Sum of Gaps 3

We have $\sum_{k=1}^n (-1)^{k+1}x^{k-1} \; = \; \frac{1 - (-x)^n}{1+x} \hspace{2cm} 0 < x < 1$ and hence $\sum_{k=1}^n \frac{(-1)^{k+1}}{k} \; = \; \int_0^1 \frac{1 - (-x)^n}{1+x}\,dx \; = \; \ln2 - \int_0^1 \frac{(-x)^n}{1+x}\,dx$ so that $S_N \; = \; \sum_{n=1}^N \left(\sum_{k=1}^n \frac{(-1)^{k+1}}{k} - \ln2\right) \; = \; -\sum_{n=1}^N \int_0^1 \frac{(-x)^n}{1+x}\,dx \; = \; \int_0^1 \frac{x(1 - (-x)^N)}{(1+x)^2}\,dx$ for all integers $N \ge 1$ . Thus the infinite sum is $S \; = \; \lim_{N \to \infty} \int_0^1 \frac{x(1 - (-x)^N)}{(1+x)^2}\,dx \; = \; \int_0^1 \frac{x}{(1+x)^2}\,dx \; = \; \ln2 - \tfrac12$ making the answer $\lfloor 10000(\ln2 - \tfrac12)\rfloor = \boxed{1931}$ .

Amazing problem! I'll be proving a more general case, but I won't be bothering with rigor (e.g. issue of convergence etc):

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Lemma: Let $\displaystyle A = \sum_{n=0}^{\infty} a_n$ and $\displaystyle f(x) = \sum_{n=0}^{\infty} a_n x^n$ . Then

$\displaystyle \sum_{n=0}^{\infty} \left( A - \sum_{k=0}^{n} a_k \right) = \lim _{ x \rightarrow 1 }{ \frac{A - f(x)}{1-x} }$

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Proof:

$\begin{aligned} \sum_{n=0}^{\infty}x^n\sum_{k=0}^n a_k &= \sum_{n=0}^{\infty}\sum_{k=0}^n x^n a_k \\ &= \sum_{0\le k\le n\le \infty} x^n a_k \quad \quad \text{Substitute } n=k+j \\ &= \sum_{0\le k\le k+j\le \infty} x^n a_k \\ &= \sum_{k=0}^{\infty}\sum_{j=0}^{\infty} x^{k+j} a_k \\ &= \left(\sum_{k=0}^{\infty}x^k a_k \right) \left(\sum_{j=0}^{\infty} x^j\right) \\ &= \frac{f(x)}{1-x} \end{aligned}$

$\begin{aligned} \sum_{n=0}^{\infty} x^n \left( A - \sum_{k=0}^{n} a_k \right) &= \frac{A}{1-x} - \frac{f(x)}{1-x} \\ &= \frac{A - f(x)}{1-x} \end{aligned}$ As such, without proving that the limit exists:

$\displaystyle \sum_{n=0}^{\infty} \left( A - \sum_{k=0}^{n} a_k \right) = \lim _{ x \rightarrow 1 }{ \frac{A - f(x)}{1-x} }$

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The question above is simply a special case of this lemma.