Question

$\displaystyle \rm \red{ \lim_{n \to \infty } \left( \frac{1}{ \sqrt{n} \sqrt{n + 1} } +\frac{1}{ \sqrt{n} \sqrt{n + 2} } + \dots +\frac{1}{ \sqrt{n} \sqrt{n + n} } \right)}$​

Answer 1

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

Given expression is

$\rm \: \displaystyle \rm { \lim_{n \to \infty } \left( \frac{1}{ \sqrt{n} \sqrt{n + 1} } +\frac{1}{ \sqrt{n} \sqrt{n + 2} } + \dots +\frac{1}{ \sqrt{n} \sqrt{n + n} } \right)}$

can be rewritten as

$\rm \: =  \: \displaystyle \rm { \lim_{n \to \infty }\displaystyle\sum_{k=1}^n\rm \left( \frac{1}{ \sqrt{n} \sqrt{n + k} } \right)}$

can be further rewritten as

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

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

Now, using Limit as sum of definite integrals

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

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

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

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

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

So, given expression can be rewritten as

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

$\rm \: =  \: \displaystyle\int_0^1\rm {\bigg(x + 1 \bigg) }^{\dfrac{-1}{2} } \: dx \:$

We know,

$\boxed{\sf{  \:\displaystyle\int\rm {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: }}$

So, using this result, we get

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

$\rm \: =  \: 2( \sqrt{1 + 1} - \sqrt{1 + 0})$

$\rm \: =  \: 2( \sqrt{2} - 1)$

Hence,

$\\ \rm \: \displaystyle \rm { \lim_{n \to \infty }\left( \frac{1}{ \sqrt{n} \sqrt{n + 1} } +\frac{1}{ \sqrt{n} \sqrt{n + 2} } + \dots +\frac{1}{ \sqrt{n} \sqrt{n + n} } \right)} \\ \\ \rm \: =  \: \: 2\bigg( \sqrt{2} - 1\bigg)$

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Additional Information :-

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