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∑n=0∞120152n−2015−(2n)\displaystyle\sum_{n=0}^{\infty} \dfrac{1}{2015^{2^{n}} - 2015^{-(2^{n})}} n=0∑∞20152n−2015−(2n)1
If the closed form of the series above is in the form ab \frac a b ba, where aaa and bbb are positive coprime integers, then find b−a.b - a.b−a.
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