# Tag Info

39

If we can change the order of summation, we obtain \begin{align} 1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\eta(k) &= 1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^k} \\ &= 1 + \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=1}^\infty \frac{(-1)^k}{(2n)^k}\\ &= 1 + \sum_{n=1}^\infty (-1)^{n+1} ... 26 In the comments section to Willie Wong's answer, the following Dirichlet series came up: the Riemann \zeta-function, Dirichlet L-functions, and Ramanujan's series \sum_{n \geq 1}\tau(n) n^{-s}, where \tau(n) is the coefficient of q^n in \Delta(q) = q\prod_{n=1}^{\infty} (1-q^n)^{24}. First note that the \zeta-function is a special case of a ... 15 Trivial zeros Numerically I found that for n any positive integer (replacing your \,\gamma+\gamma\, constant by \,\log\,\pi\,) :\tag{1}\frac{\zeta ''(-2\;n)}{2\,\zeta '(-2\;n)}+\log (n)<\log(\pi)$$with the limit approaching \log\,\pi\, as \,n\to \infty. I obtained too the following asymptotic expansion as \,n\to\infty :$$\frac{\zeta ...

14

This question is an opportunity to showcase Mellin transforms and harmonic sums, where we first compute the Mellin transform of the sum and subsequently invert it, obtaining an asymptotic expansion about zero/infinity. Consider $$g(x) = \frac{1}{1+x}.$$ The Mellin transform $g^*(s)$ of $g(x)$ is given by $$g^*(s) = \mathfrak{M}(g(x); s) = \int_0^\infty ... 11 Notice\color{blue}{^{[1]}}$$\sum_{k=1}^\infty \left( \frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right) = \sum_{\substack{k=-\infty\\ k\ne 0}}^\infty \frac{1}{(6k-1)^3} = 1 - \frac{1}{6^3}\sum_{k=-\infty}^\infty \frac{1}{(\frac16-k)^3}$$Recall the infinite product expansion of \sin x$$\sin x = x \prod_{k=1}^\infty \left( 1 - \frac{x^2}{k^2\pi^2}\right)$$... 9 The Dirichlet eta function is given by \eta(s)=\sum_{n=1}^{\infty}(-1)^{n-1}n^{-s}, but this converges only for s with positive real part, and you are proposing to use its behavior for negative integers. A globally convergent series for \eta can be derived using the Riemann zeta function (cf. here):$$ ...

7

Approximate $G(z)$ by the first two terms, $g(z) := e^{-1} + e^{-4}/2^z$. The zeroes of $g$ are $$- \frac{3}{\log 2} + (2k+1) \frac{\pi}{\log 2} i \approx - 4.328 + (2k+1) 4.532 i \ \mathrm{for} \ k \in \mathbb{Z}.$$ In particular, the zeroes of $g$ are perfectly regularly spaced. Your zeroes are pretty close to these points, so I suspect that the ...

7

You could look at Serre's article Modular forms of weight one and Galois representations in the 1975 Durham proceedings (also in his collected works). My memory is that he gives several examples of Artin $L$-functions (including examples relating them to Hecke $L$-functions in the case when the Artin representation is induced from a character). As for ...

7

Your original observation is easily explained: $d(n)$ is odd if and only if $n$ is a square.

6

The idea needed here is Dirichlet convolution. The convolution identity states that if $$\left(\sum_{j=1}^\infty \frac{a_j}{j^s}\right)\left(\sum_{k=1}^\infty \frac{b_k}{k^s}\right)=\sum_{\ell=1}^\infty \frac{c_\ell}{\ell^s}$$ then $$\sum_{k\,\mid \,n} a_k b_{n/k} = c_n$$ where the sum ranges over all positive divisors of $n$. More specifically, your ...

6

$g$ will always have a half-plane free from zeroes (cf. e.g. Titchmarsh, Theory of functions, Section 9.6). This means a functional equation would be sufficient to guarantee a critical strip (essentially calling the other zeroes trivial by definition).

5

For example, for any Dirichlet character $\chi$, the sums $\sum_{n=1}^\infty \chi(n)$ can be summed by analytic continuation of $\sum_n \chi(n)/n^s$, which has a meromorphic continuation. Similarly, for many other arithmetical functions (such as coefficients of modular forms) the analogous sum has a meromorphic continuation, so is summable in this sense. ...

5

(1) is a correct computation. In general, to treat primes of the form $kn+m$, you would have a linear combination of $\phi(k)$ sums, each of which runs over the zeros of a different Dirichlet $L$-function (of which the Riemann $\zeta$ function is a special case). And yes, assuming the generalized Riemann hypothesis, all of the terms including those sums over ...

5

Your Euler product factorization is incorrect. It should proceed like the following: $$\sum_{n=1}^\infty\frac{(-1)^{n-1}\chi(n)}{n^s}=\left(1-\frac{2\chi(2)}{2^s}\right)\sum_{n=1}^\infty\frac{\chi(n)}{n^s}=\frac{1-\chi(2)2^{1-s}}{1-\chi(2)2^{-s}}\prod_{p>2}\left(\frac{1}{1-\chi(p)p^{-s}}\right),$$ or simply $(1-\chi(2)2^{1-s})L(s,\chi)$. Compare to ...

