Number theory Multiplicative Functions A function $ s_1 (n) = \sum_d_n \frac 1d   $ or the sum of the reciprocal of the positive divisors of n is defined. Show that if $p$ is prime then
$$ s_1 (p^k) = \frac {p^{k+1} - 1}{p^k(p-1)} $$
I am proving by induction and since this function is multiplicative,
$$s_1(p^{k+1}) = s_1(p^k) s_1(p) = \frac {p^{k+1} - 1}{p^k(p-1)} * \frac {p^2 - 1}{p(p-1)}$$
but trying to reduce that to 
$$ s_1 (p^{k+1}) = \frac {p^{k+2} - 1}{p^{k+1}(p-1)} $$
proves unfruitful. Where am I going wrong?
 A: Hint: The reciprocals of the divisors of $p^k$ are $1,\frac{1}{p},\dots,\frac{1}{p^k}$.
The sum is a (finite) geometric series. There is a standard formula for the sum.
Another way: Well, not all that different! Let $S$ be our sum. Then $p^kS=1+p+p^2+\cdots+p^k$. Another finite geometric series. 
Remark: Please note that the function $f$ is multiplicative if $f(ab)=f(a)f(b)$ whenever $a$ and $b$ are relatively prime. In our case, it is not true, for example, that $f(p^2)=f(p)f(p)$.
A: Not widely known, Ramanujan considered these. This one is called $\sigma_{-1}(n).$ If you multiply it by $n$ itself, you are summing $n/d.$ That is, the result is simply $\sigma_1(n)$ added in reverse order. So, your
$$ \sigma_{-1}(n) = \frac{\sigma_1(n)}{n}.    $$ Most often, $\sigma_1(n)$ is written as just $\sigma(n).$
See  https://mathoverflow.net/questions/136940/sum-of-powers-of-divisors-function  and  https://mathoverflow.net/questions/137865/estimate-term-in-ramanujan-lost-notebook-classic-analytic-number-theory 
