Upperbound of a sum [closed]

For natural numbers $n$ how can I bound the sum $\sum_{n\leq X}\frac{n}{\phi(n)}$, where $X$ is sufficiently large and $\phi(n)$ is the Euler's totient function.

Sums of this nature are common in number theory.

Solution 1:

Hint: Use $$\underset{n\leq x}{\sum}\frac{1}{\varphi\left(n\right)}=C\left(\log\left(x\right)+O\left(1\right)\right)$$ where $$C=\frac{315\zeta\left(3\right)}{2\pi^{4}},$$ and apply Abel summation. This yields $$ \sum_{n\le X} \frac{n}{\phi(n)}=\frac{315\zeta\left(3\right)}{2\pi^{4}}X-\frac{1}{2}\log(X)+O(\log(X)^{2/3}) $$

Reference: R. Sitaramachandrarao. On an error term of Landau II, Rocky Mountain J. Math. 15 (1985), 579-588.

Solution 2:

For an answer not appealing to known results for $\sum _n1/\phi (n)$ and just going for an upper bound, note that \[ \frac {n}{\phi (n)}=\sum _{d|n}g(d)\] where $g$ is the multiplicative function defined by \[ g(p)=\frac {1}{p-1}\] and $0$ on prime powers. Essentially $g(n)\ll 1/n$ and you'll be done, but let's be precise: For squarefree $n$ we have (if it's ok to use $\omega (n)\ll \log n/\log \log n$) \[ g(n)=\prod _{p|n}\frac {1}{p-1}\leq \prod _{p|n}\frac {2}{p}=2^{\omega (n)}\prod _{p|n}\frac {1}{p}\leq \frac {e^{C\log n/\log \log n}}{n}=n^{C/\log \log n-1}\ll n^{-1/2}\hspace {5mm}\text {(say)}\] and this holds trivially if $n$ isn't squarefree. So \[ \sum _{n\leq x}\frac {n}{\phi (n)}=\sum _{d\leq x}g(d)\sum _{n\leq x\atop {d|n}}1\leq x\sum _{d\leq x}\frac {g(d)}{d}\ll x\sum _{d=1}^\infty \frac {1}{d^{3/2}}\ll x.\]

(If you ask "where should the first equality come from?" the answer is "from multiplicativity". Any multiplicative $f$ has a $g$ so that $f=g\star 1$ and therefore $g=f\star \mu $.)