Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ 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?
 A: 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.
A: A proof without words by Matt Hudleson 

A: 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 $x \to 1$.
A: $\sum_{k=1}^{n} ( \frac{1}{2k-1}-\frac{1}{2k} ) = \sum_{k=1}^{n} ( \frac{1}{2k-1}+\frac{1}{2k} ) - 2 \sum_{k=1}^{n} \frac{1}{2k} = \sum_{k=1}^{2n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k} = \sum_{k=n+1}^{2n} \frac{1}{k}$.
$\ln(2) \overset{n\to\infty}{\leftarrow} \ln(2) + \ln(\frac{2n+1}{2n+2}) = \ln(2n+1)-\ln(n+1)$
$= \int_{n+1}^{2n+1} \frac{1}{x}\ dx \le \sum_{k=n+1}^{2n} \frac{1}{k} \le \int_{n}^{2n} \frac{1}{x}\ dx$
$= \ln(2n)-\ln(n) = \ln(2)$.
So by squeeze theorem we are done.
A: 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 $\gamma-\gamma+\ln(2)$. 
A: 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. 
A: 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,
$$
\begin{align}
\lim_{n\to\infty}\left(1-\frac12+\frac13-\frac14+\dots-\frac1{2n}\right)
&=\lim_{n\to\infty}\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)\\[6pt]
&=\log(2)\tag{3}
\end{align}
$$
A: Here is another proof, based on the formula
$$\frac{1}{1+x}=\frac{(-1)^nx^n}{1+x}+\sum_{k=0}^{n-1}(-1)^kx^k.$$
Integrating both sides over $[0,t]$ gives
$$\ln(1+t)=\int_0^t\frac{(-1)^nx^n}{1+x}\,dx+\sum_{k=1}^n(-1)^{k+1}\frac{t^k}{k}.$$
Setting $t=1$ shows that the partial sums $s_n$ of the alternating harmonic series are given by
$$s_n=\ln2+(-1)^n\int_0^1\frac{x^n}{1+x}\,dx.$$
But on $[0,1]$, we have $0\leq x^n(1+x)^{-1}\leq x^{n}$, so
$$0\leq\int_0^1\frac{x^n}{1+x}\,dx\leq\int_0^1x^{n}\,dx=\frac{1}{n+1}.$$
Hence $$\ln2-\frac{1}{n+1}\leq s_n\leq\ln2+\frac{1}{n+1}.$$ So $s_n\to\ln 2$ as $n\to\infty$.
A: I want to use the infinite series expansion of the integral of a function to compute the sum. If you repeat the process of integration by parts over and over again:
$$\int f(x) dx = xf(x)-\int xf'(x)dx = xf(x) - \frac{x^2}{2}f'(x)+\int \frac{x^2}{2}f''(x)dx$$
$$\Rightarrow \int f(x) dx = xf(x) - \frac{x^2}{2}f'(x) + \frac{x^3}{6}f''(x)-\int \frac{x^3}{6}f''(x)dx$$
$$\Rightarrow \int f(x) dx = xf(x) - \frac{x^2}{2}f'(x) + \frac{x^3}{6}f''(x)-\frac{x^4}{24}f'''(x)+ \int \frac{x^4}{24}f'''(x)dx$$
Continuing this pattern we can prove using Mathematical Induction (for all $n\ \epsilon\ \mathbb{Z}^+$)that:
$$\int f(x)dx = \sum_{k=0}^{n}\left[\frac{(-1)^kf^{(k)}(x)x^{k+1}}{(k+1)!}\right]+\frac{(-1)^n}{n!}\int x^n f^{(n)}(x)dx \rightarrow (1)$$
Limiting both sides of equation (1) as $n \rightarrow \infty$ we can write the infinite series expansion:
$$\int f(x)dx = \sum_{k=0}^{\infty}\left[\frac{(-1)^kf^{(k)}(x)x^{k+1}}{(k+1)!}\right] + C=C+xf(x) - \frac{x^2}{2}f'(x) + \frac{x^3}{6}f''(x)-+...\rightarrow (2)$$
We can certainly remove the integration constant $C$ from both sides by taking the definite integral:
$$\int_{0}^{x} f(t)dt =xf(x) - \frac{x^2}{2}f'(x) + \frac{x^3}{6}f''(x)-\frac{x^4}{24}f'''(x)+-...\rightarrow (3)$$
We can prove the Taylor series expansion for functions using this idea. Now let $f(x)=\frac{1}{1+x}$ and we get:
$$\ln(1+x)=\frac{1}{1+\frac{1}{x}}+\frac{1/2}{(1+\frac{1}{x})^2}+\frac{1/3}{(1+\frac{1}{x})^3}+\frac{1/4}{(1+\frac{1}{x})^4}+\frac{1/5}{(1+\frac{1}{x})^5}+...\rightarrow (4)$$
$$y=\frac{x}{x+1} \Rightarrow \ln (\frac{1}{1-y})=y+\frac{y^2}{2}+\frac{y^3}{3}+\frac{y^4}{4}+\frac{y^5}{5}+...\rightarrow (5)$$
By letting $y=-1$ and multiplying both sides of equation $(5)$ by $-1$ we get the sum we desired:
$$\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5}-\frac{1}{6}+-...$$
P.S. You can also use $(4)$ to get another interesting expansion for $\ln 2$ by putting $x=1$:
$$\ln 2 = \frac{1}{1 \cdot 2^1}+\frac{1}{2 \cdot 2^2}+\frac{1}{3 \cdot 2^3}+\frac{1}{4 \cdot 2^4}+\frac{1}{5 \cdot 2^5}+...$$
A: Grant Sanderson, aka 3Blue1Brown, has a good explanation of this in one of his Lockdown Math videos. His explanation, summarized:

