Limit symbol through power series "in the extreme of the convergente interval" I am solving an exercise that asks us to prove that if $0 < \alpha \leq 1$, then we can write
$$\ln(1+\alpha) = \sum\limits_{k=1}^{\infty}(-1)^{k+1}\dfrac{\alpha^k}{k}.$$ I could prove it for $\alpha < 1$ using the formula for the sum of the terms of a geometric progression. I am pretty sure I could simply pass the limit to show that it works also for $\alpha = 1$, but I don't know how to justify this - since $\alpha = 1$ would be in the extreme of the convergence interval.
I would appreciate any ideas, since this same problem appeared in other exercises as well.
 A: There are $2$ options :
$\bullet$ You prove the equality when $\alpha=1$ separately.
$\bullet$ You take the limit as $\alpha\rightarrow 1$, but it requires some justifications.
For the last option, you could use Abel theorem or, if you don't know this theorem, you can do it by hand. For $\alpha<1$,
$$ 0\leqslant \sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k}-\sum_{k=0}^{\infty}\frac{(-1)^{k+1}\alpha^k}{k} \leqslant \sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}-\sum_{k=0}^{n}\frac{(-1)^{k+1}\alpha^k}{k}+\sum_{k=n+1}^{+\infty}\frac{(-1)^{k+1}}{k} $$
Let $\varepsilon>0$, take $n$ such that $\displaystyle\sum_{k=n+1}^{+\infty}\frac{(-1)^{k+1}}{k}\leqslant\varepsilon$ (such a $n$ exists because the sum converges), now for this choice of $n$, we have
$$ \sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k}-\sum_{k=0}^{\infty}\frac{(-1)^{k+1}\alpha^k}{k} \leqslant \sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}-\sum_{k=0}^{n}\frac{(-1)^{k+1}\alpha^k}{k}+\varepsilon $$
Since $n$ is fixed, $\displaystyle\alpha\mapsto\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}-\sum_{k=0}^{n}\frac{(-1)^{k+1}\alpha^k}{k}$ is continuous and therefore there exists $\delta<1$ such that for all $\alpha\in(\delta,1)$,
$$ \sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}-\sum_{k=0}^{n}\frac{(-1)^{k+1}\alpha^k}{k}\leqslant\varepsilon $$
In the end,
$$ \forall\alpha\in(\delta,1),0\leqslant\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k}-\sum_{k=0}^{+\infty}\frac{(-1)^{k+1}\alpha^k}{k}\leqslant 2\varepsilon $$
which means that
$$ \lim\limits_{\alpha\rightarrow 1^-}\sum_{k=0}^{+\infty}\frac{(-1)^{k+1}\alpha^k}{k}=\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k} $$
Now using the expression of the sum you've found for $\alpha<1$, you get that
$$ \sum_{k=0}^{+\infty}\frac{(-1)^{k+1}}{k}=\lim\limits_{\alpha\rightarrow 1^-}\log(1+\alpha)=\log 2 $$
