The function $f(x) = 10x^2 \arctan(x^5)$ is represented as a power series $$\sum_{n=0}^\infty = c_n x^n $$
Q: What is the lowest term $n$ with a nonzero coefficient?
What I've tried so far:
$$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$ $$\frac{1}{1+x^2} = \sum_{n=0}^\infty (-x^2)^n= \sum_{n=0}^\infty (-1)^n x^{2n}$$ $$...$$ Integrate both sides, replace $x$ with $x^5$, then multiply by $10x^2$ to get $$f(x) = 10x^2 \arctan(x^5) = \sum_{n=0}^\infty 10\frac{(-1)^n} {2n+1}x^{10n+7}$$
Now, I'm assuming $c_n = 10\frac{(-1)^n} {2n+1}$, and is the coefficent of $x^n$. Then when $n = 0$, $c_n = 10$, which is nonzero. However, this is incorrect. Where did I go wrong?
$$\arctan(x) = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}x^{2k+1}$$
$$\arctan(x^5) = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}x^{10k+5}$$
$$10x^2\arctan(x^5) = \sum_{k=0}^\infty \frac{(-1)^k10}{2k+1}x^{10k+7}$$
The lowest term with a non-zero coefficient is $10x^7$ which is obtained when we substitute $k=0$.