I came across the series axpansion of the arctangent function shown here: $$ \tag{1}\arctan(x)=\sum_{k=0}^\infty\frac{4^kx^{2k+1}}{(2k+1)\binom{2k}{k}\left(1+x^2\right)^{k+1}}. $$ Is it possible to rearrangenge the equation above as the series expansion $$ \tag{2}\arctan(x)=\frac{i}2\sum_{k=1}^\infty\frac{x^n}{n}\left(\frac{1}{(x+i)^n}-\frac{1}{(x-i)^n}\right) $$ that was derived in this paper? In other words, can we derive equation (2) directly from equation (1)?
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You should have $(2)$ sum over $k$ and $(2)$ can be split into logarithms using the Taylor series of $\ln(x)$ if it helps – Тyma Gaidash Jun 17 '23 at 23:55