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  4. The value of undersett arrow 1- l i m (1 - t) displaystyle ∑ r = 1∈ fty(tr/1 + tr) is

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Q. The value of $\underset{t \rightarrow 1^{-} \, }{l i m} \left(1 - t\right)\displaystyle \sum _{r = 1}^{\in fty}\frac{t^{r}}{1 + t^{r}}$ is

NTA AbhyasNTA Abhyas 2022

Solution:

$\underset{t \rightarrow 1^{-}}{l i m}\left(1 - t\right)\displaystyle \sum \left\{\underset{G . P .}{\underbrace{\underset{︸}{t^{r} - t^{2 r} + t^{3 r} - t^{4 r} + \ldots \, u p t o \infty }}}\right\}$
$=\underset{t \rightarrow 1^{-}}{l i m}\left(1 - t\right)\left[\underset{G . P .}{\underbrace{\displaystyle \sum t^{r}}} - \underset{G . P .}{\underbrace{\displaystyle \sum t^{2 r}}} + \underset{G . P .}{\underbrace{\displaystyle \sum t^{3 r}}} - \underset{G . P .}{\underbrace{\displaystyle \sum t^{4 r}}} + \ldots u p t o \infty \right]$
$=\underset{t \rightarrow 1^{-}}{l i m}\left(1 - t\right)\left[\frac{t}{1 - t} - \frac{t^{2}}{1 - t^{2}} + \frac{t^{3}}{1 - t^{3}} - \frac{t^{4}}{1 - t^{4}} + \ldots \, u p t o \infty \right]$
$=\underset{t \rightarrow 1^{-}}{l i m}\left[t - \frac{t^{2}}{1 + t} + \frac{t^{3}}{1 + t + t^{2}} - \ldots \, u p t o \infty \right]$
$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots upto\infty $
$=ln2$