- #1
sparkster
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If f is a continuous decreasing non-negative function on [1, oo) and
[itex]a_n = \sum_{k=1}^nf(k) - \int_1^nf(t)dt[/itex].
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 [itex]a_n - a_{n+1}[/itex] I get
[itex]\int_1^{n+1}f(t)dt - \int_1^nf(t)dt - f(n+1)[/itex]
but I don't know how to show this is bigger than zero. I did an example with [itex]f(x)=\frac{1}{x^2}[/itex] 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.
[itex]a_n = \sum_{k=1}^nf(k) - \int_1^nf(t)dt[/itex].
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 [itex]a_n - a_{n+1}[/itex] I get
[itex]\int_1^{n+1}f(t)dt - \int_1^nf(t)dt - f(n+1)[/itex]
but I don't know how to show this is bigger than zero. I did an example with [itex]f(x)=\frac{1}{x^2}[/itex] 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.
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