The proof about the integral test:
Suppose $f (x) $ is nonnegative monotone decreasing over $[1,\infty)$, then the positive series $\sum_{n=1}^{\infty}f(n)$ is convergent if and only if $\lim_{A\rightarrow +\infty}f(x)dx$ exists.
Proof:
Since $f(x)$ is monotone decreasing, we can get $f(n+1)<\int_{n}^{n+1}f(x)\Bbb{dx}< f(n)$, sum them up and get $\sum_{k=1}^{n+1}f(x)-f(1)<\int_{1}^{n+1}f(x)\Bbb{dx}<\sum_{k=1}^{n}f(k)$, when the series is convergent, the integral is bounded, since $f(x)$ is nonnegative, the integral is monotone increasing, the $\lim_{A\rightarrow +\infty}f(x)dx$ exists. When $\lim_{A\rightarrow +\infty}f(x)dx$ exists, the positive series is bounded, so it is convergent.
When reading the integral test for positive series, I confused about the condition
- "Monotone decreasing": this condition is used to get $f(n+1)<\int_{n}^{n+1}f(x)\Bbb{dx}< f(n)$ and following that $\sum_{k=1}^{n+1}f(x)-f(1)<\int_{1}^{n+1}f(x)\Bbb{dx}<\sum_{k=1}^{n}f(k)$, but I am confused because when the condition is monotone increasing we can just change the direction of the sign of the inequality like $f(n+1)>\int_{n}^{n+1}f(x)\Bbb{dx}> f(n)$, so why the condition "monotone decreasing" is necessary?
- "Continuous": the condition says that $f (x) $ is continuous. But when proves it seems that the condition that be used is the integrable. Is the condition continuous not necessary?
I think there is something wrong with my understanding. Could you find and point it? Thank you!
Best Answer
With $S_n = \sum_{k=1}^n f(k)$ we get the inequalities
$$\tag{*}S_n - f(1) < \int_1^n f(x) \, dx < S_n - f(n)$$
The desired conclusion for the integral test is that the ser1es and improper integral converge or diverge together.
If $S_n$ converges, then it is necessary that $f(n)= S_n - S_{n-1} \to 0$ as $n \to \infty$. The hypothesis that $f$ is nonincreasing is required to proceed. The right-hand inequality of (*) shows that $\int_1^n f(x) \, dx$ is bounded and, as it is monotone increasing, it must converge.
The reverse implication assumes that the improper integral converges. Again, this cannot be the case if $f$ is nondecreasing. Do you see why? The left-hand inequality of (*) shows that $S_n$ must be bounded and, as it is monotone increasing, it must converge.