# Thread: Proving the integral test

1. ## Proving the integral test

Integral Test: Let $\sum$ $a_{k}$ from k=1 to infinity be a series with positive terms and define a function f on [1, $\infty$) so that f(k)= $a_{k}$ for each positive integer k. Suppose that f is continuous and decreasing on [1, $\infty$) and that the limit of f(x) as x approaches infinity = 0. Then the series $\sum$ $a_{k}$ from k=1 to infinity converges if and only if the improper integral $\int$ f from 1 to infinity converges. Furthermore, if S is the sum of the series and $s_{n}$ is the nth partial sum of the series, then 0 < S- $s_{n}$ < $\int$ f from n to infinity for each positive integer n.

Prove the integral test.

2. Think about breaking the area under the curve up into rectangles.

Afterall, that is what we are doing when we integrate.

The area of each rectangle can be represented by, say, $s_{n}$.

Then, $f(1)=s_{1}, \;\ f(2)=s_{2}, ...., \;\ f(n)=s_{n}$

Compare the area under the curve to that of the rectangles.

$S_{n}=\int_{1}^{n+1}f(x)dx

3. Originally Posted by friday616
Integral Test: Let $\sum$ $a_{k}$ from k=1 to infinity be a series with positive terms and define a function f on [1, $\infty$) so that f(k)= $a_{k}$ for each positive integer k. Suppose that f is continuous and decreasing on [1, $\infty$) and that the limit of f(x) as x approaches infinity = 0. Then the series $\sum$ $a_{k}$ from k=1 to infinity converges if and only if the improper integral $\int$ f from 1 to infinity converges. Furthermore, if S is the sum of the series and $s_{n}$ is the nth partial sum of the series, then 0 < S- $s_{n}$ < $\int$ f from n to infinity for each positive integer n.

Prove the integral test.
Seriously buddy, you need to post some of your own work. What are you getting stuck at? Make an observation (similar to galactus's post below but mroe along the lines of Riemann-Stieltjes integration) about how to bound the function with the integral. If you have more problems report back, but come with your current workings.

4. this can be done analitically too.

let $x\in[j,j+1]$ for $j\in\mathbb N.$

since this set is compact and $f$ was given as decreasing, then we get $f(j)\ge f(x)\ge f(j+1),$ besides $f$ is integrable and then $a_j\ge \int_j^{j+1}f\ge a_{j+1},$ now sum this for $1\le j\le n-1$ and then take limits.