Find a power series expansion of the function $f(x)=\dfrac1{1+x^7}$ about $x=0$. Use it to evaluate $\int_0^1 \dfrac1{1+x^7} dx$. Your answer should be expressed in the form of numerical series (a series, but not a power series).
So I have this question here. I found the power series representation which is:
$$\sum_{n=0}^\infty (-1)^n x^{7n}$$
So I would be integrating:
$$\int_0^1 \sum_{n=0}^\infty (-1)^n x^{7n} \, dx$$
$$\sum_{n=0}^\infty (-1)^n \int_0^1 x^{7n} \, dx$$
Which I think is correct.
What does it mean by "You answer should be expressed in the form of a numerical series (a series, but not a power series)." though? Is there some closed form expression in which I have to represent the integral as?
Use the fact that the integral is a definite integral! $$\int_0^1{}x^{7n}dx = \frac{1}{7n+1}$$
Therefore the summation then equals $$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{7n+1}$$