Inspired by Brian Charlesworth!

$\displaystyle \sum_{n=0}^{\infty}{x^n} = \frac{1}{1-x}$

$\displaystyle \sum_{n=0}^{\infty}{ x^{n-\frac{3}{4} } } = \frac{x^{-\frac{3}{4} } } {1-x}$

$\displaystyle \int{\sum_{n=0}^{\infty}{ x^{n-\frac{3}{4} } }dx} =\sum_{n=0}^{\infty}{\int{ x^{n-\frac{3}{4} }} dx}=\int{ \frac{x^{-\frac{3}{4} } } {1-x} dx}$

$\displaystyle \sum_{n=0}^{\infty}{ \frac { 1 }{ n+\frac { 1 }{ 4 } } { x }^{ n+\frac { 1 }{ 4 } }} =\int{ \frac{x^{-\frac{3}{4} } } {1-x} dx}$

$\displaystyle \sum_{n=0}^{\infty}{ \frac { 1 }{ n+\frac { 1 }{ 4 } } { x }^{ n}} =x^{-\frac{1}{4}}\int{ \frac{x^{-\frac{3}{4} } } {1-x} dx}\quad\quad \color{#3D99F6}{x=t^4}$

$\displaystyle \sum_{n=0}^{\infty}{ \frac { 1 }{ n+\frac { 1 }{ 4 } } { x }^{ n}} =4x^{-\frac{1}{4}}\int{ \frac{1} {1-t^4} dt}$

$\displaystyle \sum_{n=0}^{\infty}{ \frac { 1 }{ n+\frac { 1 }{ 4 } } { x }^{ n}} =4x^{-\frac{1}{4}}[\frac{1}{2}(\tan^{-1}{t}+\tanh^{-1}{t})]$

$\displaystyle \sum_{n=0}^{\infty}{ \frac { 1 }{ n+\frac { 1 }{ 4 } } { x }^{ n}} =4x^{-\frac{1}{4}}[\frac{1}{2}(\tan^{-1}{x^{\frac{1}{4}}}+\tanh^{-1}{x^{\frac{1}{4}}})]\quad\quad \color{#3D99F6}{x=\frac{1}{2}}$

$\displaystyle \sum_{n=0}^{\infty}{ \frac { 1 }{ 2^n(n+\frac { 1 }{ 4 }) } } =2^{\frac{5}{4}}[\tan^{-1}{\sqrt { \frac { 1 }{ \sqrt { 2 } } } }+\tanh^{-1}{\sqrt { \frac { 1 }{ \sqrt { 2 } } } }]$

Well, this uses two of the important taylor series.

$\sum_{n=0}^{\infty}{\frac{{x}^{2n+1}}{2n+1}} = arctanh(x)$

The above can be derived easily by adding other 2 series -

$\sum_{n=0}^{\infty}{\frac{{x}^{n}}{n}} = -ln(1 -x)$

$\sum_{n=0}^{\infty}{\frac{{(-1)}^{n}{x}^{n}}{n}} = ln(1 + x)$

And we know that $ln(1+x) - ln(1-x) = 2arctanh(x)$ [If anyone wants its derivation , you can ask in the comment section below]

The second sum will be- $\sum_{n=0}^{\infty}{\frac{{-1}^{n}{x}^{2n+1}}{2n+1}} = arctan(x)$ [This is taylor series of arctangent]

Adding the 1st and 2nd sums, we get -

$2\sum_{n=0}^{\infty}{\frac{{x}^{4n+1}}{4n+1}} = arctan(x) + arctanh(x)$

Substituting $x = {y}^{\frac{1}{4}}$ ,

$2\sum_{n=0}^{\infty}{\frac{{y}^{n}{y}^{\frac{1}{4}}}{4n+1}} = arctan({y}^{\frac{1}{4}}) + arctanh({y}^{\frac{1}{4}})$

$2{y}^{\frac{1}{4}}\sum_{n=0}^{\infty}{\frac{{y}^{n}}{4n+1}} = arctan({y}^{\frac{1}{4}}) + arctanh({y}^{\frac{1}{4}})$

$\frac{1}{2}{y}^{\frac{1}{4}}\sum_{n=0}^{\infty}{\frac{{y}^{n}}{n+\frac{1}{4}}} = arctan({y}^{\frac{1}{4}}) + arctanh({y}^{\frac{1}{4}})$

$\sum_{n=0}^{\infty}{\frac{{y}^{n}}{n+\frac{1}{4}}} = \frac{2}{{y}^{\frac{1}{4}}}(arctan({y}^{\frac{1}{4}}) + arctanh({y}^{\frac{1}{4}}))$

Now substituting $y = \frac{1}{2}$ completes our job.

$= {2}^{\frac{5}{4}}(arctan(\sqrt{\frac{1}{\sqrt{2}}}) + arctanh(\sqrt{\frac{1}{\sqrt{2}}}))$

And hence, the answer becomes -

$\frac{2 \times \frac{5}{4}}{\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}} = \boxed{5}$