Is my divergence test correct?

Question

This idea came to me while looking at the following graph of $f=\frac{1}{x}$:

Now, the definite integral of $f$ from $1$ to $n$ is smaller than $f(1)+f(2)...f(n)$, from the graph above. But since the definite integral from $1$ to $n$ is $\log(n)$, which as we take the limit of $n$ to $\infty$, $\log(n)$ goes to $\infty$ as well. Hence the infinte series $f(1)+f(2)+f(3)...$ goes to $\infty$, that is diverges.

Thinking about this gave me an idea, to extend this to any function $h$, not discontinuous on an infinte number of points, satisfying the following conditions:

• $h(n)\geq h(x) \text{ whenever } n>x>n+1 $
• $h(x) > 0$
• $\lim_{x\to\infty}\int^x_a h(x)\, dx = \infty $

Then, $\sum_{i=a}^{\infty}h(i)$ diverges, where is $a$ and $n$ are natural numbers and $n\geq a$. The problem is I cannot prove it, I have not studied analysis or series. I don't know how to prove it without analysis, using proven theorems.

So, I ask here that is my test correct? I hope so, but there are so weird continuous functions that it doesn't give me any confidence. If you have a not-so-tough proof, it would be very appreciated. If it is so, how can I generalize this or relax the conditions?

Answer 1

Yes. This is called the integral test, and proofs of it are in almost any calculus book, but are little more than what you've mentioned.

Answer 2

You have a stronger theorem, saying that if $f$ is positive and decreasing, then the series $$ \sum \left\{f(n) - \int_n^{n+1}f (t)dt\right\} $$converges.

Then there is convergence of the original series iff the primitives of $f$ have a finite limit as $x\to\infty$.

Let me write the (simple) proof here: $$ f(n+1) \le\int_n^{n+1}f(t)dt \le f(n) $$ $$ f(n) - f(n+1)\ge f(n) - \int_n^{n+1}f(t)dt \ge 0 $$ so, as $f$ has a limit when $t\to\infty$, $\sum \{f(n) - f(n+1)\}$ and by comparison principle: $$ \sum \left\{f(n) - \int_n^{n+1}f (t)dt\right\}\le \infty $$