# Can someone help me to evaluate or at least to approximate the following sum.

Can someone help me to evaluate the following sum. It appeared while calculating the number of elements of intersections of some sets: $$\sum_{d|n} \frac{d}{\phi(d)^2}.$$

-
It looks to be a multiplicative function, and the result for prime powers is somewhat tractable. Have you already looked at it to that extent? – hardmath Jul 7 '12 at 23:28
To extend on hardmath's finding I got indeed $f(p)=\frac{p^2-p+1}{(p-1)^2}$ (the 'distribution' obtained is very regular with a mean value of $\approx 3.39063$) – Raymond Manzoni Jul 7 '12 at 23:34

Since $\dfrac{n}{\phi(n)^2}$ is a multiplicative function, we know that $$\varphi(n)=\sum_{d|n}\frac{d}{\phi(d)^2}\tag{1}$$ is also multiplicative. On the power of a prime, $p^j$ and $j>0$, $\phi(p^j)=p^j\frac{p-1}{p}$, so \begin{align} \varphi(p^k) &=1+\sum_{j=1}^k\frac{p^2}{p^j(p-1)^2}\\ &=1+\frac{p}{(p-1)^2}\sum_{j=0}^{k-1}\frac{1}{p^j}\\ &=1+\frac{p^k-1}{p^{k-2}(p-1)^3}\tag{2} \end{align} Thus, for $n$ whose prime factorization is $$n=\prod_jp_j^{k_j}\tag{3}$$ we have $$\varphi(n)=\prod_j\left(1+\frac{p_j^{k_j}-1}{p_j^{k_j-2}(p_j-1)^3}\right)\tag{4}$$ Estimation:
Note that for $k>0$, $$1+\frac1p\left(\frac{p}{p-1}\right)^2\le1+\frac{p^k-1}{p^{k-2}(p-1)^3}\lt1+\frac1p\left(\frac{p}{p-1}\right)^3\tag{5}$$ For larger primes, $(5)$ might help to approximate the term in $(4)$.
An equivalent for $f(p^k)$ is $1 + \frac{p^2}{(p-1)^3} (1 - p^{-k})$, which makes it clearer what the limit is as $k$ tends to $\infty$. – hardmath Jul 8 '12 at 0:18