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I need to prove the following: $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+(-1)^{n+1}\frac{1}{n}+\cdots=\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}=\ln(2)$$

Method 1:)

The series $\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}$ is an alternating series, thus it is convergent, say to $l$. Therefore, both $s_{2n}$ and $s_n$ are convergent to the same limit $l$.

$$ \begin{align} s_{2n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{2n} & =\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2n}\right) \\[10pt] & =\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} \end{align} $$

It is an easy exercise to prove that: $$\lim_{n \to \infty }s_{2n}=\lim_{n \to \infty }s_n =\lim_{n \to \infty }\left [ \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} \right ]=\ln(2)$$ which implies that the given alternating series converges to $l=\ln 2$

However, I am interested to see a proof of this problem using the definition of the Riemann Integral as a sum of infinitely many rectangles of widths tending to zero. I tried to come up with a proof for this, but I couldn't. Can anyone share please?

Also, I am interested to see other methods of solving this problem (other than my method and the Riemann method). If anyone of you is aware of any other methods, please share :)

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@leonbloy: Oops! I see that I have answered all three. At least the answers were somewhat different. –  robjohn Jul 7 '12 at 14:10
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5 Answers 5

up vote 7 down vote accepted

Usually, to use the Riemann integral, alternating terms cause a problem. We want to have the finer partitions converge nicely, but an alternating series does not allow this. So, as far as I can see, we pretty much have to use the $\zeta$ trick you employ in your method: $$ \begin{align} &\frac11-\frac12+\frac13-\frac14+\dots+\frac1{2n-1}-\frac1{2n}\\ &=\left(\frac11+\frac12+\dots+\frac1{2n}\right)-2\left(\frac12+\frac14+\dots+\frac1{2n}\right)\\ &=\frac1{n+1}+\frac1{n+2}+\frac1{n+3}+\dots+\frac1{2n}\\ &=\sum_{k=n+1}^{2n}\frac{n}{k}\frac1n\tag{1} \end{align} $$ Then to use $(1)$ as a Riemann sum (with $x=k/n$ and $\mathrm{d}x=1/n$) for $$ \int_1^2\frac1x\,\mathrm{d}x=\log(2)\tag{2} $$

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Or $\frac1{n+1}+\frac1{n+2}+\frac1{n+3}+\dots+\frac1{2n} =\frac{1}{n} \sum_{k=1}^n \frac{1}{1+\frac{k}{n}}$ –  N. S. Jul 5 '12 at 18:20
    
@N.S.: Indeed. That essentially changes variables to shift the integral to $$\int_0^1\frac{1}{1+x}\,\mathrm{d}x=\log(2)$$ –  robjohn Jul 5 '12 at 18:30
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Here's another method by the Riemann integral, but not by definition: $$ \sum_{n=1}^{\infty }(-1)^{n+1}\frac{1}{n}= \lim_{m\to\infty}\sum_{n=1}^{m}(-1)^{n+1}\frac{1}{n}= $$ $$ \lim_{m\to\infty}\int_0^1(1-x+\ldots+(-1)^{m-1}x^{m-1})\,dx= \lim_{m\to\infty}\int_0^1\frac{1-(-x)^m}{1+x}\,dx= \int_0^1\frac{dx}{1+x}=\ln2. $$

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Ah, but the second to last step is the trick - why can you bring the limit inside the integral? –  Thomas Andrews Jul 5 '12 at 17:51
    
@ThomasAndrews since the region of integration is $[0,1]$ and the only "bad point" is at $x=-1$, one may apply the Dominated Convergence Theorem. en.wikipedia.org/wiki/Dominated_convergence_theorem –  nullUser Jul 5 '12 at 17:55
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@ThomasAndrews not nessesary to bring, $\lim_{m\to\infty}\int_0^1\frac{|(-x)^m|}{1+x}\,dx\le\lim_{m\to\infty}\int_0^1 x^m\,dx=0\;$. –  Andrew Jul 5 '12 at 17:56
    
Yeah, I knew it could be clarified and completed. However, it is always risky when writing out a line like this - the beginner might believe he can "always" do this, where, by "this," I mean, move a limit inside an integral. –  Thomas Andrews Jul 5 '12 at 18:03
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As you asked for other methods of solving this problem, here is one using Taylor series.

Now that you know you are looking for the RHS of $\ln(2)$, expand $f(x) = \ln(x)$ into Taylor series around $x=1$:

$f^{(n)}(x) = (-1)^{n+1} x^{-n} (n-1)!,$ for $n>0, x \neq 0$

so choosing to evaluate at $x=1$, using $f(1) = \ln(1) = 0$, we get

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

and using $x=2$, $\ln 2 = \sum_{n=1}^\infty (-1)^{n+1}/n$ as desired.

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While this idea can be used to obtain the result, it is much more subtle than this, and needs more details. Keep in mind that the theory of Taylor series (or you can use the general power series starting with the geometric series and integration) only guarantees that $\ln(x)= \sum_{n=1}^\infty (-1)^{n+1} \frac{(x-1)^n}{n}$ on the interval $(0,2)$. The equality at $x=2$ needs more effort ;) –  N. S. Jul 6 '12 at 14:02
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However one should note that, $$\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}$$
is not absolutely convergent as $$\sum_{n=1}^\infty |(-1)^{n+1}\frac{1}{n}|=\sum_{n=1}^\infty \frac{1}{n}$$ is divergent. (Such a series is said to be conditionally convergent)

Hence there is no fixed value for the series. (Refer to the Riemann Series Theorem) $\ln 2$ is just a value that the series would take for the particular permutation of $S_{2n}$.

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You can use the following geometric series $$\sum_{n=0}^\infty x^n=\frac{1}{1-x},|x|<1.$$ Then integrate both sides to get $$\sum_{n=1}^\infty \frac{x^n}{n}=\int_0^x\frac{1}{1-t}dt=-\ln (1-x),|x|<1.$$ Since the LHS converges at $x=-1$, letting $x\to-1$ will give us $$\sum_{n=1}(-1)^{n+1}\frac{1}{n}=\ln2.$$

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