# If $f:\Bbb{R}\to\Bbb{R}$ be a monotone decreasing function, $f(x)\to 0$ as $x\to+\infty$. Prove,$\sum f(n)$ converges if $\int_1^\infty f$converges

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?

• $f$ decreasing and $f(x) \to 0$ implies that $f(x) \ge 0$. May 10, 2018 at 17:23
• @MartinR, intuitively it looks like $f(x) \ge 0$ but can you prove it rigorously? Even I am also trying to prove your comment. May 10, 2018 at 17:28
• If $f(x_0) < 0$ for some $x_0$, can the limit still be zero? May 10, 2018 at 17:31
• @MartinR, ok, if $f$ gives a negetive value at any point, then the limit which is actually infimum of $f$ on $\Bbb{R}$ will be negetive. Am I right? May 10, 2018 at 17:34

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 $f(x) < 0$.

• I can't understand how the line "$|f(x)|\ge|f(x^-)| \forall x\ge x^-$" comes? May 10, 2018 at 17:54
• Because $f(x^-) < 0$. Then $f(x) \le f(x^-) < 0$ for a \ge x^-$. May 10, 2018 at 19:10 If$f $is decreasing at$[1,+\infty) \$ then

$$\lim_{x\to +\infty}f (x)=\inf_{[1,+\infty)}f=0$$

thus $$\forall x\ge 1 \;\; f (x)\ge 0$$