# What is the closed form of :$\sum_{n=2}^{\infty }\frac{\tau(2^n-1)}{\phi(2^n-1)}$?

I would like to know if this series :$$\sum_{n=2}^{\infty }\frac{\tau(2^n-1)}{\phi(2^n-1)}$$ w'd be convergent or no which it's terms related to the euler totiont function $\phi$ and $\tau(n)$ the number of divisors of $n$ .

As show here in wolfram alpha for it's partial sum equal's $2$.

My question here is: what is the closed form of :$$\sum_{n=2}^{\infty }\frac{\tau(2^n-1)}{\phi(2^n-1)}$$ if it is a convergent series

Thank you for any help

• Appears to converge to approx. 2.1857485878042160, which Newcastle's inverse symbolic calculator does not associate with anything. There isn't always any nice closed form. – Matthew Conroy Nov 7 '16 at 21:18
• A version has also been posted to MO, mathoverflow.net/questions/254174/… – Gerry Myerson Nov 7 '16 at 22:15

You can show that $$\phi(2^n-1) \ge C_1 \frac{ 2^n}{\log n}$$ for some constant $C_1$, $n$ sufficiently large, and $$\log \tau(2^n-1) \le C_2 \frac{n}{\log n}$$ for some $C_2$, $n$ sufficiently large. Hence, the terms of your series are less than $$C_3 \frac{(\log n) e^{C_2 \frac{n}{\log n}}}{2^n} = C_3 (\log n) e^{n \left(\frac{C_2}{\log n}-\log 2 \right)} < C_4 (0.6^n) \log n, \text{ say,}$$ for some $C_3$, $C_4$, and $n$ sufficiently large, and so the series converges.

Calculating the first $100$ terms, I find the sum to be approximately $$2.18574858780421606023125753...$$ which is not found to match anything by the inverse symbolic calculator at Newcastle.

(A good reference for these kind of bounds is Gerald Tenenbaum's Introduction to Analytic and Probabilistic Number Theory, Chapter 1.5.)

• Thanks for that , is it fatser in the side of convergece than 1/n² ? – zeraoulia rafik Nov 7 '16 at 21:39
• You're welcome! Yes, it converges like $n/2^n$. – Matthew Conroy Nov 7 '16 at 21:44
• Correction: I misread Tenenbaum: the bound I thought was on tau is actually on log tau, so the argument needed a little adjustment. – Matthew Conroy Nov 7 '16 at 23:38
• The convergence is (still) faster than $1/n^2$. – Matthew Conroy Nov 7 '16 at 23:38

It is convergent because $\tau(N)\leq2^{\log_3N}$ for $N$ odd and $3\phi(N)\geq N^{\log_54}$ for $N$ odd, so that

$$\frac{\tau(2^n-1)}{\phi(2^n-1)}\leq3\cdot(2^{\log_32-\log_54})^n.$$ (Note that $\log_32-\log_54<0$.)

I personally don't hope for a closed form.

• Let $f(N) = \frac{\tau(N)}{\phi(N)}$. It is multiplicative, and $f(p^k) = \frac{k+1}{p^k-p^{k-1}}$. $f(2^k) \le \frac{4}{2^{k/2}}, f(3^k) \le \frac{3}{3^{k/2}}, f(p^k) \le \frac{1}{p^{k/2}} \implies f(N) \le \frac{12}{N^{1/2}}$ – reuns Nov 7 '16 at 21:27