Re: A problem about convergence of series.

On Wed, 14 Nov 2007 07:20:12 -0600, David C. Ullrich
<ullrich@xxxxxxxxxxxxxxxx> wrote:

On 13 Nov 2007 21:39:52 -0500, hrubin@xxxxxxxxxxxxxxxxxxxx (Herman
Rubin) wrote:

In article <1194986239.224691.238050@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<131208@xxxxxxxxx> wrote:
f is twice differentiable on [0,1] (that is, f'' exists).

f'' is continuous on [0,1].

Then prove that

sum_{n = from 1 to infinite} f(1/n) converges iff f(0)=f'(0)=0.

All one needs is that f and f' exist on [0,1] (even a little
less can do) and f''(0) exists.

So for example if f(x) = 1 for all x then the sum of f(1/n)
converges?

I read Herman Rubin's reply as claiming this:

Let f : [0,1] -> R. If f is differentiable on [0,1], and f''(0)
exists, then

sum_{n = from 1 to infinity f(1/n) converges iff f(0) = f'(0) = 0.

quasi
.

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