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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!

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  • $\begingroup$ You say that proving convergence is fairly easy. However, when $0 \leq x <1$, the series diverges. Perhaps you are looking for the limit $\lim_{x \to 1^+} S_x$? $\endgroup$
    – Jonah
    Feb 7, 2022 at 19:59
  • $\begingroup$ @Jonah I believe you've made a mistake. The series converges between 0,1. As I have given a full proof in my post, series stays within the range 0,0.5. $\endgroup$
    – Aryan
    Feb 7, 2022 at 20:11
  • $\begingroup$ The general term does not tend to 0, therefor the series diverges. It is bounded, yes, but it does not converge. $\endgroup$
    – Jonah
    Feb 7, 2022 at 20:13
  • $\begingroup$ The latter series included in my solution is increasing and bounded from above, hence it converges! $\endgroup$
    – Aryan
    Feb 7, 2022 at 20:19
  • $\begingroup$ @jonah I think I got where you misunderstood, I fixed the problem so that nobody gets confused $\endgroup$
    – Aryan
    Feb 7, 2022 at 20:33

1 Answer 1

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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

$$ \lim_{x \to 1^-} \sum_{n=1}^{\infty} (-1)^{n-1} f(x^n) = \frac{f(1) - f(0)}{2}. $$

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  • $\begingroup$ Beatiful solution. Congratulations! $\endgroup$ Feb 8, 2022 at 8:29

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