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## New answers tagged dirichlet-series

0

With the definition of the Dirichlet's $\eta$ as in your post $\sum_{k=1}^\infty \eta(k+1)$ the whole expression is divergent because the $\eta(k+1)$ converge to $1$. But if you rewrite your formula $$\sum_{k=1}^\infty (\eta(k+1)-1)$$ this shall converge because $\eta(1+k)-1$ converges quickly to zero when $k$ increases. Finally we get for this ...

0

there is the "approximate functional equation" https://en.wikipedia.org/wiki/Riemann%E2%80%93Siegel_formula which is described in the Tichmarsh and on the french wikipedia of zeta-Riemann function but not on the english one. someone should add a link to that Riemann Siegel formula to the wiki.

2

You haven't exploited the main feature of the multiplicativity of $\text{rad}(n)$: the Euler's product. $$\sum_{n\geq 1}\frac{\text{rad}(n)}{n^s}=\prod_{p}\left(1+p\cdot p^{-s}+p\cdot p^{-2s}+\ldots\right)=\prod_p\left(1+\frac{p}{p^s-1}\right).$$ By approximating $\frac{p}{p^s-1}$ with $\frac{1}{p^{s-1}}$, we have: $$\sum_{n\geq ... 0 Instead of converting a sequence A=(a_n)_{n=0}^{\infty} into some kind of function, I would just specify what transformation you want to do on the series itself. For example, a translation by m could be written T_m(A) =(0,0,...(m\ 0's), a_0, a_1, ...)  or, via the indices, T_m(A)_i =0 \ if\ i < m; \ otherwise\ A_{m-i} . For the "dilation", ... 1 There are two key steps: bounding the partial sums of the Dirichlet character, and determining the region of analyticity of its L-function. First, consider the sum$$S(a)=\sum_{n=1}^a \chi(n).$$Since \chi is not the trivial or principal character, there exists an integer b between 1 and a such that b is coprime to a and \chi(b)\neq 1. ... 1 Let L=\lim a_n. If the sequence a_n has finitely many rational numbers,$$\lim [a_n+D(a_n)]=L+1$$If it has finitely many irrational numbers,$$\lim [a_n+D(a_n)]=L$$Otherwise, a_n+D(a_n) does not converge, because it has a subsequence that converges to L and another subsequence that converges to L+1. 2 I assume that L_{p,1}(s) is the Dirichlet L-function associated with the principal character \pmod{p}. About the first identity:$$\begin{eqnarray*}\left(1-\frac{1}{2^{s-1}}\right)\zeta(s) &=& -\zeta(s)+2\left(1-\frac{1}{2^s}\right)\zeta(s)=-\zeta(s)+2\prod_{p\neq 2}\left(1-\frac{1}{p^s}\right)^{-1}\\&=&-\sum_{n\geq ...

2

Assuming $\text{Re}(s)>0$ we have: $$\begin{eqnarray*} f(s)=\sum_{n\geq 1}\frac{(-1)^n}{(2n+1)^s}&=&\sum_{n\geq 1}\frac{(-1)^n}{\Gamma(s)}\int_{0}^{1}(-\log x)^s x^{2n}\,dx\\&=&-\frac{1}{\Gamma(s)}\int_{0}^{1}(-\log x)^s \frac{x^2}{1+x^2}\,dx\\&=&-\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{t^s}{e^t(1+e^{2t})}\,dt\end{eqnarray*}$$ ...

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