[Math] Estimating error in binomial series

Question

I am having trouble with estimating the error in this exercise:

Estimate the value of the following integral with error less than 0.0001
$$
\int_{0}^{0.2}\sqrt{1+x^3} \, dx
$$
So here im using binomial expansion as my teacher told me and i got this:
$$
\int_0^{0.2}\sqrt{1+x^3}=\sum_{k=0}^\infty \binom{0.5}{k}\frac{1}{(k+1)5^{k+1}}
$$
Now the problem is how do I calculate how many terms I need to get a good approximation?
I know the how to do this for alternating series and for series that i know the n+1 derivative and can use lagrange remainder, but in this case, i have no idea what to do. What can i do?

Answer 1

HINT:

The series is an alternating series since

$$\binom{1/2}{k}=\binom{2k}{k}\frac{(-1)^{k+1}}{4^k(2k-1)}$$

HINT 2:

The expansion is on $x^3$ and $\int_0^{0.2}x^{3n}\,dx=\frac{1}{(3n+1)5^{3n+1}}$

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SPOILER ALERT: Scroll over the highlighted area to reveal the solution

$$\begin{align}\int_0^{0.2}\sqrt{1+x^3}\,dx&=\int_0^{0.2}\sum_{n=0}^\infty \binom{2n}{n}\frac{(-1)^{n+1}}{4^n(2n-1)}x^{3n}\,dx\\\\&=\sum_{n=0}^\infty \binom{2n}{n}\frac{(-1)^{n+1}}{4^n(2n-1)}\frac{1}{(3n+1)5^{3n+1}}\end{align}$$Since the series alternates, we need only find the value of $n$ so that the absolute value of the $n$'th term of the series is less than 0.001. The result is $n=2$.