sum of series $\displaystyle \frac{8}{5}+\frac{16}{65}+\cdots \cdots +\frac{128}{2^{18}+1}$
I have calculate $\displaystyle a_{n} = \frac{2^{n+2}}{4^{2n-1}+1}$
could some help me with this, thanks
[Math] Sum of series $\frac85+\frac{16}{65}+\cdots \cdots +\cdots +\frac{128}{2^{18}+1}$
sequences-and-series
Related Solutions
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{{4 \over 10} + {4 \cdot 7 \over 10 \cdot 20} + {4 \cdot 7 \cdot 10 \over 10 \cdot 20 \cdot 30} + \cdots = \sum_{n = 1}^{\infty}{\prod_{k = 1}^{n}\pars{3k + 1} \over \prod_{k = 1}^{n}10k}:\ {\LARGE ?}}$.
\begin{align} &\bbox[10px,#ffd]{\ds{\sum_{n = 1}^{\infty} {\prod_{k = 1}^{n}\pars{3k + 1} \over \prod_{k = 1}^{n}10k}}} = \sum_{n = 1}^{\infty} {3^{n}\prod_{k = 1}^{n}\pars{k + 1/3} \over 10^{n}n!} = \sum_{n = 1}^{\infty} {\pars{3/10}^{n} \over n!}\,\pars{4 \over 3}^{\large\overline{n}} \\[5mm] = &\ \sum_{n = 1}^{\infty} {\pars{3/10}^{n} \over n!}\,{\Gamma\pars{4/3 + n} \over \Gamma\pars{4/3}} = \sum_{n = 1}^{\infty} \pars{3 \over 10}^{n}\,{\pars{n + 1/3}! \over n!\pars{1/3}!} \\[5mm] = &\ \sum_{n = 1}^{\infty}\pars{3 \over 10}^{n}\,{n + 1/3 \choose n} = \sum_{n = 1}^{\infty}\pars{3 \over 10}^{n} \bracks{{-4/3 \choose n}\pars{-1}^{n}} \\[5mm] = &\ \sum_{n = 1}^{\infty} {-4/3 \choose n}\pars{-\,{3 \over 10}}^{n} = \bracks{1 + \pars{-\,{3 \over 10}}}^{-4/3} - 1 \\[5mm] = &\ \bbx{\pars{10 \over 7}^{4/3} - 1} \approx 0.6089 \end{align}
That term in the denominator is also called the double factorial, namely: $$(2n+1)!!=\frac{(2n+1)!}{2^n n!}$$ And our series can be rewritten as: $$S=\sum_{n=0}^\infty \frac{1}{(2n+1)!!}$$ Now from this link see (21), is given that: $$\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!!}=\sqrt{\frac{\pi}{2}}\text{erf}\left(\frac{x}{\sqrt 2} \right) e^{\frac{x^2}{2}}$$ $$\Rightarrow S=\sqrt{\frac{e\pi}{2}}\text{erf}\left(\frac{1}{\sqrt 2}\right)$$ Where $\text{erf(z)}$ is the error function, defined as: $\displaystyle{\text{erf}(z)=\frac{2}{\sqrt{\pi}}\int_0^z e^{-t^2}dt}$
Best Answer
We can write $\displaystyle a_{n} = \frac{2^{n+2}}{4^{2n-1}+1} $ as $\displaystyle T_{n} = \frac{16.2^{n}}{2^{4n}+4} $
$\displaystyle T_{n} = \frac{4}{2^{2n}-2.2^{n}+2}-\frac{4}{2^{2n}+2.2^{n}+2} $
Adding upto n terms we get $\displaystyle S_{n} = 2-\frac{4}{2^{2n}+2.2^{n}+2} $ as it is telescopic
Solving n=5
$\displaystyle S_{5} = 2-\frac{4}{1090}=\frac{1088}{545} $