At the first look, I thought that this can be easly done using "Cauchy-Maclaurin Integral Test". But after recalling the statement of the theorem, I found myself wrong. Cauchy Maclaurin Test asserts that
Let $f$ be a positive, decreasing function on $[1,+\infty]$. Then the series $\displaystyle \sum_{k=1}^{+\infty} f(k)$ converges if and only if the improper integral $\displaystyle \int_1^{+\infty} f(t) dt$ converges.
But in the given problem one property of the function $f$ is removed, which is $f$ is positive. And a new condition has been added which is $\displaystyle \lim_{x \to +\infty}f(x)=0$. And note that we are just asked to prove the "if" part only.
Even it intuitively looks like $f$ is monotone decreasing and $f(x)\to 0$ as $x\to+\infty$ impliy $f(x)\ge 0 \forall x \in \Bbb{R}$. But can I prove it rigorously?
Can anybody give me a proof of the statement given in my title part?
Thanks for your assistance in advance.
If $f(x) \to 0$ as $x \to \infty$, then, for any $c > 0$ there is an $x(c)$ such that $|f(x)| < c$ for $x > x(c)$.
If $f$ is monotone decreasing, then $x_2 > x_1$ implies that $f(x_2) \le f(x_1)$.
Suppose there is an $x^-$ such that $f(x^-) < 0$. Then $f(x) \le f(x^-)$ for all $x > x^-$. Therefore $|f(x)| \ge |f(x^-)|$ for all $x \ge x^-$ which contradicts $f(x) \to 0$.
Therefore there is no $x$ such that $f(x) < 0$.