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

Question

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

Answer 1

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 $(4)$.