Question

$\huge\displaystyle \sf\sum_{n = 0}^{ \infty } \frac{1}{(3n + 1) { \varphi}^{3n + 1} }$​

Answer 1

Answer:

Step-by-step explanation:

Note that,

                  $\int\limits^{\frac{1}{\varphi}}_0 {x^{3n}} \, dx = \frac{1}{(3n+1)\varphi^{3n+1}}$

Now,

        $\huge\displaystyle \sf \sum_{n = 0}^{ \infty } \frac{1}{(3n + 1) { \varphi}^{3n + 1} }\\\\ = \huge\displaystyle \sf \sum_{n = 0}^{ \infty } \int\limits^{\frac{1}{\varphi}}_0 {x^{3n}} \, dx \\\\= \huge\displaystyle \sf \int\limits^{\frac{1}{\varphi}}_0 \sum_{n = 0}^{ \infty } {x^{3n}} \, dx \\\\= \huge\displaystyle \sf \int\limits^{\frac{1}{\varphi}}_0 ( 1+x^3+x^6+x^9+\ldots)\,dx\\\\= \huge\displaystyle \sf \int\limits^{\frac{1}{\varphi}}_0 \frac{1}{1-x^3} \,dx$

       $= -\huge\displaystyle \sf \int\limits^{\frac{1}{\varphi}}_0 \dfrac{1}{x^3-1}\,\mathrm{d}x\\\\= - {\huge\displaystyle \sf \int\limits^{\frac{1}{\varphi}}_0}\dfrac{1}{\left(x-1\right)\left(x^2+x+1\right)}\,\mathrm{d}x\\\\=-{\huge\displaystyle \sf \int\limits^{\frac{1}{\varphi}}_0}\left(\dfrac{1}{3\left(x-1\right)}-\dfrac{x+2}{3\left(x^2+x+1\right)}\right)\mathrm{d}x$  

        $= {\huge \displaystyle \sf \int\limits_0^\frac{1}{\varphi} \bigg({\dfrac{1}{2}\bigg(\dfrac{2x+1}{x^2+x+1}\bigg) + {\dfrac{3}{2}}} \bigg(\dfrac{1}{x^2+x+1}\bigg)\,\bigg) - \frac{1}{3(x-1)}\bigg)dx$

       $= \bigg[\dfrac{\ln\left(x^2+x+1\right)}{6}+\dfrac{\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)}{\sqrt{3}}-\dfrac{\ln\left(\left|x-1\right|\right)}{3} \bigg]_0^{\frac{1}{\varphi}} \\\\=\dfrac{\ln\left(\frac{{\varphi}^2+{\varphi}+1}{{\varphi}^2}\right)}{6}+\dfrac{\tan^{-1}\left(\frac{\sqrt{3}{\varphi}+2\sqrt{3}}{3{\varphi}}\right)}{\sqrt{3}}-\dfrac{\ln\left(-\frac{{\varphi}-1}{{\varphi}}\right)}{3}-\dfrac{{\pi}}{2{\cdot}3^\frac{3}{2}}$