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Prove $$\lim_{x\rightarrow 1}\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{x}}=\ln2.$$


Of course $$\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}}=\ln2,$$ but we can not use the Proposition :

If a sequence of functions that are continuous on a set converges uniformly on that set ,then the limit function is continuous on the set. because for $\forall\delta >0$,we have $\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{x}} $converges on$\left(1-\delta ,1+\delta \right), $but nonuniformly .


In fact ,for every ${x}^{'},{x}^{"}\in \left(1,1+\delta \right),$in other words $\left| {x}^{'}-{x}^{"}\right|<\delta $,fixing the point $ {x}^{'}, $let ${x}^{"}\rightarrow {1}^{+},$we can get paradox:$$1\geq |\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{{x}^{'}}}-(+\infty)|=+\infty>1.$$ So $\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{x}}$ converge nonuniformly on $\left(1-\delta ,1+\delta \right)$.


My question is how can we get $\lim_{x\rightarrow 1}\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{x}}=\ln2.$

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    $\begingroup$ Please: Write $\ln2$, not $ln2$. It is coded as \ln 2. The backslash results in proper spcing in things like $2\ln3$ and in non-italicization of $\ln$. Some other aspects of your way of writing MathJax code are appalling; they look as if they came from one those web-based applications supposedly intended to help with MathJax. I've cleaned it up somewhat. $\endgroup$ Aug 7 '14 at 16:43
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Let

$$f_N(x) = \sum_{n=1}^N \frac{(-1)^{n-1}}{n^x},\quad R_N(x) = \sum_{n=N+1}^\infty \frac{(-1)^{n-1}}{n^x}.$$

It is clear that for every $N$ you have $\lim_{x\to 1} f_N(x) = f_N(1)$. It remains to find a bound on

$$\lvert R_N(x)\rvert.$$

The fact that the series is alternating should help with that. (And that shows that actually the convergence of the series is uniform on $(1-\delta,1+\delta)$ for $0 < \delta < 1$.)

Namely, for $x > 0$, we have $\frac{1}{n^x} > \frac{1}{(n+1)^x}$, and thus by Leibniz' criterion for alternating series, it follows that $$\lvert R_N(x)\rvert = \left\lvert \sum_{k=0}^\infty \frac{(-1)^k}{(N+1+k)^x}\right\rvert \leqslant \frac{1}{(N+1)^x}.$$ Since $x \mapsto \frac{1}{(N+1)^x}$ is monotonically decreasing on $(0,+\infty)$ for every $N\in\mathbb{N}\setminus \{0\}$, we thus have $$\lvert R_N(x)\rvert \leqslant \frac{1}{(N+1)^x} \leqslant \frac{1}{(N+1)^{1-\delta}}$$ for all $N$ and $x \geqslant 1-\delta$. It follows that the convergence of $f_N$ is uniform on every $[1-\delta,+\infty)$ for $0 < \delta < 1$. Thus the proposition cited in the question can be used.

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  • $\begingroup$ $\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{x}}$ converge nonuniformly on $\left(1-\delta ,1+\delta \right)$.why @Daniel Fischer get the convergence of the series is uniform on $\left(1-\delta ,1+\delta \right).$I can't understand this. $\endgroup$
    – Jay
    Aug 7 '14 at 17:15
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    $\begingroup$ Why do you think the convergence is not uniform? The convergence is not absolute (for $x \leqslant 1$), so the Weierstraß test fails. Yet, the convergence is uniform, since you can find a uniform bound on $\lvert R_N(x)\rvert$ on $(1-\delta,1+\delta)$ as long as $0 < \delta < 1$. $\endgroup$ Aug 7 '14 at 17:19
  • $\begingroup$ In fact ,for every ${x}^{'},{x}^{"}\in \left(1,1+\delta \right),$in other words $\left| {x}^{'}-{x}^{"}\right|<\delta $,fixing the point $ {x}^{'}, $let ${x}^{"}\rightarrow {1}^{+},$we can get paradox:$$1\geq |\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{{x}^{'}}}-(+\infty)|=+\infty>1.$$ So $\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{x}}$ converge nonuniformly on $\left(1-\delta ,1+\delta \right)$. $\endgroup$
    – Jay
    Aug 7 '14 at 17:20
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    $\begingroup$ I don't see where you take the $+\infty$ from there. $\endgroup$ Aug 7 '14 at 17:22
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    $\begingroup$ @Jay It is. Somewhere your reasoning went astray, and I'm trying to figure out where. Should I fill in the details I left out so far before we explore that further, or shall we explore first? $\endgroup$ Aug 7 '14 at 17:26
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You may observe that $$ \sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{x}}=\frac{1}{\Gamma(x)}\int_0^\infty\frac{u^{x-1}}{e^u+1} \mathrm{d}u, \quad x>1. $$ Letting $ x \rightarrow 1^+$ gives $$ \lim_{x\rightarrow 1^+}\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{x}}=\int_0^\infty\frac{1}{e^u+1} \mathrm{d}u=\ln 2 . $$

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  • $\begingroup$ Do you konw $x\rightarrow {1}^{+}$ is different from $x\rightarrow {1}?$ $\endgroup$
    – Jay
    Aug 7 '14 at 17:28
  • $\begingroup$ @Jay I realize that I did not fully understand your question. $\endgroup$ Aug 7 '14 at 20:23
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{}$ \begin{align}&\color{#66f}{\large\lim_{x\ \to\ 1}\sum_{n = 1}^{\infty}% {\pars{-1}^{n - 1} \over n^{x}}} =-\lim_{x\ \to\ 1}\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{x}} =-\lim_{x\ \to\ 1}{\rm Li}_{x}\pars{-1}=-\,{\rm Li}_{1}\pars{-1} \\[3mm]&=-\braces{-\ln\pars{1 - \bracks{-1}}} =\color{#66f}{\Large\ln\pars{2}} \approx {\tt 0.6931} \end{align}

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