<s>[INST] Limit of a series with interchanging signs Let $S_x=\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}(1-\frac{1}{x^n+1})$. The problem is to find the following limit: $\displaystyle\lim_{x\to1^{-}}S_x$.
I've tried to tackle the problem in several methods so far. Proving  that this series is convergent is fairly easy. For evaluating the limit, I've tried bounding the sum and found out that answer is between $0,\frac12$. Here is a sketch of the method:
$$\forall x<1 : S_x=\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}(1-\frac{1}{x^n+1})=\displaystyle\sum_{n=1}^{\infty}\bigg(\frac{1}{x^{2n}+1}-\frac{1}{x^{2n-1}+1}\bigg)$$$$=\displaystyle\sum_{n=1}^{\infty}\frac{x^{2n-1}-x^{2n}}{(x^{2n-1}+1)(x^{2n}+1)}=(1-x)\displaystyle\sum_{n=1}^{\infty}\frac{x^{2n-1}}{(x^{2n-1}+1)(x^{2n}+1)}$$
Hence, $S(x)$ is clearly positive. For the upper bound:
$$\frac{x^{2n-1}}{(x^{2n-1}+1)(x^{2n}+1)}\le x^{2n-1}$$
$$\implies \displaystyle\sum_{n=1}^{\infty}\frac{x^{2n-1}}{(x^{2n-1}+1)(x^{2n}+1)}\le\displaystyle\sum_{n=1}^{\infty}x^{2n-1}=1+x\cdot\displaystyle\sum_{n=1}^{\infty}x^{2n}$$$$=1+x\cdot(\frac{1}{1-x^2}-1)=(1-x)+\frac{x}{1-x^2}$$
$$\implies S_x\le(1-x)\displaystyle\sum_{n=1}^{\infty}\frac{x^{2n-1}}{(x^{2n-1}+1)(x^{2n}+1)}$$$$\le(1-x)\cdot((1-x)+\frac{x}{1-x^2})=(1-x)^2+\frac{x}{1+x}$$
Which approaches $\frac12$ as $x$ gets closer to $1$.
But clearly, this is not enough to evaluate the desired limit. I'm guessing (by the numerical tests) that the answer is zero. Either I shall take another approach, or the bound shall be refined. I couldn't take the problem any further from here.
Any help would be appreciated!
 [/INST] Write
$$ S_x = \frac{x}{1+x} \sum_{n=1}^{\infty} \underbrace{\frac{x^{2n-2} - x^{2n}}{(1+x^{2n-1})(1 + x^{2n})}}_{=T_{n,x}}. $$
Then we find that
$$ T_{n,x} \geq \frac{x^{2n-2} - x^{2n}}{(1+x^{2n-2})(1 + x^{2n})} = \frac{1}{1+x^{2n}} - \frac{1}{1+x^{2n-2}} $$
and
$$ T_{n,x} \leq \frac{x^{2n-2} - x^{2n}}{(1+x^{2n-1})(1 + x^{2n+1})} = \frac{1}{x}\left( \frac{1}{1+x^{2n+1}} - \frac{1}{1+x^{2n-1}} \right). $$
Summing these for $n = 1, 2, \ldots$, both bounds telescope and hence we obtain
$$ \frac{x}{1+x} \left( 1 - \frac{1}{2} \right) \leq S_x \leq \frac{1}{1+x} \left(1 - \frac{1}{1+x} \right).$$
Therefore, by the squeezing lemma,
$$ \lim_{x \to 1^-} S_x = \frac{1}{4}. $$

More generally, for any $C^1$-function $f$ on $[0, 1]$, we get
$$ \lim_{x \to 1^-} \sum_{n=1}^{\infty} (-1)^{n-1} f(x^n) = \frac{f(1) - f(0)}{2}. $$
</s>