# Need help with convergence of this series

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• Nov 11th 2008, 10:29 AM
Hweengee
Need help with convergence of this series
Hi guys I've been doing some problems on series and I ran into this one,

$\displaystyle \sum _{n=1}^{\infty } \left( \int _{0}^{\frac{1}{n}}\!{\frac { \sqrt {x}}{1+{x}^{2}}}{dx} \right)$

which I have no idea how to go about doing.

Any help would be appreciated.

Thanks.
• Nov 11th 2008, 11:31 AM
Laurent
Quote:

Originally Posted by Hweengee
Hi guys I've been doing some problems on series and I ran into this one,

$\displaystyle \sum _{n=1}^{\infty } \left( \int _{0}^{\frac{1}{n}}\!{\frac { \sqrt {x}}{1+{x}^{2}}}{dx} \right)$

which I have no idea how to go about doing.

Any help would be appreciated.

Thanks.

Think about what the integral looks like when $\displaystyle n$ is large: then you integrate on a very small interval $\displaystyle [0,\frac{1}{n}]$ the function $\displaystyle \frac{\sqrt{x}}{1+x^2}$, which is almost equal to $\displaystyle \sqrt{x}$ since $\displaystyle x$ is small... and you can integrate $\displaystyle \int_0^{1/n}\sqrt{x}\,dx$ explicitly. It gives $\displaystyle \frac{2/3}{n^{3/2}}$, which is the general term of a convergent series.

This justifies why the majoration $\displaystyle \frac{\sqrt{x}}{1+x^2}\leq \sqrt{x}$ will allow you to conclude.
• Nov 11th 2008, 11:47 AM
Hweengee
So when n is large the interval [0,$\displaystyle \frac{1}{n}]$ becomes small. And this implies that the area bounded by the curve becomes smaller and smaller, and so the series is likely to converge? Hence the comparison with a convergent series?
• Nov 11th 2008, 11:51 AM
Mathstud28
Quote:

Originally Posted by Hweengee
Hi guys I've been doing some problems on series and I ran into this one,

$\displaystyle \sum _{n=1}^{\infty } \left( \int _{0}^{\frac{1}{n}}\!{\frac { \sqrt {x}}{1+{x}^{2}}}{dx} \right)$

which I have no idea how to go about doing.

Any help would be appreciated.

Thanks.

You could also do it this way

$\displaystyle f(n)=\int_0^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^2}dx$

It should be clear that since the interval of integration is positive and so is the integrand that $\displaystyle f(n)>0~\forall{n}\in\mathbb{N}$

We can also see that it is continuous (Ill try to prove this if you need be). And finally $\displaystyle f(n)\in\downarrow$. So we may apply the integral test

Therefore the series shares the convergence/divergence of

$\displaystyle \int_1^{\infty}\int_0^{\frac{1}{n}}\frac{\sqrt{x}} {1+x^2}dxdn$

Now switching the order of integration gives

$\displaystyle \int_0^1\int_1^{\frac{1}{x}}\frac{\sqrt{x}}{1+x^2} dndx$

$\displaystyle =\int_0^1\bigg[\frac{1}{\sqrt{x}(x^2+1)}+\frac{1}{\sqrt{x}}+\frac {1}{2}\bigg]dx$

All of which converge (for the justification of the first just consider that $\displaystyle \frac{1}{\sqrt{x}(x^2+1)}\sim\frac{1}{\sqrt{x}}$. Therefore the series converges.
• Nov 11th 2008, 11:51 AM
Laurent
Quote:

Originally Posted by Hweengee
So when n is large the interval [0,$\displaystyle \frac{1}{n}]$ becomes small. And this implies that the area bounded by the curve becomes smaller and smaller, and so the series is likely to converge? Hence the comparison with a convergent series?

Not only is the integral smaller and smaller, but it is like $\displaystyle \frac{2/3}{n^{3/2}}$. The idea is to neglect the denominator since it becomes close to 1, so that we obtain a function that we know how to integrate. Actually I almost wrote the full proof. Just use the majoration I gave and integrate it.
• Nov 11th 2008, 11:52 AM
Hweengee
Ok thanks alot guys.