5

The orthogonality relations for Dirichlet characters imply that$$\frac1{\phi(q)} \sum_{\chi\pmod q} L(s,\chi) = \sum_{n\equiv1\pmod q} n^{-s}.$$ Therefore$$H(s) = \zeta(s) \sum_{n\equiv1\pmod q} n^{-s} = \sum_{n=1}^\infty n^{-s} \sum_{\substack{d\mid n \\ d\equiv1\pmod q}} 1,$$ from which you can answer your particular question.

5

There is a closed form for your series: $$\sum_{k=1}^{\infty}\left( \zeta(2k)-\beta(2k)\right)=\frac{1}{2}+\frac{1}{2}\ln2. \tag1$$ Proof. Using absolute convergence of the series, you may write \begin{align} \sum_{k=1}^{\infty}\left( \zeta(2k)-\beta(2k)\right) & = \sum_{k=1}^{\infty}\left( ... 5 We have: \zeta(s)=\prod_{p}\left(1-\frac{1}{p^s}\right)^{-1},\qquad \zeta(s)^2=\sum_{n\geq 1}\frac{d(n)}{n^s} $$hence, assuming that the sequence \{a(n)\}_{n\in\mathbb{N}^*} is the sequence of coefficients of the Dirichlet series associated with f(s):$$ f(s)=\zeta(s)^2\left(1-\frac{1}{2^{s-1}}\right)\left(1-\frac{1}{3^{s-1}}\right)=\sum_{n\geq ...

4

See my response here.

4

Let $D_f(s)$ denote the Dirichlet generating series: $$D_f(s) = \sum_{n =1}^{\infty} \frac{f(n)}{n^s}.$$ When the series convergences absolutely, we can write $D_f(s)$ as an Euler product: $$D_f(s) = \prod_p \left(1+f(p)p^{−s} +f(p^2)p^{−2s} \dots \right)$$ where the product ranges over all the primes $p$. Looking at the Euler product for ...

4

The "divisor convolution" of two arithmetic functions $a_n$ and $b_n$ is the arithmetic function $(a\star b)(n) = \sum_{d \mid n} a_db_{n/d}$. If $\sum a(n) n^{-s} = L(a, s)$ is the Dirichlet series of $a$, then we have the relation $$L(a, s)L(b, s) = L(a \star b, s).$$ In particular, the Möbius transform is $$L(\mu \star a, s) = L(a, s)/\zeta(s).$$ We ...

4

I was hoping someone else more knowledgeable would give a more detailed answer, but: the answer is the Euler-Maclaurin formula (applied to the difference between the two sums). The points at which the digits don't match correspond to the part of the formula involving a sum over Bernoulli numbers. The values of $N$ which most have this property are powers of ...

4

Approach, which uses Fourier series. Denote $$S = \sum_{k=1}^\infty \left[\frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right].$$ Consider function $$f(x) = \dfrac{\pi x(\pi-x)}{8}, \qquad x\in[0,\pi];\tag{1}$$ construct the odd extension of $f(x)$ to the interval $[−\pi, \pi]$: $f(-x)=-f(x), x\in [0,\pi]$; and make it $2\pi$-periodic: copy to each segment ...

4

Possible keywords include Dirichlet L-functions or the generalized Riemann hypothesis. For the case where $c_k$ are the values of a Dirichlet character, in certain situations the existence of a critical strip is known. See the Wikipedia article for references.

4

We can, without loss of generality, assume $s = 0$ (otherwise replace $a_n$ by $b_n = a_n\cdot n^{-s}$). We prove the convergence of $\sum\limits_{n=1}^\infty \frac{a_n}{n^z}$ for $\operatorname{Re} z > 0$ under the slightly weaker assumption that the sequence $$P_n = \sum_{k=1}^n a_k$$ of partial sums is bounded. To prove the convergence of ...

3

I have worked out a partial answer to my own question. Define $A(s) = \sum\limits_{n=1}^\infty\frac{a_n}{n^s}$. First, assume that $s = 1$. WLOG we can also assume that $a = 1$. Observe that for some $N$ and all $n > N$, $\Re(a_n) > 1/2$. It follows that $A(1)$ diverges. I found a pertinent result in Knopp, Konrad. Infinite Sequences and Series. ...

3

Hint: $e^{ikx}+e^{i(-k)x}=2cos(kx)$, for all $1\le k\le n$.

3

Let $S(t)=\sum_{2\leq n\leq t}\chi(n)$, and suppose that $s>1.$ Using integration by parts, we have that $$\sum_{n=2}^{\infty}\chi(n)n^{-s}=\int_{2}^{\infty}t^{-s}d\left(S(t)\right)=s\int_{2}^{\infty}S(t)t^{-s-1}dt,$$ and by applying the trivial bound $|S(t)|\leq q,$ it follows that \left|\sum_{n=2}^{\infty}\chi(n)n^{-s}\right|\leq q2^{-s}.\ \ \ \ \ \ \ ...

3


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