User:Virginia-American/Sandbox/Ramanujan sum

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[edit] Sums

Let

T_q(n) = c_q(1) + c_q(2)+\dots+c_q(n)\mbox{ and }
U_q(n) = T_q + \tfrac12\phi(q).


Then


\sigma_{-s}(1)+
\sigma_{-s}(2)+
\dots+
\sigma_{-s}(n)
=
\zeta(s+1)
\left(
n+
\frac{T_2(n)}{2^{s+1}}+
\frac{T_3(n)}{3^{s+1}}+
\frac{T_4(n)}{4^{s+1}}
+\dots
\right)
=
\zeta(s+1)
\left(
n+\tfrac12+
\frac{U_2(n)}{2^{s+1}}+
\frac{U_3(n)}{3^{s+1}}+
\frac{U_4(n)}{4^{s+1}}
+\dots
\right)-
\tfrac12\zeta(s)
,



d(1)+
d(2)+
\dots+
d(n)
=
-\frac{T_2(n)\log2}{2}
-\frac{T_3(n)\log3}{3}
-\frac{T_4(n)\log4}{4}
-\dots
,

d(1)\log1+
d(2)\log2+
\dots+
d(n)\log n
=
-\frac{T_2(n)(2\gamma\log2-\log^22)}{2}
-\frac{T_3(n)(2\gamma\log3-\log^23)}{3}
-\frac{T_4(n)(2\gamma\log4-\log^24)}{4}
-\dots
,

r_2(1)+
r_2(2)+
\dots+
r_2(n)
=
\pi
\left(
n
-\frac{T_3(n)}{3}
+\frac{T_5(n)}{5}
-\frac{T_7(n)}{7}
+\dots
\right)
.