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This is just a quick query to confirm that I have this right and that my way of constructing it is correct. We were asked to prove:

$a=\frac{-1}{log(1-\theta)}$ for $p(x)=a\frac{\theta^x}{x}$ for $x=1,2,...$

Using the axiom, $P(S) = \sum_{i=1}^{\infty}f(x)=1$

So, end up with:

$\sum_{i=1}^{\infty}a\frac{\theta^x}{x}=1$

Then, using the Maclaurin Series where $\log(1-x)=-\sum_{i=1}^{\infty}\frac{x^n}{n}$

Then putting this all together to get:

$-a\log(1-\theta)=1$

And then rearrange to get the required answer.

Glen_b
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Syzorr
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1 Answers1

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The approach looks fine, but there's one thing I'd regard as perhaps "missing" in the argument there: checking the radius of convergence of the Maclaurin series covers the required range of $\theta$ (which is unstated, but should be $0<\theta<1$).

[Generally speaking, where possibe it's best to avoid questions where there's a potential that the answer will be simply "yes", since we're not supposed to give one-sentence answers, let alone single word ones. If possible, it's also better to try to find a question or a phrasing that will help more people, even if it becomes a bit indirect for your problem.]

Glen_b
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  • Yes, fair. I'll try to make it more general in future. I struggle to find sufficient resources to feel comfortable with my answers because I shoot for 100% but my maths isn't as good as I'd like it to be. – Syzorr Sep 03 '15 at 02:14