Power Series and Functions — Question 8

Question 8 – Solution

Step 1: Expand the integrand.

For $|t|<1$, $$\frac1{1-t^4}=\sum_{n=0}^{\infty}t^{4n}.$$ On every closed subinterval inside $(-1,1)$, this power series converges uniformly, so it may be integrated term by term.

Step 2: Integrate.

$$F(x)=\sum_{n=0}^{\infty}\int_0^x t^{4n}\,dt=\boxed{\sum_{n=0}^{\infty}\frac{x^{4n+1}}{4n+1}}.$$

Step 3: Test the endpoints.

Integration preserves radius $R=1$. At $x=1$, the series is $\sum1/(4n+1)$, which diverges by Limit Comparison with $1/n$. At $x=-1$, it is the negative of the same divergent positive series.

Conclusion.

The interval is $\boxed{(-1,1)}$.