Proof of integral test

Question

$f:[1,\infty)\rightarrow \mathbb{R}$ monotone decreasing and $f(x)\ge 0$ and $\int_1^{\infty} f(x) dx$ exists $\Rightarrow$ $\sum_{n=1}^{\infty} f(n)$ is convergent
I need to prove this statement with these marks.
I know that $\sum_{k=2}^n f(k) \le \int_1^n f(x) dx \lt \int_1^{\infty} f(x) dx \lt \infty $.How can i use this?

Answer 1

Hint: $\sum_{k=2}^{\infty}f(k)=\lim_{n}\sum_{k=2}^{n}f(k)$ so (you can obtain this by using results about sequences) $\sum_{k=2}^{\infty}f(k)\leq\int_{1}^{\infty}f(x)dx$.

(If you are using Reimann integration use the definition of $\int_{1}^{\infty}f(x)dx$ and if you are using Lebesgue integration, you would need to use the monotone convergence theorem to make use of the fact that the inequality you have guarantees $\sum_{k=2}^{\infty}f(k)\leq\int_{1}^{\infty}f(x)dx$.)

Answer 2

Hint: Consider $\int_2^n f(x)dx \leq \sum_{k=2}^{n} f(k) \leq \int_1^n f(x)dx $, apply the Comparison Test.