I am trying to figure out the following problem: Show $\sum_{n=2}^{N} \frac{1}{n^a}≤ \int_1^{N}\frac{1}{x^a} dx$, and use this to prove the convergence of the series for $a>1$.
My work: I have $\int_1^{N}\frac{1}{x^a} dx = [\frac{-1}{(1-a)x^{a-1}}]_1^N=\frac{-1}{(1-a)N^{a-1}}+\frac{1}{(1-a)}$. I do not know how to show the inequality from here. Does one express $\sum_{n=2}^{N} \frac{1}{n^a}$ since it is geometric and show the integral expression is larger? But I do not see how that works. I presume that once we have that inequality we can claim convergence of the series for large N.
Thanks in advance!
The inequality is only true for $a \geq 0$ so I will focus on that case. Then $\frac{1}{x^a}$ is non-increasing on $x>0$. Therefore, $$\int_{n-1}^{n} \frac{1}{x^a} dx \geq \frac{1}{n^a}$$ For any $n > 1$ due to the geometry (the area of the integral entirely covers the single height rectangle). Since this applies to each term in the sum, the overall integral must be greater that or equal to the overall sum.