# Show that the improper integral $\int_1^\infty f(x) \ dx$ exists iff $\sum_1^\infty a_n$ converges.

The assignment:

Let $(a_n)_{n\in\mathbb{N}}$ be a sequence of real numbers and $f: [1, \infty) \rightarrow \mathbb{R}$ be a function, defined by $f(x) = a_n$, for $x \in [n,n+1).$ Show that: $$\lim_{b\to\infty} \int_1^b f(x) \ dx$$ exists if and only if $$\sum_{n=1}^\infty a_n$$converges. Similarly show that the equivalence also holds for the absolute improper integral and that if we have convergence the equality $\int_1^\infty f(x) \ dx = \sum_1^\infty a_n$ is true.

I think the comparison tests for both series and integrals might be helpful but I don't know which inequality to use to get from the series to the integral and vice versa, since I need either two integrals or two series to use a comparison test.

Any help would be appreciated.

• – user122283
Commented Apr 30, 2014 at 1:03
• There shouldn't be a $\lim_{b\to\infty}$ in the first integral.
– user122283
Commented Apr 30, 2014 at 1:04
• @SanathDevalapurkar f is neither necessarily non-negative nor decreasing.
– Nhat
Commented Apr 30, 2014 at 1:06
• Doesn't $f$ needs to be monotone? Commented Apr 30, 2014 at 1:46

Hint: Note that $\displaystyle \int _1^\infty f=\sum \limits_{n=1}^\infty \left(\int _{n}^{n+1}f\right)=\sum \limits_{n=1}^\infty \left(a_n\right)$.
• @KitKat Simply because $$\displaystyle \int _{n}^{n+1}f(x)\mathrm dx=\int _n^{n+1}f_{\mid [n, n+1[}(x)\mathrm dx=\int _n^{n+1}a_n\mathrm dx=a_n(n+1-n).$$ And please don't take my comment the wrong way. I said it because it's obvious you can handle this, you just happened to miss it. It happens to every one. Commented Apr 30, 2014 at 11:39