Calculating the sum of $\sum_{k=1}^{\infty}{\frac{(-1)^k}{k}}$ I am trying to find the sum of
$$\sum_{k=1}^{\infty}{\frac{(-1)^k}{k}}$$
I've proven that this converges using the Leibniz test, since
$a_n > 0$ and $\lim_{n\to\infty}{a_n} = 0$.
I am not sure how to go about summing this series up though. Every example I've seen up to now does some manipulation so that a geometric series pops out. I've been at it for a bit and I don't see how I could convert this to a geometric series to sum it up.
 A: Using the Taylor series for the natural logarithm,
$$
\ln(x+1)=\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}{n}
$$
for $-1<x\le 1$. Abel's theorem guarantees that
$$
\ln2=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}.
$$
Hence,
$$\sum_{n=1}^\infty\frac{(-1)^{n}}{n}=-\ln2.
$$
A: Since an alternating sum is the difference of the non-alternating sum and twice the sum of the even terms, we get
$$
\begin{align}
\sum_{k=1}^{2n}\frac{(-1)^{k-1}}k
&=\sum_{k=1}^{2n}\frac1k-2\sum_{k=1}^n\frac1{2k}\\
&=\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\\
&=\sum_{k=n+1}^{2n}\frac1k\tag{1}
\end{align}
$$
By comparison to an integral, we have
$$
\log\left(\frac{2n+1}{n+1}\right)=\int_{n+1}^{2n+1}\frac1x\,\mathrm{d}x\le\sum_{k=n+1}^{2n}\frac1k\le\int_n^{2n}\frac1x\,\mathrm{d}x=\log(2)\tag{2}
$$
Apply the Squeeze Theorem to $(2)$ and incorporate $(1)$ and we get
$$
\sum_{k=1}^\infty\frac{(-1)^{k-1}}k=\log(2)\tag{3}
$$
A: Hint:
The geometric series is very very close.
Replace $-1$ by $x$ and derive term-wise:
$$\left(\sum_{k=1}^\infty\frac{x^k}k\right)'=\sum_{k=1}^\infty\left(\frac{x^k}k\right)'=\sum_{k=1}^\infty x^{k-1}.$$
Now you can use the geometric series formula and integrate from $0$ to $-1$ to get the solution
$$-\left.\ln(1-x)\right|_0^{-1}.$$
A: You can find the sum using only basic calculus -- without invoking termwise integration or differentiation of infinite Taylor series or Abel's theorem. 
Begin with the finite geometric sum
$$\sum_{k=0}^{n-1} (-x)^k = \frac{1 - (-x)^n}{1+x}.$$
Rearranging, we have
$$\sum_{k=0}^{n-1} (-1)^kx^k - \frac{1}{1+x}= \frac{(-1)^nx^n}{1+x}.$$
Integrate both sides over $[0,1]$, noting that termwise integration of the finite sum requires no special justification,
$$\sum_{k=0}^{n-1} (-1)^k\int_0^1x^k \, dx - \int_0^1\frac{1}{1+x} \, dx= \int_0^1\frac{(-1)^nx^n}{1+x} \, dx.$$
Hence,
$$\sum_{k=0}^{n-1} \frac{(-1)^k}{k+1} - \ln 2= (-1)^n\int_0^1\frac{x^n}{1+x} \, dx,$$
and
$$0 \leqslant \left|\sum_{k=1}^{n} \frac{(-1)^{k+1}}{k} - \ln 2 \right| = \left|\int_0^1\frac{x^n}{1+x} \, dx\right| \\ \leqslant \int_0^1\left|\frac{x^n}{1+x}\right| \, dx \leqslant \int_0^1 x^n \, dx = \frac{1}{n+1}.$$
By the squeeze theorem, it follows that
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = \ln 2.$$
