Limit of an infinite Summation!

Let $f(a) = \left[ \displaystyle \sum_{n=0}^\infty \dfrac{(2n+1)a^2 + (-1)^n}{(2n+1)e^{(2n+1)a}} \right]$ .

Then $f(a) = a^2(e^{-a} + e^{-3a} + ^{-5a} + \ldots) + \left(e^{-a} -\dfrac13 e^{-3a} + \dfrac15 e^{-5a} - \ldots \right)$

Since $e^{-a} < 1$ , we can sum the series to get:

$\DeclareMathOperator{\csch}{csch} f(a) = a^2 \left( \dfrac{e^{-a}}{1-e^{-2a}} \right) + \tan^{-1} (e^{-a}) = \dfrac12 a^2 \csch (a) + \tan^{-1}(e^{-a})$

Then:

$\DeclareMathOperator{\csch}{csch} f'(a) = a \csch (a) - \dfrac12 a^2 \csch (a) \coth (a) - \dfrac{e^{-a}}{1+e^{-2a}}$ $\DeclareMathOperator{\sech}{sech} = a \coth (a) \sech (a) - \dfrac12 a^2 \sech (a) \coth^2 (a) - \dfrac12 \sech (a)$ $\DeclareMathOperator{\sech}{sech} = - \left[\dfrac12 \sech (a) \right] (a^2 \coth^2 (a) - 2a \coth (a) + 1) = - \left[\dfrac12 \sech (a) \right](a \coth (a) - 1)^2 < 0$

Hence $f$ is strictly decreasing on $(0, \infty)$ . Using L'Hopital Rule, we get:

$\displaystyle \lim_{a \to 0^+} f(a) = \lim_{a \to 0^+} \left( \dfrac{2a}{e^a + e^{-a}} \right) + \tan^{-1}(1) = \dfrac{\pi}{4}$

So, $A=1, B=4, C=1, A+B+C = \boxed{6}$ . Also $f(a) < \dfrac{\pi}{4}$ for $a>0$ .