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Conclude if the given series converges or diverges and justify your conclusion

by RoRi

Test the following series for convergence or divergence. Justify the decision.

    \[ \sum_{n=1}^{\infty} \int_0^{\frac{1}{n}} \frac{\sqrt{x}}{1+x^2} \, dx. \]

First, for n \geq 2, we have

    \[ \int_0^{\frac{1}{n}} \frac{\sqrt{x}}{1+x^2} \, dx < \frac{1}{n} \cdot \frac{\sqrt{\frac{1}{n}}}{1+\left( \frac{1}{n} \right)^2}}. \]

(We know this since \frac{\sqrt{x}}{1+x^2} is decreasing on the interval \left[ 0, \frac{1}{3} \right], and so the area of the rectangle of width \frac{1}{n} and height f \left( \frac{1}{n} \right) is greater than the area under the curve f(x).) Then, this implies

    \[ \int_0^{\frac{1}{n}} \frac{\sqrt{x}}{1+x^2} \, dx \leq \frac{1}{n} \cdot \frac{\sqrt{\frac{1}{n}}}{1+ \left( \frac{1}{n} \right)^2} = \frac{1}{n^{\frac{3}{2}} + \frac{1}{\sqrt{n}}} < \frac{1}{n^{\frac{3}{2}}}. \]

But, we know \sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}} converges; hence,

    \[ \sum_{n=1}^{\infty} \int_0^{\frac{1}{n}} \frac{\sqrt{x}}{1+x^2} \, dx \]

also converges.

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