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Adventure with infinite series, a curiosity

Discussion in 'Mathematics' started by T. Amdeberhan, Oct 8, 2018.

  1. It is easily verifiable that $$\sum_{k\geq0}\binom{2k}k\frac1{2^{3k}}=\sqrt{2}.$$ It is not that difficult to get $$\sum_{k\geq0}\binom{4k}{2k}\frac1{2^{5k}}=\frac{\sqrt{2-\sqrt2}+\sqrt{2+\sqrt2}}2.$$


    Question. Is there something similarly "nice" in computing $$\sum_{k\geq0}\binom{8k}{4k}\frac1{2^{10k}}=?$$ Perhaps the same question about $$\sum_{k\geq0}\binom{16k}{8k}\frac1{2^{20k}}=?$$

    NOTE. The powers of $2$ are selected with a hope (suspicion) for some pattern.

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