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$\lim_{n \to \infty} \left [ \frac{1}{(1+n)} + \frac{1}{(2+n)} + - - - - - + \frac{1}{(n+n)} \right ]$

a) $log 2$                    b) $2$

c) $\frac{1}{2}$                          d) $\Pi /4$

3 Answers

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$\lim_{n \to \infty} \left [ \frac{1}{(1+n)} + \frac{1}{(2+n)} + - - - - - + \frac{1}{(n+n)} \right ]$

$\lim_{n \to \infty}\sum_{r=1}^n \frac{1}{n+r}$  = $\lim_{n \to \infty}\sum_{r=1}^n \frac{\frac{1}{n}}{1+\frac{r}{n}}$

$\lim_{n \to \infty}\frac{1}{n}\sum_{r=1}^n \frac{1}{1+\frac{r}{n}}$

Which is a Riemann sum for the integral

=$\int\limits_{0}^{1}\frac{1}{1+x}\, dx$

=$[log|x+1| ]_0^1$

=$log2 - log1$

=$log2$

Alternate Method:

We may use Botez-Catalan identity and immediately get that 

$\lim_{n \to \infty} \left [ \frac{1}{(1+n)} + \frac{1}{(2+n)} + - - - - - + \frac{1}{(n+n)} \right ]$=$\lim_{n \to \infty} \left [ 1- \frac{1}{2} +\frac{1}{3}-\frac{1}{4}+ ......... +(-1)^{2n+1} \frac{1}{2n} \right ]$

=$\lim_{n \to \infty} \left [ 1- \frac{1}{2} +\frac{1}{3}-\frac{1}{4}+ ......... +(-1)^{n+1} \frac{1}{n} \right ]$

=Log2

Hence,Option(A)$log2$ is the correct choice.

edited by
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a) log 2 ?

 

because , if we do , 1/n common , then the series becomes

( 1 + 1/2 + 1/3 + .... ) which is i guess log 2
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Summation of this series(1+1/2+1/3+.....+1/K) =logK

similary for given series it will be log(n+n)=log(2n)

which can be further splitted as log2 and log n now apply the limit and we have the ans as log2

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