by Isaac
Last Updated January 16, 2018 08:20 AM

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

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

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

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.

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.

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 $x \to 1$.

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

Then

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

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

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

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 $$ \exp\left[\lim\limits_{n\to\infty}\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)\right]=2\tag{2} $$ That is, $$ \begin{align} \lim_{n\to\infty}\left(1-\frac12+\frac13-\frac14+\dots-\frac1{2n}\right) &=\lim_{n\to\infty}\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)\\[6pt] &=\log(2)\tag{3} \end{align} $$

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

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

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

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

So by squeeze theorem we are done.

Here is another proof, based on the formula

$$\frac{1}{1+x}=\frac{x^n}{1+x}+\sum_{k=0}^n(-x)^k$$

Integrating both sides over $[0,t]$ gives

$$\ln(1+t)=\int_0^t\frac{x^n}{1+x}\,dx+\sum_{k=1}^n\frac{(-t)^{k+1}}{k}$$

Setting $t=1$ shows that the partial sums $s_n$ of the alternating harmonic series are given by

$$s_n=\ln2-\int_0^1\frac{x^n}{1+x}\,dx$$

But on $[0,1]$, we have $0\leq x^n(1+x)^{-1}\leq x^{n-1}$, so

$$0\leq\int_0^1\frac{x^n}{1+x}\,dx\leq\int_0^1x^{n-1}\,dx=\frac{1}{n}$$

Hence $s_n\to\ln 2$ as $n\to\infty$.

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