*

*$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots = f(1)$, where $f(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} -\ldots$

*$\frac{df}{dx} = 1 - x + x^2- x^3 + x^4 - \ldots = \frac{1}{1+x}$ (when $-1 ≤ x ≤ 1$)

*$f(x) = \ln(1+x)$

*$\therefore f(1) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots = \ln(2)$
A: Proving that this series converges can be done using the alternating series test: any series that alternates forever between positive and negative terms, where each term is smaller than the preceding term, and the terms approach a limit of 0 converges.
Why this test works:

*

*The odd partial sums decrease forever. Every odd partial sum (except the first) is less than the one before it: $S_{2n+1} = S_{2n-1} + \left(-\frac{1}{2n} + \frac{1}{2n+1} \right)$ (where $S_n$ is the $n$th partial sum) and $-\frac{1}{2n} + \frac{1}{2n+1} < 0$ for all $n$.

*Similarly, the even partial sums increase forever.

*Every odd partial sum is greater than the (even) partial sum before it, since all the terms at odd positions are positive.

*Similarly, every even partial sum is less than the (odd) partial sum before it.

*Since the odd partial sums decrease forever but never go below the first even partial sum $\left( 1 - \frac{1}{2} \right)$, they must approach a limit.

*Similarly, since the even partial sums increase forever but never go above the first term $\left( 1 \right)$, they must approach a limit.

*The odd partial sums and the even partial sums must approach the same limit, since the difference between each odd partial sum and the even partial sum before it approaches 0: $\lim\limits_{n \to \infty}\frac{1}{n} = 0$. Similarly, the difference between each even partial sum and the odd partial sum before it approaches 0.

*∴ the series converges.

A: Here is one proof, using how ${ \log (1+x) }$ is "sandwiched" between its Taylor polynomials on ${ [0,1] }.$
Consider ${ \log(1+x) }$ on ${ (-1, \infty) }.$ Its Taylor polynomials about $0$ are ${ S _n (x) = x - \frac{x ^2}{2} + \ldots + (-1) ^{n+1} \frac{x ^n}{n}. }$
Notice errors ${ \varepsilon _n (x) := \log(1+x) - \left( x - \frac{x ^2}{2} + \ldots + (-1) ^{n+1} \frac{x ^n}{n} \right) }$ satisfy ${ \varepsilon _n (0) = 0, }$ and ${ \varepsilon _n ' (x) }$ ${ = \frac{1}{1+x} -(1-x + \ldots + (-1) ^{n+1} x ^{n-1}) }$ ${ = \frac{1 - (1+ (-1) ^{n+1} x ^n)}{1+x} }$ ${ = \frac{ (-1) ^n x ^n}{1+x} }$ is $\geq 0$ for ${ \lbrace  x \in [0, \infty); n \text{ even} \rbrace }$ and $\leq 0$ for ${ \lbrace x \in [0, \infty); n \text{ odd} \rbrace }.$
So ${ \varepsilon _{n} (x) }$ is ${ \geq 0 }$ for ${ \lbrace x \in [0, \infty); n \text{ even} \rbrace }$ and ${ \leq 0 }$ for ${ \lbrace x \in [0, \infty); n \text{ odd} \rbrace }.$
So ${ S _{2n-1} (x) \geq \log(1+x) \geq S _{2n} (x) }$ for ${ x \in [0, \infty), n \in \mathbb{Z} _{\gt 0} }.$
Further ${ S _{2(n+1) -1} (x) - S _{2n-1} (x) }$ ${ = - \frac{x ^{2n}}{2n} + \frac{x ^{2n+1}}{2n+1} }$ ${ = x ^{2n} ( \frac{x}{2n+1} - \frac{1}{2n}) \leq 0}$ if ${ \underline{ x \in [0,1] } },$ ie sequence ${ (S _{2n-1} (x)) _{n \geq 1} }$ is decreasing if ${ x \in [0,1] }.$
Similarly ${ S _{2(n+1)} (x) - S _{2n} (x) }$ ${ = \frac{x ^{2n+1}}{2n+1} - \frac{x ^{2n+2}}{2n+2} }$ ${ = x ^{2n+1} (\frac{1}{2n+1} - \frac{x}{2n+2}) \geq 0 }$ if ${ \underline{ x \in [0,1] } },$ ie sequence ${ (S _{2n} (x) ) _{n \geq 1} }$ is increasing if ${ x \in [0,1] }.$
Fix ${ x \in [0,1] }.$
Now sequence ${ (S _{2n} (x) ) _{n \geq 1} }$ is increasing and ${ (S _{2n-1} (x) ) _{n \geq 1} }$ decreasing, with ${ S _{2n-1} (x) \geq \log(1+x) \geq S _{2n} (x) }.$
So say ${ \lim _{n \to \infty} S _{2n} (x) = \ell _e }$ and ${ \lim _{n \to \infty} S _{2n-1} (x) = \ell _o }.$ But ${ S _{2n-1} (x) - S _{2n} (x) = \frac{x ^{2n}}{2n} \leq \frac{1}{2n} \to 0 }$ as ${ n \to \infty }.$ So ${ \ell _o - \ell _e = 0 },$ ie ${ \ell _o = \ell _e }.$
This gives ${ \ell _o = \ell _e = \log(1+x), }$ making ${ \log(1+x) = \lim _{j \to \infty} S _j (x). }$

Especially ${ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} +  \ldots }$ converges, and to ${ \log(2) }.$

