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I need to show that

$1+ \frac{1}{p} + \frac{1}{p^2} + ... + \frac{1}{p^k} = \frac{p^{k+1}-1}{p^k(p-1)}$ , where $p$ is a prime number.

I have started by doing the following:

$$1 + p^{-1} + p^{-2} + ... p^{-k} = 1 + (p + p^2 + ... + p^k)^{-1} = 1 + [p(1+p+p^2+...+p^{k-1})] ^{-1}$$

= $$1 + [p(\frac{p^k-1}{p-1})]^{-1} = 1 + \frac {p-1}{p^{k+1}+p} =\frac {p^{k+1} - 1} {p^{k+1} - p} $$

I have $p$ instead of $p^k$ in the last denominator.

UPD.

Mr.SrJenea
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    Note that this is a geometric series (finite) not only when $p$ is prime, but whenever $p$ is nonzero. In your penultimate line you've made a mistake in taking a reciprocal "outside the parentheses". – hardmath Feb 13 '18 at 01:54
  • Your first guess is wrong. For instance, $1 + 3^{-1} = \frac43$ is not the same as $(1 + 3)^{-1} = \frac14$ – NovaDenizen Feb 13 '18 at 02:14

2 Answers2

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This sounds like a job for induction!

Base case: $k=0$ $$\frac{p^{k+1}-1}{p^k(p-1)}= \frac{p-1}{p-1}=1 $$ It works so we assume that the identity holds for k $$1 + \frac{1}{p}+... + \frac{1}{p^k} = \frac{p^{k+1}-1}{p^k(p-1)}$$ And try to prove the identity true for k+1 $$1 + \frac{1}{p}+... + \frac{1}{p^k} + \frac{1}{p^{k+1}} = \frac{p^{k+1}-1}{p^k(p-1)} + \frac{1}{p^{k+1}} = \frac{p(p^{k+1}-1)+ (p-1)}{p^{k+1}(p-1)}= $$ $$= \frac{p^{(k+1) +1}-1}{p^{k+1}(p-1)} $$ QED Notice that nowhere in the proof did I use the fact that p is prime. As long as p is not zero or one (denominator is not zero) the theorem holds.
Daniel Gendin
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You might be knowing the sum of GP starting with any number say $a$ with common ratio $r$ and number of terms $n$ is $$\frac {a(r^n-1)}{r-1}$$

In your case $a=1$ , $r=\frac {1}{p}$ , $n=k+1$

Just substitute the values to get the answer.

Rohan Shinde
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