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Question Number 43546 by maxmathsup by imad last updated on 11/Sep/18

$$let{S}_n=\sum _{k=1}^n\frac{1}{\sqrt{k^2+n^2}}calculate{lim}_{n\to +\mathrm{\infty }}{S}_n$$

Answered by behi83417@gmail.com last updated on 12/Sep/18

$$S_n=\sum _1^n\frac{1}{n}.\frac{1}{\sqrt{1+{\left(\frac{k}{n}\right)}^2}}={\int }_0^1\frac{dx}{\sqrt{1+x^2}}=$$$$=ln{(x+\sqrt{1+x^2})}_0^1=ln(\sqrt{2}+1).$$

Commented by maxmathsup by imad last updated on 12/Sep/18

$$cha7gementx=tan\theta give{\int }_0^1\frac{dx}{\sqrt{1+x^2}}={\int }_0^{\frac{\pi }{4}}\frac{1+{tan}^2\theta }{\sqrt{1+{tan}^2\theta }}d\theta$$$$={\int }_0^{\frac{\pi }{4}}\sqrt{1+{tan}^2\theta }={\int }_0^{\frac{\pi }{4}}\frac{d\theta }{cos\theta }{=}_{tan\left(\frac{\theta }{2}\right)=u}{\int }_0^{\sqrt{2}-1}\frac{1}{\frac{1-t^2}{1+t^2}}\frac{2du}{1+u^2}$$$$={\int }_0^{\sqrt{2}-1}\{\frac{1}{1-u}+\frac{1}{1+u}\}du={[ln\mid \frac{1+u}{1-u}\mid ]}_0^{\sqrt{2}-1}=ln\mid \frac{\sqrt{2}}{2-\sqrt{2}}\mid$$$$=ln\left(\frac{1}{\sqrt{2}-1}\right)=ln(1+\sqrt{2})\Rightarrow {lim}_{n\to +\mathrm{\infty }}{S}_n=ln(1+\sqrt{2})$$$$anotherway{\int }_0^1\frac{dx}{\sqrt{1+x^2}}={\left[argsh\left(x\right)\right]}_0^1={\left[ln(x+\sqrt{1+x^2})\right]}_0^1=ln(1+\sqrt{2}).$$

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