which we will call $\Psi(x)$. Important properties (from that web page) are: $\Psi(x)$ is analytic in the complex plane except at the nonpositive integers where it has simple poles. $\Psi(x+1)-\Psi(x) = 1/x$. $\Psi(x) > 0$ for $x>2$. Asymptotics: $$\Psi(x) = \log x - \frac{1}{2x} - \frac{1}{12x^2} + \frac{1}{120x^4} + O(x^{-6}) \qquad\text{as } x \to \infty .$$ So, define $T(z) ={}$ $$-\sum_{k = 1}^{\infty} \Biggl[\Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) - z + 1\Biggr) + \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + z\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + 1\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr)\Biggr)\Biggr]$$ For any fixed $z$, only finitely many preliminary terms involve $\Psi$ evaluated at a nonpositive argument, and the asymptotics of the remaining terms are computed (from the asymptotics given above) as $$\Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) - z + 1\Biggr) + \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + z\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + 1\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr)\Biggr)$$
$=z(1-z)/(k^2\pi^2)$ =z(1-z)/(k^2\pi^2) + o(k^{-2})$as$k \to \infty$. So the series converges absolutely except when we are at a pole of one of the preliminary terms. Now, because of absolute convergence, we may subtract term-by-term and simplify to get $$T(z+1)-T(z) = \sum_{k=1}^\infty\Biggl[\frac{8z}{(-\pi+2\pi k-2z)(-\pi+2\pi k+2z)}\Biggr] = \tan z .$$ 3 added 2 characters in body I add more details for the solution in the distinguished answer due to Anixx. First, we need the digamma function http://en.wikipedia.org/wiki/Digamma_function which we will call$\Psi(x)$. Important properties (from that web page) are:$\Psi(x)$is analytic in the complex plane except at the nonpositive integers where it has simple poles.$\Psi(x+1)-\Psi(x) = 1/x$.$\Psi(x) > 0$for$x>2$. Asymptotics: $$\Psi(x) = \log x - \frac{1}{2x} - \frac{1}{12x^2} + \frac{1}{120x^4} + O(x^{-6}) \qquad\text{as } x \to \infty .$$ So, define$T(z) ={}$$$-\sum_{k = 1}^{\infty} \Biggl[\Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) - z + 1\Biggr) + \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + z\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + 1\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr)\Biggr)\Biggr]$$ For any fixed$z$, only finitely many preliminary terms involve$\Psi$evaluated at a nonpositive argument, and the asymptotics of the remaining terms are computed (from the asymptotics given above) as $$\Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) - z + 1\Biggr) + \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + z\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + 1\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr)\Biggr)$$$=z(1-z)/(k^2\pi^2)$as$k \to \infty$. So the series converges absolutely except when we are at a pole of one of the preliminary terms. Now, because of absolute convergence, we may subtract term-by-term and simplify to get $$T(z+1)-T(z) = \sum_{k=1}^\infty\Biggl[\frac{8z}{(\pi+2\pi k-2z)(\pi+2\pi sum_{k=1}^\infty\Biggl[\frac{8z}{(-\pi+2\pi k-2z)(-\pi+2\pi k+2z)}\Biggr] = \tan z .$$ 2 added 168 characters in body; deleted 14 characters in body; added 32 characters in body; deleted 1 characters in body; edited body I add more details for the solution in the distinguished answer due to Anixx. First, we need the digamma function http://en.wikipedia.org/wiki/Digamma_function which we will call$\Psi(x)$. Important properties (from that web page) are:$\Psi(x)$is analytic in the complex plane except at the nonpositive integers where it has simple poles.$\Psi(x+1)-\Psi(x) = 1/x$.$\Psi(x) > 0$for$x>2$. Asymptotics: $$\Psi(x) = \log x - \frac{1}{2x} - \frac{1}{12x^2} + \frac{1}{120x^4} + O(x^{-6}) \qquad\text{as } x \to \infty .$$ So, define$T(z) ={}$$$-\sum_{k = 1}^{\infty} \Biggl[\Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) - z + 1\Biggr) + \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + z\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + 1\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr)\Biggr)\Biggr]$$ For any fixed$z$, only finitely many preliminary terms involve$\Psi$evaluated at a nonpositive argument, and the asymptotics of the remaining terms are computed (from the asymptotics given above) as $$\Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) - z + 1\Biggr) + \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + z\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + 1\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr)\Biggr)$$$=z(1-z)/(k^2\pi^2)$as$k \to \infty\$. So the series converges absolutely except when we are at a pole of one of the finitely many preliminary terms. Now, because of absolute convergence, we may subtract term-by-term and simplify to get
$$T(z+1)-T(z) = \sum_{k=1}^\infty\Biggl[\frac{8z}{(\pi+2\pi k-2z)(\pi+2\pi k+2z)}\Biggr] = \tan z .$$