Re: divergence of improper integral implies divergence of series

SusanP wrote:
> SusanP wrote:
> > Let f: [0, oo)-> [0, oo) and [sum](a_k) be given. Assume
> > that for all sufficiently large k and all x in [k, k+1),
> > f(x)<=(a_k). Prove that divergence of the improper
> > integral (from 0 to oo) of f(x) implies divergence of
> > [sum](a_k). (Do I use the comparison test here?
>
> RG Vickson wrote:
>
> Well...what do you think? Yes, or no? Try giving an answer first, then
> maybe if you have made an error, people will be willing to help. One
> small hint: draw a picture.
>
> R,G. Vickson
>
> Ok, so I think I need to compare (a_k) with the
> integral (k to k+1) f(x)dx.
> What does the inequality f(x) <= a_k on [k,k+1) tell me
> about this integral? I think it says that:
> If f(x) <= a_k on [k,k+1), then
> integral(k to k+1)f(x)dx <= integral(k to k+1)(a_k)dx.

Right. And what is the integral of a_k from k to k+1? Hint: it's very
easy.

The integral of (a_k) from k to k+1 is the [sum](a_k)?
I really have no idea...

and how would I represent this by a picture?

>
> Then the integral(1 to oo)f(x)dx=
> [sum](1 to oo)integral (1 to k+1) f(x)dx.
>
> I got this far... but I don't see how I can prove that
> f(x) diverges and thus that this proves that [sum](a_k)
> diverges....
.

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