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Floor series

Let \left \lfloor \cdot \right \rfloor denote the floor function. Evaluate the series

    \[\mathcal{S} = \sum_{k=1}^{\infty} \frac{(-1)^{\left \lfloor \sqrt{k} + \sqrt{k+1} \right \rfloor}}{k(k+1)}\]


First of all we note that 4k+3 and 4k+2 are never squares. Thus, there exists a positive integer m such that

    \[m^2 \leq \sqrt{4k+1} < \sqrt{4k+2}< \sqrt{4k+3} < \left ( m+1 \right )^2\]

It is easy to see that \sqrt{4x+1} \leq\sqrt{x} + \sqrt{x+1} < \sqrt{4x+3} and thus we conclude that

    \[\left \lfloor \sqrt{k} + \sqrt{k+1} \right \rfloor = \left \lfloor 4k+1 \right \rfloor\]

Now \left \lfloor 4k+1 \right \rfloor is equal to the even number 2n if-f

    \[\left ( 2n \right )^2 \leq 4k+1 < \left ( 2n+1 \right )^2 \Leftrightarrow k \in \left [ n^2, n^2+n \right )\]

Hence, since the series is absolutely convergent we can rearrange the terms and by noting that the finite sums are telescopic , we get that:

    \begin{align*} \sum_{k=1}^{\infty} \frac{(-1)^{\left \lfloor \sqrt{k} + \sqrt{k+1} \right \rfloor}}{k(k+1)} &= \sum_{n=1}^{\infty} \bigg ( \sum_{k=n^2}^{n^2+n-1} \left ( \frac{1}{k} - \frac{1}{k+1} \right )- \\ &\quad \quad \quad  \quad \quad - \sum_{k=n^2+n}^{\left ( n+1 \right )^2-1}\left ( \frac{1}{k} - \frac{1}{k+1} \right ) \bigg )\\ &=\sum_{n=1}^{\infty} \left ( \frac{1}{n^2} -\frac{1}{n^2+n}-\frac{1}{n^2+n} +\frac{1}{(n+1)^2} \right ) \\ &=\sum_{n=1}^{\infty} \frac{1}{n^2} + \sum_{n=1}^{\infty} \frac{1}{(n+1)^2} - 2 \sum_{n=1}^{\infty} \frac{1}{n^2+n} \\ &=\frac{\pi^2}{6} + \frac{\pi^2}{6}-1 -2 \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} \right ) \\ &=\frac{\pi^2}{3} -1-2 \\ &= \frac{\pi^2}{3}-3 \end{align*}

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