I have the following function (pdf of geometric distribution) $f(k)=(1-p)^{k-1}p$ for $k=1,2,3...$ and $p\in[0,1]$. I'm looking for a $p\in[0.25,1]$ which maximizes $\sum_3 f(k)$. Ignore the $3$ in the sum (...it's related to calculating a p-value). All I'd like to know if, by inspection, one can see that $p=0.25$ maximizes the sum of $(1-p)^{k-1}p$ for $k$ going possibly to infinity?

$\sum\limits_{k=1}^\infty p(1-p)^{k-1} = p\sum\limits_{k=0}^\infty (1-p)^k = p\dfrac1{1-p}$. Now use calculus.

Doesn't look right. This function is increasing on $[0,1)$.

Oh, I didn't see the first part.

You're saying the $3$ has nothing to do with starting $k$ with $3$. So, doesn't look right.

@TedShifrin Sorry for being unclear, but it kind of clarified it.

Does the $3$ mean I should start the summation at $k=3$?

Oh, duh. I have a typo. Hold on.

What I typed is incorrect.

::smack::

$\sum\limits_{k=1}^\infty p(1-p)^k = p\sum_{k=0}^\infty (1-p)^k = p\cdot\dfrac1{1-(1-p)} = 1$, independent of the value of $p$.

(otherwise it wouldn't be a pdf)

So I don't understand the question.

I'll clarify it.

Sometimes I wish I could remove my own garbage.

@user85795 Duly smacked.

:-)

@Astyx: Indeed. I wasn't thinking of context. In fact, I certainly taught this stuff in my probability course.

I'm calculating (the p-value) $P(X\geq 3| H_0)$, where $H_0$ is a null hypothesis stating $p\geq0.25$. My observation is $X=3$. When calculating p-values, you'd like to reject it when it's small, so I'd like to determine for which $p\in H_0$, $P(X\geq 3| H_0)$ is biggest, so that all the other $p$'s in $H_0$ will be smaller than that and I can know if I can reject it.

$P(X\geq 3| H_0)$ is the sum we've been talking about.

OK, so you do need to start the sum at $k=3$. So then you have $(1-p)^2$ for the answer, and this function is obviously decreasing on $[.25,1)$.

if the sum starts from $1$, I don't see that $p=0.25$ maximizes it

It's constant, of course, if the sum starts from $1$.

@TedShifrin So the sum starting at $k=3$ is $(1-p)^2$?

@schn mentioned "going possibly to infinity", so I was considering finite sums

Yeah, for finite sums we need to do a derivative computation, I think.

Yes, @schn; you need to know how to do this.

@TedShifrin By your trick above with $k=1$?

Yes, re-indexing makes it easier. But you should know how to do $\sum\limits_{k=n}^\infty a^k$ for $|a|<1$. Yes.

Yeah, that's a common sum.

@TedShifrin Thanks!

Sure :)