Question

Evaluate the following

$lim_{n \to \: \infty }( \frac{1}{1 + \sqrt{1} } + \frac{1}{2 + \sqrt{2n} } + \frac{1}{3 + \sqrt{3n} } + - - - + \frac{1}{n + \sqrt{ {n}^{2} } } )$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \displaystyle\lim_{n \to \infty }\rm \bigg( \frac{1}{1 + \sqrt{1} } + \frac{1}{2 + \sqrt{2n} } + \frac{1}{3 + \sqrt{3n} } + - - - + \frac{1}{n + \sqrt{ {n}^{2} } }\bigg)$

can be rewritten as

$\rm \: = \: \displaystyle\lim_{n \to \infty }\displaystyle\sum_{r=1}^{n}\rm \frac{1}{r + \sqrt{rn}}$

can be rewritten as

$\rm \: = \: \displaystyle\lim_{n \to \infty }\displaystyle\sum_{r=1}^{n}\rm \frac{1}{n\bigg(\dfrac{r}{n} + \dfrac{ \sqrt{rn} }{n} \bigg)}$

$\rm \: = \: \displaystyle\lim_{n \to \infty }\displaystyle\sum_{r=1}^{n}\rm \frac{1}{n\bigg(\dfrac{r}{n} + \sqrt{ \dfrac{r}{n} } \bigg)}$

Now, using limit as a sum of definite integrals, we have

$\rm \:\dfrac{r}{n} \: changes \: to \: x$

$\rm \:\dfrac{1}{n} \: changes \: to \: dx$

$\rm\:\displaystyle \sf \lim_{n \to \infty} \sum \: changes \: to \: \int$

$\rm \: \: lower \: limit \: a \: = \: \displaystyle \sf \lim_{n \to \infty}\frac{1}{n} = 0$

$\rm \: \: upper \: limit \: b \: = \: \displaystyle \sf \lim_{n \to \infty}\frac{n}{n} = \: 1$

So, above expression can be rewritten as

$\rm \: = \: \displaystyle\int_0^1\rm \frac{1}{x + \sqrt{x} } \: dx$

$\rm \: = \: \displaystyle\int_0^1\rm \frac{1}{ \sqrt{x} \times \sqrt{x} + \sqrt{x} } \: dx$

$\rm \: = \: \displaystyle\int_0^1\rm \frac{1}{\sqrt{x}( \sqrt{x} + 1) } \: dx$

$\rm \: = \: \displaystyle\int_0^1\rm \frac{2}{2\sqrt{x} \: ( \sqrt{x} + 1) } \: dx$

$\rm \: = \: 2\displaystyle\int_0^1\rm \frac{1}{2\sqrt{x} \: ( \sqrt{x} + 1) } \: dx$

We know,

$\boxed{ \rm{ \:\displaystyle\int\rm \frac{f'(x)}{f(x)} \: dx \: = \: log |f(x)| + c \: }}$

So, using this identity, we get

$\rm \: = \: 2 \: \bigg(log | \sqrt{x} + 1 | \bigg)_0^1$

$\rm \: = \: 2 \: \bigg(log | \sqrt{1} + 1 | - log |1| \bigg)$

$\rm \: = \: 2 \: log2$

Hence,

$\color{green}\boxed{ \rm{\rm lim_{n \to \infty }\bigg( \frac{1}{1 + \sqrt{1} } + \frac{1}{2 + \sqrt{2n} } + \frac{1}{3 + \sqrt{3n} } + - - - + \frac{1}{n + \sqrt{ {n}^{2}}}\bigg ) = 2log2}}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$