Find the sum of the $\sum_{m=k}^{+\infty}\binom{m}{k}(1-p)^k\cdot p^{m-k}$

Question

Let $0<p<1$,Find the sum $$\sum_{m=k}^{+\infty}\binom{m}{k}(1-p)^k\cdot p^{m-k}$$

Answer 1

Use the fact that for $|x|<1, \:k\geqslant0$ $$ \frac1{(1-x)^{k+1}}=\sum_{m=k}^{+\infty}\binom{m}{k}x^{m-k} $$ which can be proved by differentiating $k$ time on both sides of following $$ \frac1{1-x}=\sum_{m=0}^{+\infty}x^m $$ Thus \begin{align} \sum_{m=k}^{+\infty}\binom{m}{k}(1-p)^k\cdot p^{m-k}&=(1-p)^k\sum_{m=k}^{+\infty}\binom{m}{k}p^{m-k} \\ &=(1-p)^k\frac1{(1-p)^{k+1}} \\ &=\frac1{1-p} \end{align}

Answer 2

In this ansswer, it is shown that $$ \begin{align} \binom{m}{k} &=\binom{m}{m-k}\\ &=(-1)^{m-k}\binom{-k-1}{m-k} \end{align} $$ Plug this into $$ \begin{align} \sum_{m=k}^\infty\binom{m}{k}(1-p)^kp^{m-k} &=\sum_{m=k}^\infty(-1)^{m-k}\binom{-k-1}{m-k}(1-p)^kp^{m-k}\\ &=\sum_{m=0}^\infty(-1)^m\binom{-k-1}{m}(1-p)^kp^m\\ &=(1-p)^k(1-p)^{-k-1}\\[4pt] &=\frac1{1-p} \end{align} $$ which converges for $\left|p\right|\lt1$.

Answer 3

Say you have a coin which lands heads up with probability $p$, and you keep flipping it until it lands tails-up $k+1$ times, recording the number of flips this takes as $X$. Then:

$$\mathbb{P}(X=m+1)=\mathbb{P}(k \text{ tails in the first } m \text{ flips, then another tails})=\binom{m}{k}(1-p)^{k+1}p^{m-k}$$

Given that $k+1\le X < \infty$ almost surely, we see that

$$1=\sum_{m\ge k}\mathbb{P}(X=m+1)=\sum_{m=k}^{\infty}\binom{m}{k}(1-p)^{k+1}\cdot p^{m-k}$$

which gives the desired sum as $\frac{1}{1-p}$

Appendix: $X < \infty \text{ a.s.}$

Proof: $\mathbb{P}(X\ge a)=\mathbb{P}(\le k \text{ tails in first }a \text{ flips})=\sum_{0\le j\le k} \binom{a}{j}p^{a-j}(1-p)^j=O(a^kp^a)\to 0 \text{ as } a \to \infty$