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Prove that there are no real numbers $x$ such that

$$\sum_{n\,=\,0}^\infty \frac {(-1)^{n + 1}} {n^x} = 0$$

Can I have a hint please?

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    $\begingroup$ What do you know about the error estimates in an alternating series? $\endgroup$ – Daniel Fischer Apr 24 '14 at 13:14
  • $\begingroup$ for $x\le0$ the sum diverges, for $x>0$, hmm ... use sqeeze theorem. $\endgroup$ – Santosh Linkha Apr 24 '14 at 13:14
  • $\begingroup$ @DanielFischer Nothing, I only know about the alternating series test. $\endgroup$ – Superbus Apr 24 '14 at 13:24
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    $\begingroup$ This is the Dirichlet $\eta$ function. $\endgroup$ – Lucian Apr 24 '14 at 13:34
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    $\begingroup$ The $n=0$ term is problematic... Did you mean $\sum_{n=1}^\infty$? $\endgroup$ – Noam D. Elkies Apr 29 '14 at 4:30
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For $x \leqslant 0$, the terms of the series don't converge to $0$, hence the series diverges then. Therefore, we need only consider $x > 0$.

For $x > 0$, the sequence $\left(\frac{1}{n^x}\right)_{n\in\mathbb{Z}^+}$ is strictly decreasing and converges to $0$, thus by Leibniz' criterion

$$\eta(x) := \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^x}$$

converges.

Now, if $(a_n)_{n\in\mathbb{Z}^+}$ is any monotonically non-increasing sequence converging to $0$, we can consider the partial sums of an odd and an even number of terms separately,

$$s_{2p} = \sum_{n=1}^{2p} (-1)^{n+1} a_n;\qquad s_{2p+1} = \sum_{n=1}^{2p+1} (-1)^{n+1} a_n.$$

We find

$$\begin{align} s_{2p+3} - s_{2p+1} &= (-1)^{2p+3}a_{2p+2} + (-1)^{2p+4}a_{2p+3} = a_{2p+3} - a_{2p+2} \leqslant 0,\\ s_{2p+2} - s_{2p} &= (-1)^{2p+3} a_{2p+2} + (-1)^{2p+2} a_{2p+1} = a_{2p+1} - a_{2p+2} \geqslant 0,\\ s_{2p+1} - s_{2p} &= (-1)^{2p+2}a_{2p+1} = a_{2p+1} \geqslant 0. \end{align}$$

So

  • the sequence of partial sums of an odd number of terms is monotonically non-increasing,
  • the sequence of partial sums of an even number of terms is monotonically non-decreasing, and
  • the partial sum of an odd number of terms is never smaller than the partial sum of an even number of terms.

If - like for the specific sequence under consideration - the sequence $(a_n)$ is strictly monotonically decreasing, all inequalities above are strict.

It is straightforward to deduce from that that $\eta(x) > 0$ for all $x > 0$.

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{n = \color{#f00}{\LARGE 1}}^{\infty}{\pars{-1}^{n + 1} \over n^{x}} = 0:\ {\large ?}}$

\begin{align} &\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n^{x}} =\sum_{n = 1}^{\infty}\bracks{{1 \over \pars{2n - 1}^{x}} - {1 \over \pars{2n}^{x}}} \\[3mm]&={1 \over 2^{x}}\sum_{n = 1}^{\infty} \bracks{{1 \over \pars{n - 1/2}^{x}} - {1 \over n^{x}}} > 0\quad \mbox{when}\quad x > 0.\qquad\qquad\mbox{So ?...} \end{align}

$\ds{x \leq 0}$ cases are not considered for obvious reasons.

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This is an observation which is too long for a comment, but which someone will probably appreciate. Remark that

$$\sum_{n=1}^\infty \frac{(-1)^n}{n^s} = \zeta(s) - 2\times 2^{-s}\zeta(s) = (1-2^{1-s})\zeta(s).$$

The Riemann zeta function is equal to the Euler product $\prod_p (1-p^{-s})^{-1}$ in the right half-plane $\Re s>1$. The Euler product converges in the sense of convergence for infinite products. A convergent infinite product is never zero, so $\zeta(s)\neq 0$ for $\Re s > 1$. Moreover, it's easy to see that $1-2^{1-s} \neq 0$ for $\Re s> 1$. So we get your result for $\Re s >1$.

For $0<\Re s \leq 1$, it's a little tricky because the Euler product no longer converges. However, your sum is still equal to $(1-2^{1-s})\zeta(s)$ when $0<\Re s \leq 1$, essentially by analytic continuation. At $s=1$, $(1-2^{1-s})\zeta(s) \neq 0$ because the zero of $1-2^{1-s}$ is cancelled by the simple pole of $\zeta(s)$.

Thus, the proofs that Felix and Daniel have given show that the Riemann zeta function doesn't vanish for real $s$ between $0$ and $1$!

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