Editing Divisor function (section)

==Series relations==
Two [[Dirichlet series]] involving the divisor function are: {{sfnp|Hardy|Wright|2008|pp=326-328|loc=§17.5}}

:<math>\sum_{n=1}^\infty \frac{\sigma_{a}(n)}{n^s} = \zeta(s) \zeta(s-a)\quad\text{for}\quad s>1,s>a+1,</math>

where <math>\zeta</math> is the [[Riemann zeta function]]. The series for ''d''(''n'')&nbsp;=&nbsp;''&sigma;''<sub>0</sub>(''n'') gives: {{sfnp|Hardy|Wright|2008|pp=326-328|loc=§17.5}}

: <math>\sum_{n=1}^\infty \frac{d(n)}{n^s} = \zeta^2(s)\quad\text{for}\quad s>1,</math>

and a [[Ramanujan]] identity{{sfnp|Hardy|Wright|2008|pp=334-337|loc=§17.8}}

:<math>\sum_{n=1}^\infty \frac{\sigma_a(n)\sigma_b(n)}{n^s} = \frac{\zeta(s) \zeta(s-a) \zeta(s-b) \zeta(s-a-b)}{\zeta(2s-a-b)},</math>

which is a special case of the [[Rankin–Selberg method|Rankin–Selberg convolution]].

A [[Lambert series]] involving the divisor function is: {{sfnp|Hardy|Wright|2008|pp=338-341|loc=§17.10}}

:<math>\sum_{n=1}^\infty q^n \sigma_a(n) = \sum_{n=1}^\infty \sum_{j=1}^\infty n^a q^{j\,n} = \sum_{n=1}^\infty \frac{n^a q^n}{1-q^n} = \sum_{n=1}^\infty \operatorname{Li}_{-a}(q^n)</math>

for arbitrary [[complex number|complex]] |''q''|&nbsp;≤&nbsp;1 and&nbsp;''a'' (<math>\operatorname{Li}</math> is the [[polylogarithm]]).  This summation also appears as the [[Eisenstein series#Fourier series|Fourier series of the Eisenstein series]] and the [[Weierstrass elliptic functions#Invariants|invariants of the Weierstrass elliptic functions]].

For <math>k>0</math>, there is an explicit series representation with [[Ramanujan sum]]s <math> c_m(n) </math> as :<ref>{{cite book |author=E. Krätzel |title=Zahlentheorie |publisher=VEB Deutscher Verlag der Wissenschaften |place =Berlin |year=1981 |pages=130}} (German)</ref>
:<math>\sigma_k(n) = \zeta(k+1)n^k\sum_{m=1}^\infty \frac {c_m(n)}{m^{k+1}}.</math>
The computation of the first terms of <math>c_m(n)</math> shows its oscillations around the "average value" <math>\zeta(k+1)n^k</math>:
:<math>\sigma_k(n) = \zeta(k+1)n^k \left[ 1 + \frac{(-1)^n}{2^{k+1}} + \frac{2\cos\frac {2\pi n}{3}}{3^{k+1}} + \frac{2\cos\frac {\pi n}{2}}{4^{k+1}} + \cdots\right]</math>