calculate-lim-n-k-1-n-kn-k-2-n-2-

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 \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 \dots

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