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I was messing around in Mathematica with infinite sums until I tried taking the sum of the following: $$\sum_{n = 1}^{\infty} \frac{(-1)^n}{n}$$ Mathematica spat out -Log[2]. Can somebody give me proof explaining this answer?

Mathime
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    Note how this is exactly the log's Taylor series at z = 2 times -1. Thus you obtain $-\log (2)$. – Piotr Benedysiuk Aug 14 '16 at 14:23
  • Hint: Try a Maclaurin expansion of $\log{(1+x)}$ and see what you get. – ekkilop Aug 14 '16 at 14:24
  • @PiotrBenedysiuk I wasn't aware of that? Is there any other way to reach an answer? – Mathime Aug 14 '16 at 14:26
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    $-\log 2 = -\int_0^1 \frac{1}{1+x} dx = -\int_0^1 \sum_{k\geq 0} (-x)^k dx = \sum_{n\geq 1} (-1)^n/n$ (if we don't worry about convergence, and not really different from using Maclaurin of log) – H. H. Rugh Aug 14 '16 at 14:28
  • I'd say that every argument you are making here will be based on power/Taylor series. – Piotr Benedysiuk Aug 14 '16 at 14:29
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    Warning: it is silly in this context for someone to say not to worry about convergence issues -- that is whole point! Any answer using the Taylor series should be saying the words "Abel's Theorem," or giving some argument like Olivier Oloa's. – symplectomorphic Aug 14 '16 at 15:01

2 Answers2

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One may start with the standard geometric series identity, $$ 1-x+x^2-\cdots+(-1)^Nx^{N}=\frac{1+(-1)^Nx^{N+1}}{1+x},\quad x \neq-1, $$ giving $$ \int_0^1\left(1-x+x^2-\cdots+(-1)^Nx^N\right)dx=\int_0^1\frac{1+(-1)^Nx^{N+1}}{1+x}\:dx $$ or $$ \sum_{n = 1}^{N+1} \frac{(-1)^{n-1}}{n}=\int_0^1\frac1{1+x}\:dx+\int_0^1\frac{(-1)^Nx^{N+1}}{1+x}\:dx=\ln2+\int_0^1\frac{(-1)^Nx^{N+1}}{1+x}\:dx $$ then, letting $N \to \infty$, one may use $$ \left|\int_0^1\frac{(-1)^Nx^{N+1}}{1+x}\:dx\right|\le \int_0^1x^{N+1}dx=\frac1{N+2} \to 0, $$ which gives

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

Olivier Oloa
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A formal argument (ignoring convergence questions)

Start with the Geometric series $$\frac 1{1-x}=1+x+x^2+\cdots$$

Integrate to obtain $$-\ln(1-x)=x+\frac {x^2}2+\frac {x^3}3+\cdots$$

Now evaluate at $x=-1$ to get your result.

To be more rigorous, note that (inductively) it is easy to prove

$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$

And the right hand can then be rewritten as $$\frac{1}{n} \left[ \frac{1}{1+\frac{1}{n}}+ \frac{1}{1+\frac{2}{n}}+\cdots+\frac{1}{1+\frac{n}{n}} \right]$$

Which is the standard Riemann sum approximation to $$\int_0^1 \frac {dx}{1+x}=\ln(2)$$

lulu
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    Rather than offer a formal argument that doesn't prove anything, why don't you just say Abel's theorem justifies the (last step of) the first calculation? – symplectomorphic Aug 14 '16 at 15:03
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    @symplectomorphic Well, because I didn't expect that the OP was familiar with Abel's Theorem. To me, the formal argument (while admittedly lacking rigor) explains the connection to logs which otherwise looks quite mysterious. In any case, the second argument should be satisfactory (and more elementary than Abel). – lulu Aug 14 '16 at 15:08
  • @lulu Your second argument using the Riemann Sum is pretty much the same as Ramanujan's. It's on the first page of the 2nd chapter of his first notebook. http://www.imsc.res.in/~rao/ramanujan/notebookindex.htm – Beauty Is Truth Aug 15 '16 at 06:43
  • @BeautyIsTruth Didn't know that...thanks for the link! – lulu Aug 15 '16 at 10:08
  • @lulu No problem. ^_^ I just wish I could find a link to Ramanujan's notebooks in full color. – Beauty Is Truth Aug 15 '16 at 10:25