Suppose $f(x)$ monotonous decreases on $[0,+\infty) $ Suppose $f(x)$ monotonous decreases on $[0,+\infty) $ and $$\lim_{x\to+\infty}f(x) \text dx=0 $$
Then, proof that  $\sum_{n=1}^{\infty} f(n) $ converges if and only if $\int_{0}^{\infty}f(x)\text dx $converges.  
Actually I listed an inequality and almost thought it has been solved without the condition $$\lim_{x\to+\infty}f(x) \text dx=0$$
that is 
$$f(k)=\int_{k-1}^{k}f(k)\text dx\le\int_{k-1}^{k}f(x)\text dx\le\int_{k-1}^{k}f(k-1)\text dx=f(k-1) $$
then we have
$$\sum_{n=2}^{\infty}f(k) \le \sum_{n=2}^{\infty} \int_{k-1}^{k}f(x)\text dx \le \sum_{n=2}^{\infty} f(k-1)=\sum_{n=1}^{\infty} f(k) $$
So if  $\sum_{n=1}^{\infty} f(n) $ converges, then $\sum_{n=2}^{\infty} \int_{k-1}^{k}f(x)\text dx=\int_{1}^{\infty }f(x)\text dx$ converges. So is $\int_{0}^{\infty }f(x)\text dx$.
Also, if $\int_{0}^{\infty }f(x)\text dx$ converges, then $\int_{1}^{\infty }f(x)\text dx$ converges. Thus $\sum_{n=2}^{\infty}f(k)$ converges.  
With the process, I think I have solved this without the condition $\lim_{x\to+\infty}f(x) \text dx=0 $. But since this a question from a serious exam I am not so sure about my proof. So anyone could help confirm or show some more ways to proof this?

What you did is right. The fact that $\lim_{x\to\infty}f(x)=0$ is irrelevant for the proof that $\displaystyle\sum_{n=1}^\infty f(k)$ converges if and only if $\displaystyle\int_1^\infty f(x)\,\mathrm dx$ converges. Of course, when they converge, then $\lim_{x\to\infty}f(x)=0$.
