# Dev Blog

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# Simple Sums

$$0 < p < 1, p \in \mathbb{R}$$

\begin{align} \sum_{k=0}^{\infty} p^k = \frac{1}{1-p} \end{align}

Proof:

\begin{align} S & = \sum_{k=0}^{\infty} p^k \end{align}

\begin{align} \sum_{k=0}^{\infty} p^k & = 1 + \sum_{k=1}^{\infty} p^k \end{align}

\begin{align} p S &= \sum_{k=0}^{\infty} p^{k+1} \\ &= \sum_{k=1}^{\infty} p^k \\ \end{align}

\begin{align} S - p S &= 1 \\ S &= \frac{1}{1-p} \\ \end{align}

\begin{align} \sum_{k=0}^{s-1} p^k = \frac{1-p^s}{1-p} \\ \sum_{k=s}^{\infty} p^k = \frac{p^s}{1-p} \end{align}

Proof:

\begin{align} \sum_{k=0}^{s-1} p^k &= \sum_{k=0}^{\infty} p^k - \sum_{k=s}^{\infty} p^k \\ &= \sum_{k=0}^{\infty} p^k - p^s \sum_{k=0}^{\infty} p^k \\ &= \frac{1}{1-p} - \frac{p^s}{1-p} \\ &= \frac{1 - p^s}{1-p} \\ \end{align}

\begin{align} \sum_{k=s}^{\infty} p^k &= p^s \sum_{k=0}^{\infty} p^k \\ &= \frac{p^s}{1-p} \\ \end{align}

\begin{align} \sum_{k=0}^{\infty} k p^k = \frac{p}{(1-p)^2} \end{align}

Proof:

\begin{align} S' &= \sum_{k=0}^{\infty} k p^k \\ p S' &= \sum_{k=0}^{\infty} k p^{k+1} \\ &= \sum_{k=1}^{\infty} (k - 1) p^{k} \\ &= \sum_{k=1} k p^k - \sum_{k=1} p^k \\ &= S' - \frac{p}{1-p} \\ p S' - S' &= - \frac{p}{1-p} \\ S' (p - 1) &= - \frac{p}{1-p} \\ S' (1 - p) &= \frac{p}{1-p} \\ S' &= \frac{p}{(1-p)^2} \\ \end{align}

\begin{align} \end{align}