Convergence test for $\Sigma \frac{1}{n^\alpha}$
This is a convergence test in my textbook for $\Sigma \frac{1}{n^\alpha}$. What I don't understand is why is it that the sum is less than the integral for $\alpha>1$ but the sum is bigger than the integral for $\alpha\leq 1$?
Solution 1:
Because if $f$ is a continuous decreasing function from $[1,\infty)$ into $[0,\infty)$, then$$f(n+1)\leqslant\int_n^{n+1}f(x)\,\mathrm dx\leqslant f(n)$$and therefore$$\sum_{n=2}^\infty f(n)\leqslant\int_1^\infty f(x)\,\mathrm dx\leqslant\sum_{n=1}^\infty f(n).$$This is the basic ingredient in one of the proofs of the integral test.