The Integral Test

Recall that every series is a set of points on a continuous curve. So if the improper integral of that curve converges then we can us it to bound the infinite series.

Each rectangle has width one and so the sum of the areas of the rectangles equals the sum of the series. From the picture we see the upper bound when the improper intergal converges.

If the improper integral diverges we can use it to form a lower bound for the infinite series which guarantees that the series diverges as well. The same argument holds from above, but the area of the rectangles is larger than the area under the curve. So if the improper integral diverges the series also diverges by direct comparison.

So it follows that the infinite $p$-series

$$\displaystyle \sum^{\infty}_{n=1}\displaystyle \frac{1}{n^p}$$

converges if and only if $p>1$. So the series

$$\displaystyle \sum^{\infty}_{n=1}\displaystyle \frac{1}{n^{3/2}}$$

converges, and the series

$$\displaystyle \sum^{\infty}_{n=1}\displaystyle \frac{1}{n^{1/2}}$$

diverges.