calculate lim_(n→+∞) Σ_(k=1) ^n ((√(kn))/((k^2 +n^2 )))


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Question Number 99237 by abdomathmax last updated on 19/Jun/20

$calculate \lim_{n \to + \infty} \sum_{k = 1}^{n} \frac{\sqrt{kn}}{(k^{2} + n^{2})}$
	Answered by Ar Brandon last updated on 19/Jun/20
	$l=lim_(n→∞) Σ_(k=1) ^n ((√(kn))/((k^2 +n^2 )))=lim_(n→∞) Σ_(k=1) ^n ((√((kn^2 )/n))/((k^2 +n^2 )))=lim_(n→∞) (n/n^2 )Σ_(n→∞) ^n ((√(k/n))/([(k^2 /n^2 )+1])) =∫_1 ^2 ((√x)/(x^2 +1))dx , t^2 =x⇒2tdt=dx ⇒l=∫_1 ^(√2) ((2t^2 )/(t^4 +1))dt=∫_1 ^(√2) (((t^2 +1)+(t^2 −1))/(t^4 +1))dt=∫_1 ^(√2) ((t^2 +1)/(t^4 +1))dt+∫_1 ^(√2) ((t^2 −1)/(t^4 +1))dt =∫_1 ^(√2) ((1+(1/t^2 ))/(t^2 +(1/t^2 )))dt+∫_1 ^(√2) ((1−(1/t^2 ))/(t^2 +(1/t^2 )))dt=∫_1 ^(√2) ((1+(1/t^2 ))/((t−(1/t))^2 +2))dt+∫_1 ^(√2) ((1−(1/t^2 ))/((t+(1/t))^2 −2))dt u=t−(1/t)⇒du=(1+(1/t^2 ))dt, v=t+(1/t)⇒dv=(1−(1/t^2 ))dt ⇒l=∫_0 ^(1/(√2)) (du/(u^2 +2))+∫_2 ^(3/(√2)) (dv/(v^2 −2))=[(1/(√2))tan^(−1) ((u/(√2)))]_0 ^(1/(√2)) −[(1/(√2))tanh^(−1) ((v/(√2)))]_2 ^(3/(√2)) ⇒l=(1/(√2))[tan^(−1) ((1/2))]−(1/(√2))[tanh^(−1) ((3/2))−tanh^(−1) ((√2))] undefined { (),() :}f(x)=tanh^(−1) x⇒x∈]−1,1[$
	$l = \lim_{n \to \infty} \underset{k = 1}{\sum^{n}} \frac{\sqrt{kn}}{(k^{2} + n^{2})} = \lim_{n \to \infty} \underset{k = 1}{\sum^{n}} \frac{\sqrt{\frac{{kn}^{2}}{n}}}{(k^{2} + n^{2})} = \lim_{n \to \infty} \frac{n}{n^{2}} \underset{n \to \infty}{\sum^{n}} \frac{\sqrt{\frac{k}{n}}}{[\frac{k^{2}}{n^{2}} + 1]}$ $= \int_{1}^{2} \frac{\sqrt{x}}{x^{2} + 1} dx, t^{2} = x \Rightarrow 2 tdt = dx$ $\Rightarrow l = \int_{1}^{\sqrt{2}} \frac{{2 t}^{2}}{t^{4} + 1} dt = \int_{1}^{\sqrt{2}} \frac{(t^{2} + 1) + (t^{2} - 1)}{t^{4} + 1} dt = \int_{1}^{\sqrt{2}} \frac{t^{2} + 1}{t^{4} + 1} dt + \int_{1}^{\sqrt{2}} \frac{t^{2} - 1}{t^{4} + 1} dt$ $= \int_{1}^{\sqrt{2}} \frac{1 + \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} dt + \int_{1}^{\sqrt{2}} \frac{1 - \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} dt = \int_{1}^{\sqrt{2}} \frac{1 + \frac{1}{t^{2}}}{{(t - \frac{1}{t})}^{2} + 2} dt + \int_{1}^{\sqrt{2}} \frac{1 - \frac{1}{t^{2}}}{{(t + \frac{1}{t})}^{2} - 2} dt$ $u = t - \frac{1}{t} \Rightarrow du = (1 + \frac{1}{t^{2}}) dt, v = t + \frac{1}{t} \Rightarrow dv = (1 - \frac{1}{t^{2}}) dt$ $\Rightarrow l = \int_{0}^{\frac{1}{\sqrt{2}}} \frac{du}{u^{2} + 2} + \int_{2}^{\frac{3}{\sqrt{2}}} \frac{dv}{v^{2} - 2} = {[\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{u}{\sqrt{2}})]}_{0}^{\frac{1}{\sqrt{2}}} - {[\frac{1}{\sqrt{2}} \tanh^{- 1} (\frac{v}{\sqrt{2}})]}_{2}^{\frac{3}{\sqrt{2}}}$ $\Rightarrow l = \frac{1}{\sqrt{2}} [\tan^{- 1} (\frac{1}{2})] - \frac{1}{\sqrt{2}} [\tanh^{- 1} (\frac{3}{2}) - \tanh^{- 1} (\sqrt{2})]$ $undefined {\begin{cases} \end{cases} f (x) = \tanh^{- 1} x \Rightarrow x \in] - 1, 1 [$
	Commented by abdomathmax last updated on 19/Jun/20

