I am trying to show that if the power series $\sum (a_nx^n)$ coverges to a function f for $|x|<r$, then $$\int_{0}^{r} f(x)dx=\sum_{n=0}^{\infty} \frac{a_n}{n+1} r^{n+1}$$ provided that the series on the right side is convergent. the book tells us to apply Abel's theorem, but I have no idea how to deal with the integral. Can someone help me? Thanks in advance.
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I've never worked with Abel's theorem before, but I'll give it a try.
If the power series $$f(x) = \sum_{i=0}^{\infty} a_n x^n$$ converges for $\lvert x\rvert < r$, then the formal power series integral of $f(x)$ is $$F(x) = \sum_{i = 0}^{\infty} \frac{a_n}{n + 1} x^{n + 1} + C$$ for some $C$. You said that $F(x)$ converges at $x = r$, so then by Abel's theorem we can say that $$\lim_{t \to 1^-} F(tr) = F(r).$$ Then $$ \begin{align*}\int_0^r f(x) dx &= \lim_{t \to 1^-} F(r) - F(0) \\ &= F(r) - F(0) \\ &= \sum_{i=0}^{\infty} \frac{a_n}{n+1} x^{n+1}. \end{align*} $$
feralin
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If the power series converges for $|x| < r$ then it converges uniformly on any compact subinterval $[0,r - \epsilon]$.
This justifies switching the sum and integral as
$$\int_0^{r - \epsilon}f(x)\, dx = \sum_{n=0}^\infty \int_0^{r - \epsilon}a_nx^n\, dx = \sum_{n=0}^\infty \frac{a_n}{n+1}(r - \epsilon)^{n+1}.$$
Now apply Abel's theorem, taking limits as $\epsilon \to 0$.
RRL
- 90,707