Journal of Number Theory 94, 299--319 (2002). MR1916275 (2003c:11023)
Let g(x,n), with x
+,
be a step function for each n. Assuming certain technical hypotheses, we
give a constant
and function f such that ∑
n=1 :g(x,n)
can be written in the form
+∑0<r<x f(r),
where the summation is extended over all points in (0,x) at which some
g(,n)
is not continuous. A typical example is ∑
n=1z
n/x
=(−1)∑ :zq/(1−zq),
with the summation extending over all pairs p,q of positive
integers satisfying 0<p/q<x and gcd(p,q)=1.
We then apply such representations to prove identities such as
(z)=∑n=1(
(z)−(z,1+)),
the Lambert series for Euler's totient function, and ∑
n=0(−1)n=
,
where
(z)
and
(z,a)
are the Riemann and Hurwitz zeta functions and
z(n)=∑d
n dzd.
We also give a generalization of the Rayleigh–Beatty theorem and a new result of
a similar nature for the sequences (2n−n)n=1.
Just in case the above isn't readable on your browser, here it is again:
Let $g(x,n)$, with $x\in\R^+$, be a step function for each $n$. Assuming certain technical hypotheses, we give a constant $\alpha$ and function $f$ such that $\sum_{n=1}^\infty g(x,n)$ can be written in the form $\alpha + \sum_{0<r<x} f(r)$, where the summation is extended over all points in $(0,x)$ at which some $g(\,\cdot\,,n)$ is not continuous. A typical example is $\sum_{n=1}^\infty z^{\floor{n/x}}=\left(\tfrac1z-1\right)\sum \tfrac{z^q}{1-z^q}$, with the summation extending over all pairs $p,q$ of positive integers satisfying $0<p/q<x$ and $\gcd(p,q)=1$.
We then apply such representations to prove identities such as
$\zeta(z)=\sum_{n=1}^\infty \frac{\phi(n)}{n^z}
\left(\zeta(z)-\zeta(z,1+\tfrac1n)\right)$,
the Lambert Series for Euler's Totient function, and
$\sum_{n=0}^\infty (-1)^n \dfrac{\sigma_z(2n+1)}{2n+1} =
\dfrac{\pi}{4}\dfrac{z}{1+z^2}$,
where $\zeta(z)$ and $\zeta(z,a)$ are the Riemann and Hurwitz zeta functions and $\sigma_z(n)=\sum_{d|n} dz^d$. We also give a generalization of the
Rayleigh-Beatty Theorem, and a new result of a similar nature for the sequences
$(\floor{2n\alpha}-\floor{n\alpha})_{n=1}^\infty$.
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