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I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series

$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots ? \tag{II}$$

Does it converge? If so, what is its sum?

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$$\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}n\quad\iff\quad\sum_{n=1}^\infty\frac{(-1‌​)^{n+1}}n=-\sum_{n=1}^\infty\frac{(-1)^n}n=\ln(1-[-1])=\ln2$$ – Lucian Nov 19 '13 at 7:39
I know a non-calculus way, but you have to assume $e^x\ge x+1$. Would you like me to share it with you? – Akiva Weinberger Mar 24 '15 at 17:37
up vote 15 down vote accepted

Complementary to Mau's answer:

Call a series $a_n$ absolutely convergent if $\sum|a_n|$ converges. If $a_n$ converges but is not absolutely convergent we call $a_n$ conditionally convergent The Riemann series theorem states that any conditionally convergent series can be reordered to converge to any real number.

Morally this is because both the positive and negative parts of your series diverge but the divergences cancel each other out, one or other's canceling the other can be staggered by adding on, say, the negative bits every third term in stead of every other term. This means that in the race for the two divergences to cancel each other out, we give the positive bit something of a head-start and will get a larger positive outcome. Notice how, even in this rearranged version of the series, every term will still come up exactly once.

It is also worth noting, on the Wikipedia link Mau provided, that the convergence to $\ln 2$ of your series is at the edge of the radius of convergence for the series expansion of $\ln(1-x)$- this is a fairly typical occurrence: at the boundary of a domain of convergence of a Taylor series, the series is only just converging- which is why you see this conditional convergence type behavior.

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it is not absolutely convergent (that is, if you are allowed to reorder terms you may end up with whatever number you fancy).

If you consider the associated series formed by summing the terms from 1 to n of the original one, that is you fix the order of summation of the original series, that series (which is not the original one...) converges to $\ln(2)$ See Wikipedia.

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But rearranging terms on pretty much anything that goes out infinitely allows any possible answer, hence changing what you are summing. – Simple Art Jan 7 at 22:55
this is why I wrote "If you consider the associated series [...]" and I computed that sum. – mau Jan 8 at 12:14
I just want future readers to understand your answer better. – Simple Art Jan 8 at 22:00
and indeed my first sentence says "(that is, if you are allowed to reorder terms you may end up with whatever number you fancy)." – mau Jan 9 at 9:53

$\sum_{k=1}^{n} ( \frac{1}{2k-1}-\frac{1}{2k} ) = \sum_{k=1}^{n} ( \frac{1}{2k-1}+\frac{1}{2k} ) - 2 \sum_{k=1}^{n} \frac{1}{2k} = \sum_{k=1}^{2n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k} = \sum_{k=n+1}^{2n} \frac{1}{k}$.

$\ln(2) \overset{n\to\infty}{\leftarrow} \ln(2) + \ln(\frac{2n+1}{2n+2}) = \ln(2n+1)-\ln(n+1)$

$= \int_{n+1}^{2n+1} \frac{1}{x}\ dx \le \sum_{k=n+1}^{2n} \frac{1}{k} \le \int_{n}^{2n} \frac{1}{x}\ dx$

$= \ln(2n)-\ln(n) = \ln(2)$.

So by squeeze theorem we are done.

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In this answer, I used only Bernoulli's inequality to show that $$ \left(\frac{2n+1}{n+1}\right)^\frac{n}{n+1} \le\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)} \le\frac{2n+1}{n+1}\tag{1} $$ The squeeze theorem and $(1)$, show that $$ e^{\lim\limits_{n\to\infty}\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}=2\tag{2} $$ That is, $$ \begin{align} \lim_{n\to\infty}\left(1-\frac12+\frac13-\frac14+\dots-\frac1{2n}\right) &=\lim_{n\to\infty}\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)\\[6pt] &=\log(2)\tag{3} \end{align} $$

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Is it the change in form of the series, across the = in the line above (3), that is dependent on the ordering of the series? – Isaac Apr 16 '13 at 16:14
@Isaac: The stuff inside the parentheses is equal. There is no problem about the ordering when summing a finite number of terms. – robjohn Apr 16 '13 at 16:21
@Isaac: in the sum $\displaystyle\sum_{k=1}^\infty(-1)^{k-1}\frac1k$ the order is implied. However, we if we try to sum the numbers in the set $\{1,-\frac12,\frac13,-\frac14,\dots\}$, because no order is implied, the Riemann Series Theorem says that depending on the order, the partial sums might diverge or converge to any real number. – robjohn Apr 16 '13 at 16:44
Ahh, right—in my quick read, I hadn't thought about the difference between the limit of the finite series and the infinite series. I hadn't meant to imply that the two sides of that = weren't equal; I more meant that the infinite series related to the left side is conditionally convergent and rearrangement would make it not equal to the right side. – Isaac Apr 16 '13 at 16:49
This is really very unusual and new for me. +1. – Paramanand Singh Mar 29 at 4:19

There are actually two "more direct" proofs of the fact that this limit is $\ln (2)$.

First Proof Using the well knows (typical induction problem) equality:

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

The right side is $\frac{1}{n} \left[ \frac{1}{1+\frac{1}{n}}+ \frac{1}{1+\frac{2}{n}}+..+\frac{1}{1+\frac{n}{n}} \right]$ which is the standard Riemann sum associated to $\int_0^1 \frac{1}{1+x} dx \,.$

Second Proof Using $\lim_n \frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}-\ln (n) =\gamma$.


$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}= \left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2n-1}+\frac{1}{2n} \right]-2 \left[\frac{1}{2}+\frac{1}{4}...+\frac{1}{2n} \right] $$

$$= \left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2n-1}+\frac{1}{2n} \right]-\ln(2n) - \left[\frac{1}{1}+\frac{1}{2}...+\frac{1}{n} \right]+\ln(n) + \ln 2 \,.$$

Taking the limit we get $\gamma-\gamma+\ln(2)$.

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First equation, right-hand side: Should the first fraction be $\frac{1}{n+1}$? – Mike Spivey Jun 2 '11 at 5:09
Thank you. Fixed. – N. S. Jun 2 '11 at 10:47
I don't consider the second a proof because it is totally not apparent why that limit should exist, and it is no easier. – user21820 Apr 3 '15 at 11:35
@user21820 The existence of the limit $\lim_n \frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}-\ln (n)$ is well known and an easy consequence of the estimates of the integral test. – N. S. Apr 3 '15 at 15:05
@N.S.: Yes indeed but so is $\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\cdots$ well known and provable without even using the existence of $\gamma$ that you used. My point is that what you left unproven is as hard or even harder than the original problem. – user21820 Apr 4 '15 at 5:07

Let's say you have a sequence of nonnegative numbers $a_1 \geq a_2 \geq \dots$ tending to zero. Then it is a theorem that the alternating sum $\sum (-1)^i a_i$ converges (not necessarily absolutely, of course). This in particular applies to your series.

Incidentally, if you're curious why it converges to $\log(2)$ (which seems somewhat random), it's because of the Taylor series of $\log(1+x)$ while letting $x \to 1$.

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To add to Akhil's answer, one needs to invoke Abel's theorem (, since 1 is at the border of the interval of convergence. This is a delicate test that ensures that the numerical series converges to the number the power series predicts. – Andrés Caicedo Jun 2 '11 at 5:12
@Andres: Thanks for fixing my grammar and for the comment! – Akhil Mathew Jun 6 '11 at 13:22

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