{4r}}}+\dots upto\infty \right]$
$=\underset{t\to 1^{-}}{\lim }\left(1-t\right)\left[\frac{t}{1-t}-\frac{t^2}{1-t^2}+\frac{t^3}{1-t^3}-\frac{t^4}{1-t^4}+\dots upto\infty \right]$
$=\underset{t\to 1^{-}}{\lim }\left[t-\frac{t^2}{1+t}+\frac{t^3}{1+t+t^2}-\dots upto\infty \right]$
$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots upto\infty$
$=ln2$