Number of terms to estimate the integral within the indicated accuracy using series! $$\int_{0}^{0.4}\sqrt{1+x^4} dx$$ with |Error| less than or equal to $$\frac{0.4^9}{72}$$
So far i have broken down the sqrt to $$(1+x^4)^{1/2}$$ then made it into the series $$\sum_{n=0}^\infty \binom{1/2}{n}x^{4n}$$ then integrated it to 
$$\sum_{n=0}^\infty \binom{1/2}{n}\frac{x^{4n+1}}{4n+1}$$
From here i plug in x as 0.4 and plug in n=0,1,2,3 ... etc to try and find when $$a_{n+1}$$ equals the error, however I dont seem to find it?  Did i use the right approach/ where is my mistake? Thank you in advance. 
Note: $$\binom{1/2}{n}$$ is the binomial coefficient
 A: You need to find a value $m$ for $n$ so that 
$$\sum_{n=m}^\infty \binom{1/2}{n}\frac{x^{4n+1}}{4n+1}\leq \frac{0.4^9}{72} $$
not just a single $a_n$.
A: Note:
I added a note at the end
to show that the series
is still decreasing
even with the
$4n+1$ included.
I will show that
the series is alternating
and decreasing in
absolute value,
so the standard alternating series result
can be used.
$b_k
=\binom{\frac12}{k}
=\frac{\prod_{i=0}^{k-1}(\frac12-i)}{k!}
$
so
$\begin{array}\\
\dfrac{b_{k+1}}{b_k}
&=\dfrac{\frac{\prod_{i=0}^{k}(\frac12-i)}{(k+1)!}}{\frac{\prod_{i=0}^{k-1}(\frac12-i)}{k!}}\\
&=\dfrac{\frac12-k}{k+1}\\
&=-\dfrac{2k-1}{2k+2}\\
&=-\dfrac{2k+2-3}{2k+2}\\
&=-(1-\dfrac{3}{2k+2})\\
\end{array}
$
so
$\binom{\frac12}{n}
$
is an alternating and decreasing series.
Since $0 < x < 1$,
so is your sum.
Therefore
the sum of the series
is always between 
two consecutive partial sums.
Just compute the series
until the last term used
is less than
the error you want.
(added later)
Since what the OP wants is
$\dfrac{b_k}{4k+1}
$,
the actual ratio is
$\begin{array}\\
\dfrac{2k-1}{2k+2}\dfrac{4k+5}{4k+1}
&=\dfrac{8k^2+6k-5}{ 8k^2+10k+2}\\
&=\dfrac{8k^2+10k-4k+2-7}{ 8k^2+10k+2}\\
&=1-\dfrac{4k+7}{ 8k^2+10k+2}\\
\end{array}
$
which is still less than one.
