Find

$\underset{n\to \mathrm{\infty}}{lim}{n}^{2}\sum _{k=1}^{n}\frac{1}{({n}^{2}+{k}^{2})(\sqrt{{n}^{2}+{k}^{2}}+n)}.$

$\underset{n\to \mathrm{\infty}}{lim}{n}^{2}\sum _{k=1}^{n}\frac{1}{({n}^{2}+{k}^{2})(\sqrt{{n}^{2}+{k}^{2}}+n)}.$

Cierra Castillo
2022-07-08
Answered

Find

$\underset{n\to \mathrm{\infty}}{lim}{n}^{2}\sum _{k=1}^{n}\frac{1}{({n}^{2}+{k}^{2})(\sqrt{{n}^{2}+{k}^{2}}+n)}.$

$\underset{n\to \mathrm{\infty}}{lim}{n}^{2}\sum _{k=1}^{n}\frac{1}{({n}^{2}+{k}^{2})(\sqrt{{n}^{2}+{k}^{2}}+n)}.$

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Jaruckigh

Answered 2022-07-09
Author has **11** answers

For the last integral consider the indefinite integral

$I=\int \frac{1}{\sqrt{{x}^{2}+1}+1}dx=\int \frac{\sqrt{{x}^{2}+1}-1}{{x}^{2}}dx$

Here, you can use the hyperbolic trig substitution $x=\mathrm{sinh}t$ for the first summand in the numerator which yields

$\int \frac{\sqrt{{x}^{2}+1}}{{x}^{2}}dx=\int {\mathrm{coth}}^{2}t\text{}dt=t-\mathrm{coth}t+C$

which after substituting back yields an antiderivative for I

$I={\mathrm{sinh}}^{-1}x-\frac{\sqrt{{x}^{2}+1}-1}{x}+C$

Note that this function is, as expected, regular at $x=0$, unlike the one in the 2nd equation above.

$I=\int \frac{1}{\sqrt{{x}^{2}+1}+1}dx=\int \frac{\sqrt{{x}^{2}+1}-1}{{x}^{2}}dx$

Here, you can use the hyperbolic trig substitution $x=\mathrm{sinh}t$ for the first summand in the numerator which yields

$\int \frac{\sqrt{{x}^{2}+1}}{{x}^{2}}dx=\int {\mathrm{coth}}^{2}t\text{}dt=t-\mathrm{coth}t+C$

which after substituting back yields an antiderivative for I

$I={\mathrm{sinh}}^{-1}x-\frac{\sqrt{{x}^{2}+1}-1}{x}+C$

Note that this function is, as expected, regular at $x=0$, unlike the one in the 2nd equation above.

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