My proof that this series sums to log 2, EDIT: Please see my attempt below at a proof to show the answer is $log2$
Let $$\alpha = \lim_{n \to \infty} \sum_{j=1}^n \frac {(-1)^{j+1}}{j}$$
Part (1): Show that $\alpha$ exists.
Part (2): Show that $\alpha = log2$.
My work:
$\alpha$ exists, by the alternating series test - simple enough.  But how can I get started on part(2)?
Proof:
Start with $log(1+x)$.  Rewrite it as 
$$log(1+x) = \int \frac{1}{1+x}dx$$
Should there be a constant, c, added to the L.H.S.?  Or, should I look to make the R.H.S. a definite integral instead?
Continuing along, I get that
$$\int \frac{1}{1+x}dx$$
$$= \int [\sum_{n=0}^{\infty} (-x)^n]dx$$
with the sum valid for $|x|<1$.
Inside the interval of convergence of this sum, i.e., inside (-1,1), we have uniform convergence, and so we can integrate term-by-term, getting 
$$ \sum_{n=0}^{\infty} \int (-x)^ndx$$
$$= \sum_{n=0}^{\infty} \frac {(-x)^{n+1}}{n+1}$$
same question: do I add a constant to the result of the indefinite integration?
Now relabeling indices gives
$$ \sum_{n=1}^{\infty} \frac {(-x)^{n}}{n}$$
$$ = \sum_{n=1}^{\infty} \frac {(-1)^{n+1}x^{n}}{n}$$
but we started with $log(1+x)$, so this shows that we have the equation:
$$ log(1+x) = \sum_{n=1}^{\infty} \frac {(-1)^{n+1}x^{n}}{n}$$
and this implies that 
$$ log(1+1) = log(2) = \sum_{n=1}^{\infty} \frac {(-1)^{n+1}1^{n}}{n}$$
$$ \ => \alpha = log(2) = \sum_{n=1}^{\infty} \frac {(-1)^{n+1}}{n}$$
as required.
What do you think?  Am I close?  
My concerns are the indefinite integration that I used a couple of times.
Also, I completely neglected the interval (and boundary) of convergence for $log(1+x)$, and just plugged in $x=1$, since I arrived at an equation. Is this ok to do?
Any hints or comments are welcome.
Thanks,
 A: You can prove that for all $x\in(-1,1)$, $$\log(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^k$$ and use Abel theorem.
A: You need to use Abel partial summation theorem to prove that
$$
\log(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n
$$
is uniform convergent on $[0,1]$ and thus continuous on on $[0,1]$. Hence
$$
\lim_{x\to1}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=\log{2}
$$
Let $B_n=\sum_{k=m}^n \frac{(-1)^{k+1}}{k}$. Since $\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}$ converges, by Cauchy Criterion, for any $\epsilon>0$, there is a $N$ such that 
$$
|B_n|<\epsilon \quad\text{whenever }\quad n,m>N\tag1
$$
We have
\begin{align}
\sum_{k=m}^n \frac{(-1)^{k+1}}{k}x^k&=\sum_{k=m}^n (B_k-B_{k-1})x^k
\\
&=\sum_{k=m}^n B_kx^k -\sum_{k=m}^n B_{k-1}x^k
\\
&=\sum_{k=m}^{n-1} B_k(x^k-x^{k+1})+B_nx^n\tag{2}
\end{align}
Note: $B_{m−1}=0$.
Since $\lim_{n\to\infty}x^n$ exists and $\:x^n \downarrow$ on $[0,1]$,  for all $k>0$ and $x\in[0,1]$ we have
$$
x^k-x^{k+1}\geqslant0\:
$$
Since for all $k>m$, $-\epsilon<B_k<\epsilon$ 
$$
|B_k(x^k-x^{k+1})|<\epsilon(x^k-x^{k+1})\tag3
$$
So for all $n,m>N-1$ and $x\in[0,1]$, by $(1)$, $(2)$ and $(3)$ there is
\begin{align}
\left|\sum_{k=m}^n \frac{(-1)^{k+1}}{k}x^k\right|&\leqslant\sum_{k=m}^{n-1} |B_k(x^k-x^{k+1})|+|B_nx^n|
\\
&\leqslant\sum_{k=m}^{n-1} \epsilon\:(x^k-x^{k+1})+\epsilon \:x^n
\\
&=\epsilon \:(x^m-x^n+x^n)
\\
&=\epsilon \:x^m
\\
&\leqslant \epsilon
\end{align}
So by Cauchy Criterion, $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}x^k$ converges uniformly on $[0,1]$.
