So, I was reading about the integral test for convergence of a series from wikipedia, which says that the series converges if the integral of the monotonically decreasing function converges, but to my intuition


Simply because the function is monotonically decreasing and positive. In that case I might have integrals converging and the series diverging, all I would be able to say is that if the integral diverges, the series definitely diverges. I mean, it looks like a test of divergence more than a test of convergence. Am I going wrong somewhere? Any help is sincerely appreciated since I am not one with a formal mathematical background.

  • $\begingroup$ No capital letters, please. $\endgroup$ – Saad Jun 9 '18 at 9:30
  • $\begingroup$ @AlexFrancisco Ok, no problems....someone has already edited it $\endgroup$ – ubuntu_noob Jun 9 '18 at 9:32

Note that for any $N$, we have

$$\sum_{n=2}^{N} f(n) \leq \int_1^N f(x) \; dx\leq \sum_{n=1}^{N} f(n).$$

(where the thing to notice is the lower limit on the first sum.) If you're picturing the integral as an area and the right-hand sum as an area of 1-unit wide rectangles, then picture the left-hand sum as the same rectangles, but shifted to the left one unit (and throw away the first rectangle.)

The inequality shows that the sums and the integral converge or diverge together.


Note that since $f$ is decreasing $$ \int_{1}^{\infty}f(x)dx \ge \int_{1}^{\infty}f(\lceil x \rceil)dx = \sum_{n=2}^{\infty}f(n). $$


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