If lim n -1-1-n-1-2-2 n-1-3-3 n-1-2 n-log e k- then k -

T_r= 1r+√(rn), r=1,2, 3,... = 1n ( nr+√(rn)) = 1n ( 1rn+√( rn)) ∴lim_n →∞( 11+√(n)+ 12+√(2n)+ 13+√(3n)+ ⋯ +12n) nbsp; =∫_0^1dxx+√(x) =∫_0^1dx√(x)(√(x)+1) = ...

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Byju's Answer

Standard XII

Mathematics

Integration by Parts

If lim n →∞1/...

Question

If $lim n \to \infty (\frac{1}{1 + \sqrt{n}} + \frac{1}{2 + \sqrt{2 n}} + \frac{1}{3 + \sqrt{3 n}} + \dots + \frac{1}{2 n}) = {log}_{e} k$ , then $k =$

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Solution

$T_{r} = \frac{1}{r + \sqrt{r n}}, r = 1, 2, 3, . . .$
$= \frac{1}{n} (\frac{n}{r + \sqrt{r n}})$
$= \frac{1}{n} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{1}{\frac{r}{n} + \sqrt{\frac{r}{n}}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠$

$∴ lim n \to \infty (\frac{1}{1 + \sqrt{n}} + \frac{1}{2 + \sqrt{2 n}} + \frac{1}{3 + \sqrt{3 n}} + \dots + \frac{1}{2 n})$
$= 1 \int 0 \frac{d x}{x + \sqrt{x}}$
$= 1 \int 0 \frac{d x}{\sqrt{x} (\sqrt{x} + 1)}$
$= 2 [{log}_{e} (1 + \sqrt{x})]_{0}^{1}$
$= 2 {log}_{e} 2$
$= {log}_{e} 4$
Hence, $k = 4$

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If limn→∞(11+√n+12+√2n+13+√3n+⋯+12n)=logek, then k=

Tr=1r+√rn, r=1,2,3,... =1n(nr+√rn) =1n⎛⎜ ⎜ ⎜ ⎜⎝1rn+√rn⎞⎟ ⎟ ⎟ ⎟⎠ ∴limn→∞(11+√n+12+√2n+13+√3n+⋯+12n) =1∫0dxx+√x =1∫0dx√x(√x+1) =2[loge(1+√x)]10 =2loge2 =loge4 Hence, k=4

If $lim n \to \infty (\frac{1}{1 + \sqrt{n}} + \frac{1}{2 + \sqrt{2 n}} + \frac{1}{3 + \sqrt{3 n}} + \dots + \frac{1}{2 n}) = {log}_{e} k$ , then $k =$