	$thanks sir .$
	Commented by mathmax by abdo last updated on 21/Jun/20

	$oo yes you are right . . .$
	Commented by mathmax by abdo last updated on 21/Jun/20

	$t + \frac{1}{t} = v \Rightarrow \int_{0}^{1} \frac{1 - \frac{1}{t^{2}}}{{(t + \frac{1}{t})}^{2} - 2} dt = - \int_{2}^{\infty} \frac{dv}{v^{2} - 2} = - \frac{1}{2 \sqrt{2}} \int_{2}^{\infty} (\frac{1}{v - \sqrt{2}} - \frac{1}{v + \sqrt{2}}) dv$ $= - \frac{1}{2 \sqrt{2}} {[\ln ∣ \frac{v - \sqrt{2}}{v + \sqrt{2}} ∣]}_{2}^{\infty} = \frac{1}{2 \sqrt{2}} \ln (\frac{2 - \sqrt{2}}{2 + \sqrt{2}}) \Rightarrow$ $I = \frac{π}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}} \ln (\frac{2 - \sqrt{2}}{2 + \sqrt{2}})$
	Commented by mathmax by abdo last updated on 20/Jun/20

	$s = \lim_{n \to + \infty} \frac{1}{n} \sum_{k = 1}^{n} \frac{\sqrt{\frac{k}{n}}}{1 + {(\frac{k}{n})}^{2}} = \int_{0}^{1} \frac{\sqrt{x}}{1 + x^{2}} dx$ $=_{\sqrt{x} = t} \int_{0}^{1} \frac{t}{1 + t^{4}} (2 t) dt = 2 \int_{0}^{1} \frac{t^{2}}{1 + t^{4}} dt but$ $\int_{0}^{1} \frac{t^{2}}{1 + t^{4}} = \int_{0}^{1} \frac{1}{\frac{1}{t^{2}} + t^{2}} dt = \frac{1}{2} \int_{0}^{1} \frac{1 + \frac{1}{t^{2}} + 1 - \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} dt$ $= \frac{1}{2} \int_{0}^{1} \frac{1 + \frac{1}{t^{2}}}{{(t - \frac{1}{t})}^{2} + 2} dt (t - \frac{1}{t} = - u) + \frac{1}{2} \int_{0}^{1} \frac{1 - \frac{1}{t^{2}}}{{(t + \frac{1}{t})}^{2} - 2} (t + \frac{1}{t} = - v)$ $= - \frac{1}{2} \int_{0}^{\infty} \frac{- du}{u^{2} + 2} - \frac{1}{2} \int_{0}^{\infty} \frac{- dv}{v^{2} - 2} = \frac{1}{2} \int_{0}^{\infty} \frac{du}{u^{2} + 2} + \frac{1}{2} \int_{0}^{\infty} \frac{dv}{v^{2} - 2} we hsve$ $\int_{0}^{\infty} \frac{du}{u^{2} + 2} =_{u = \sqrt{2} α} \int_{0}^{\infty} \frac{\sqrt{2} d α}{2 (1 + α^{2})} = \frac{1}{\sqrt{2}} \times \frac{π}{2} = \frac{π}{2 \sqrt{2}}$ $\int_{0}^{\infty} \frac{dv}{v^{2} - 2} = \frac{1}{4} \int_{0}^{\infty} (\frac{1}{v - 2} - \frac{1}{v + 2}) dv = \frac{1}{4} {[\ln ∣ \frac{v - 2}{v + 2} ∣]}_{0}^{\infty} = \frac{1}{4} \times 0 = 0 \Rightarrow$ $s = \frac{π}{\sqrt{2}}$
	Commented by Ar Brandon last updated on 21/Jun/20

	$How did you obtain your limits for v ?$ $I thought it was supposed to range from 2 to - \infty$
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