Why $\sum_{i=1}^{k} \frac{1}{p^i} = \frac{p^{-k}-1}{1-p}$?

Question

I have the statement:

$$\sum_{i=1}^{k} \frac{1}{p^i} = \frac{p^{-k}-1}{1-p}$$

without a clear step explanation why.

I've tried expanding the LHS and rewriting the RHS, but I don't see it yet.

$$\sum_{i=1}^{k} \frac{1}{p^i}= ? =\frac{p^{-k}-1}{1-p}$$

I'd appreciate an intermediary step "?" to help understand the result.

Answer 1

You simply apply the formula for the sum of a geometric series: $$x+x^2+\dots+x^k=x\bigl(1+x+\dots+x^{k-1}\bigr)=x\,\frac{x^k-1}{x-1},$$ substitute $\dfrac1p$ to $x$ in this formula and simplify.

Answer 2

This is a geometric series. Hint: write $$S=a+ar+ar^2+...+ar^N$$ and then multiply it by the common factor $r$ to obtain $$rS=ar+ar^2+...+ar^{N+1}.$$ What you get if you substract one equation from the another one?

Answer 3

Quoting from wikipedia:

$$\sum_{k=1}^{n} ar^{k-1} = a\ \frac{1-r^n}{1-r}.$$

To derive this formula, first write a general geometric series as:

$$\sum_{k=1}^{n} ar^{k-1} = ar^0+ar^1+ar^2+ar^3+\cdots+ar^{n-1}.$$

We can find a simpler formula for this sum by multiplying both sides of the above equation by $1 − r,\ $ and we'll see that:

\begin{align} (1-r) \sum_{k=1}^{n} ar^{k-1} & = (1-r)(ar^0 + ar^1+ar^2+ar^3+\cdots+ar^{n-1}) \\ & = ar^0 + ar^1+ar^2+ar^3+\cdots+ar^{n-1} - ar^1-ar^2-ar^3-\cdots-ar^{n-1} - ar^n \\ & = a - ar^n \end{align}

since all the other terms cancel. If $r\neq1,\ $ we can rearrange the above to get the convenient formula for a geometric series that computes the sum of n terms:

$$\sum_{k=1}^{n} ar^{k-1} = a\ \frac{1-r^n}{1-r}.$$