Upper bound for the sum $\sum_{k=1}^N \frac{1}{\varphi(k)}$

Is there an upper bound for the sum $$\sum_{k=1}^N \frac{1}{\varphi^{\alpha}(k)}$$ where $\varphi(n)$ is the Euler totient function and $\alpha\geq 1$ a real constant? In particular, I'm interested in the cases $\alpha=1$. I don't want any asymptotic estimate with error term, since in that case there is already a good reference from Sitaramachandrarao (http://projecteuclid.org/download/pdf_1/euclid.rmjm/1250127232). I'd like to have a reasonable bound like $$\sum_{k=1}^N \frac{1}{\varphi(k)} \le f(N)\;,$$ for a certain function $f$, maybe with a range of validity like $\forall N\ge n_0$, for a certain integer $n_0$.

A trivial bound can be find in this way. Using$$\frac{1}{\varphi\left(n\right)}<2e^{\gamma}\frac{\log\left(\log\left(n\right)\right)}{n},\,\forall n>2$$ where $\gamma$ is the Euler-Mascheroni constant, we have, $\forall\alpha>1$ $$\sum_{n=1}^{N}\frac{1}{\varphi\left(n\right)^{\alpha}}<2^{\alpha}e^{\gamma\alpha}\sum_{n\geq1}\left(\frac{\log\left(\log\left(n\right)\right)}{n}\right)^{\alpha}<2^{\alpha}e^{\gamma\alpha}\zeta\left(\alpha(1-\epsilon)\right)$$ where $\zeta\left(*\right)$ is the Riemann Zeta function.
Wikipedia gives $$\sum_{k=1}^n{1\over\phi(k)}={315\zeta(3)\over2\pi^4}\left(\log n+\gamma-\sum_p{\log p\over p^2-p+1}\right)+O\left({(\log n)^{2/3}\over n}\right)$$ with a link to R. Sitaramachandrarao. On an error term of Landau II, Rocky Mountain J. Math. 15 (1985), 579-588.
• Thanks Gerry, the case $\alpha=1$ is ok, I'll check the reference to see if the case $\alpha=2$ can be derived from it. – PITTALUGA Jun 20 '14 at 9:41
• I suspect that for $a=2$ the sum converges as $n\to\infty$. – Gerry Myerson Jun 20 '14 at 9:44
• Yeah I looked at them, they deal with errors, I hoped a reasonable upper bound for the finite sum of $1/\varphi(k)$ existed but I wasn't able to find anything...thanks a lot anyway Gerry! – PITTALUGA Jun 26 '14 at 7:40
$\forall n\in \mathbb{N} |n\ge 1$ $\sqrt{\frac{n}{2}}\le\phi(n)\le n$ (which can be proved elementarily) , then , by taking the inverses, you can see by the confront principle that if $0< a\le 1$ then the series $\sum_{n\ge 1}\frac{1}{\phi(n)^a}$ diverges, so your succession is unbounded, while if a>2 , since the series $\sum_{n\ge 1}\frac{1}{\sqrt{n}^a}$ converges (a>2 => a/2>1), then the series $\sum_{n\ge 1}\frac{1}{\phi(n)^a}$ converges, and is $\le$ $2^\frac{a}{2} \sum_{n\ge 1}\frac{1}{n^\frac{a}{2}}$ (and thus so is your succession). This leads uncovered the case $1<a\le 2$.