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

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

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.

2 comments

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):