Let $m=\frac{1}{2}$ and $n=\frac{1}{2}$ be the probabilities of success and failure respectively and let p(k) be a probability mass function such that $p(k)=m\sum _{i=1}^{k}{n}^{i}+n\sum _{i=1}^{k}{m}^{i}.$

Is this a valid probability mass function?

Say we carry out this test infinitely many times, does $p(k)=1$?

We recognise that the probability mass function is a sum of two slightly shifted geometric series. Let's consider one of the two (the other will follow the exact same argument).

Say we are summing from 0 to $\mathrm{\infty}$, that is let

${S}_{0}=\sum _{i=0}^{\mathrm{\infty}}{n}^{i}.$

Then this is simply evaluated as ${S}_{0}=1+n+{n}^{2}+{n}^{3}+\cdots =\frac{1}{1-n}$

which is perfectly valid as $n=\frac{1}{2}$. If, then, I start this sequence at $i=1$, is it valid to say that

${S}_{1}=\sum _{i=1}^{\mathrm{\infty}}{n}^{i}=\frac{1}{1-n}-1=\frac{1-(1-n)}{1-n}=\frac{n}{1-n}$

and as such that, as $k\to \mathrm{\infty}$, we have something like

$\frac{mn}{1-n}+\frac{mn}{1-m}=1.$

Have I made any ridiculous logical jumps or is this valid?

Is this a valid probability mass function?

Say we carry out this test infinitely many times, does $p(k)=1$?

We recognise that the probability mass function is a sum of two slightly shifted geometric series. Let's consider one of the two (the other will follow the exact same argument).

Say we are summing from 0 to $\mathrm{\infty}$, that is let

${S}_{0}=\sum _{i=0}^{\mathrm{\infty}}{n}^{i}.$

Then this is simply evaluated as ${S}_{0}=1+n+{n}^{2}+{n}^{3}+\cdots =\frac{1}{1-n}$

which is perfectly valid as $n=\frac{1}{2}$. If, then, I start this sequence at $i=1$, is it valid to say that

${S}_{1}=\sum _{i=1}^{\mathrm{\infty}}{n}^{i}=\frac{1}{1-n}-1=\frac{1-(1-n)}{1-n}=\frac{n}{1-n}$

and as such that, as $k\to \mathrm{\infty}$, we have something like

$\frac{mn}{1-n}+\frac{mn}{1-m}=1.$

Have I made any ridiculous logical jumps or is this valid?