# Integral / Sequence question

1. Dec 29, 2005

### sparkster

If f is a continuous decreasing non-negative function on [1, oo) and
$a_n = \sum_{k=1}^nf(k) - \int_1^nf(t)dt$.

I need to show that a_n is a decreasing sequence of non-negative real numbers.

I'm working on the decreasing part. If I subtract $a_n - a_{n+1}$ I get

$\int_1^{n+1}f(t)dt - \int_1^nf(t)dt - f(n+1)$

but I don't know how to show this is bigger than zero. I did an example with $f(x)=\frac{1}{x^2}$ and of course it worked, but it didn't help me with the proof.

Any suggestions would be appreciated.

ETA: Let me clarify. Looking at this graphically, I realize that if I do the subtractions, I'm going to get a positive area. But I don't know how to formalize this.

Last edited: Dec 29, 2005
2. Dec 29, 2005

### Muzza

First, the expression for a_n - a_(n + 1) can be simplified so that it only involves a single integral. Then you can show the desired inequality by "wisely" choosing a step function for the integral.

3. Dec 29, 2005

### eok20

$\int_1^{n+1}f(t)dt - \int_1^nf(t)dt = \int_n^{n+1}f(t)dt$
and since f is strictly decreasing you know that this integral is going to be greater than any right end point approximation. With 1 rectangle, such an approximation is f(n+1).

4. Dec 29, 2005

### sparkster

Damn...I'm embarrassed that I even asked.