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My book directly writes-

$$\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\cdots=-\ln 2+1.$$

How do we prove this simply.. I am a high school student.

geek101
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  • http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29#Alternating_harmonic_series – peter.petrov Mar 21 '15 at 17:51
  • http://en.wikipedia.org/wiki/Alternating_series_test – peter.petrov Mar 21 '15 at 17:51
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    Hm, this seems not so simple to prove if only high-school math is allowed to be used :) – peter.petrov Mar 21 '15 at 17:53
  • Ohh thats bad of my book then – geek101 Mar 21 '15 at 17:54
  • Well, others may think differently but I studied this thing during my 1st university year. – peter.petrov Mar 21 '15 at 17:55
  • How am i supposed to solve this easily then-$\log_42-log_82+log_{16}2$.... – geek101 Mar 21 '15 at 17:56
  • Please help me solve the same without Taylor series – geek101 Mar 21 '15 at 17:59
  • You can actually prove this without calculus, if you let yourself assume that $e^x\ge x+1$ for all $x$. (Graph it here to see better what that means.) Specifically, let the partial sum $\frac12-\frac13+\dotsb\frac1n$ be called $S_n$; you can prove that $1-\ln(2+\frac1n)<S_n<1-\ln(2-\frac1n)$ for all $n$, and the infinite sum follows from the squeeze theorem. – Akiva Weinberger Mar 22 '15 at 01:11
  • By the way, that property above is unique to $e$; there are values of $x$ where $2^x<x+1$, for example. (Try to graph $a^x$ and $x+1$ in the above link. It'll give you an adjustable slider to set the value of $a$.) – Akiva Weinberger Mar 22 '15 at 01:16

4 Answers4

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How about the following way? Let us prove that $$\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}=\ln 2.$$

Since we have $$\begin{align}\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}&=\left(\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}+2\sum_{k=1}^{n}\frac{1}{2k}\right)-2\sum_{k=1}^{n}\frac{1}{2k}\\&=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{k}\\&=\sum_{k=1}^{n}\frac{1}{n+k}\end{align}$$ we have $$\begin{align}\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}&=\lim_{n\to \infty}\sum_{k=1}^{n}\frac{1}{n+k}\\&=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+(k/n)}\\&=\int_{0}^{1}\frac{1}{1+x}dx\\&=\ln 2.\end{align}$$

So, you'll have $$\sum_{k=2}^{\infty}\frac{(-1)^k}{k}=1-\ln 2.$$

mathlove
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$$\sum_{k=2}^{+\infty}\frac{(-1)^k}{k}=\sum_{n\geq 1}(-1)^{n-1}\int_{0}^{1}x^n\,dx = \int_{0}^{1}\frac{x dx}{1+x} = 1-\int_{0}^{1}\frac{dx}{1+x}=\color{red}{1-\log 2}.$$

Jack D'Aurizio
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  • Hi @JackD'Aurizio - this is really pretty. I just used your style to write up the log2 proof (instead of 1-log2). The only tricky part was justifying swapping the integral (with finite limits, 0 and 1) and with the infinite summation symbol. The rest is just integration using the convergent geometric series formula. So, how do you justify the swapping of integral with summation? Is it uniform convergence? Is it because...the integral has finite lower and upper limits? Thanks, – User001 Nov 15 '15 at 01:00
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    @LebronJames: the swap is allowed by the dominated convergence theorem. Over $[0,1]$, the function $\frac{1}{1+x}$ is between $\frac{1}{2}$ and $1$. – Jack D'Aurizio Nov 15 '15 at 01:17
  • Ah, perfect -- yes, agreed. Writing it first as a limit of partial sums makes it apparent; the limit of the integrals is the integral of the limit. Thanks so much @JackD'Aurizio. I really enjoyed your one-line proof of this classic question :-) – User001 Nov 15 '15 at 01:24
  • Hi @JackD'Aurizio, I just wrote up a proof in a new question on MSE that is in the same spirit as your one-line proof here. If you have time, I would love your feedback on my proof. I prefer to use this technique rather than the harder-to-memorize summation by parts / Abel's partial summation formulas. Thanks so much :-) – User001 Nov 23 '15 at 05:11
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In calculus there's this famous alternating harmonic series:

$S = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... + (-1)^{(n+1)} \cdot \frac{1}{n} + ... $

(*) It's convergent and its sum is equal to $ln2$

Your series is equal to exactly $T = -S + 1$ so it's
also convergent and its sum must be exactly $(-ln2+1)$

I realize that I didn't prove this (*) statement. I am not
aware of an elementary proof but there might be one.

peter.petrov
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Break up the series into two sub-series, depending on the parity of the denominator, then, after rewriting each of the two in terms of harmonic numbers, use the fact that $H_K~\simeq~\ln K$, and let $N\to\infty$, as follows:

$\begin{align} \color{green}{S_{2N}} &~=~\color{blue}{\dfrac11}\color{red}{-\dfrac12}\color{blue}{+\dfrac13}\color{red}{-\dfrac14}\color{blue}{+\dfrac15}\color{red}{-\dfrac16}+~\ldots~\color{blue}{+\dfrac1{2N-1}}\color{red}{-\dfrac1{2N}}~=~ \\\\ &~=~\color{blue}{\dfrac11}\color{red}{+\dfrac12}\color{blue}{+\dfrac13}\color{red}{+\dfrac14}\color{blue}{+\dfrac15}\color{red}{+\dfrac16}+~\ldots~\color{blue}{+\dfrac1{2N-1}}\color{red}{+\dfrac1{2N}}~-~ \\\\ &\color{red}{~\qquad~-\dfrac22\qquad~-\dfrac24\qquad-\dfrac26\qquad-\quad\ldots\qquad~-\frac2{2N}}~=~ \\\\ &~=~\color{blue}{\dfrac11}\color{red}{+\dfrac12}\color{blue}{+\dfrac13}\color{red}{+\dfrac14}\color{blue}{+\dfrac15}\color{red}{+\dfrac16}+~\ldots~\color{blue}{+\dfrac1{2N-1}}\color{red}{+\dfrac1{2N}}~-~ \\\\ &\color{red}{~\qquad~-\dfrac11\qquad~-\dfrac12\qquad-\dfrac13\qquad-\quad\ldots\qquad~-\frac1N}~=~ \\\\ &~=~\color{green}{H_{2N}~-~H_N~\simeq~\ln(2N)-\ln N~=~\ln\dfrac{2N}N~=~\ln2}. \end{align}$

More rigorously, we could write $H_K~=~\ln K+\gamma_k$ , where $\displaystyle\lim_{k\to\infty}\gamma_k~=~\gamma$, see Euler-Mascheroni constant for more information.

Lucian
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