P(total) = P(H) + P(TH) + P(TTH) + P(TTTH) + ...
= p + (1-p)kp + ((1-p)^2)(k^2)p + ((1-p)^3)(k^3)p + ((1-p)^4))(k^4)p + ...
= p(1/(1-k(1-p)) (as k and (1-p) are less than one, the ratio is less than one and hence this is true)
= p/(1-k(1-p))
You can check if k=1 then the chance is 1 which makes sense, as eventually she will toss a heads no matter the probability if it is unchanging, and if k=0 then the probability is p, as she can only win if she tosses a head initially, as if she doesn't she can never win.
 of showing heads, and changes its heads probability according to the rule: if the result of the $n$-th toss ($n=1,2,3,\ldots$) was tails, the heads probability on toss $n+1$ becomes $k$ times that of in toss $n$ (where $k$ is some fixed number strictly between $0$ and $1$). Find the probability Alice wins.$)
\geq 0$ and $f(0)=0$, then $f(x)\geq 0$ for $x> 0.\\\\$ Show that $\sin x-x+\frac{x^3}{6}\geq 0$ for $x> 0 )
