How can I prove that if the coefficients $\{a_k\}$ of the power series $\sum_{0}^{\infty} \{a_k\}x^k$ form a bounded sequence, then the radius of convergence is at least 1?
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3Did you try the formula for radius of convergence? – Yuval Filmus May 16 '11 at 18:10
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Hint: If the sequence $a_k$ is bounded, then there exists $M$ such that $|a_k|\leq M$ for all $k$.
Then, what can you say about $|\sum_{k=1}^\infty a_k x^k|$ and $\sum_{k=1}^\infty Mx^k$? What is the radius of convergence of the second one?
Yuval Filmus
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Eric Naslund
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If $|x| \lt 1$ then $\frac{1}{1-|x|} = \sum_{n = 0}^{\infty} |x|^{n}$ by the summation formula for the geometric series. Now use this, the triangle inequality and the assumption that $|a_{n}| \leq C$ for all $n$ to estimate your given series from above, hence it series converges absolutely for $|x| \lt 1$.
Alternatively, you can easily show that the radius of convergence $\rho^{-1} = \limsup_{n \to \infty} \sqrt[n]{|a_n|}$ satisfies $\rho^{-1} \leq 1$, since $\sqrt[n]{C} \; \xrightarrow{n \to \infty}\; 1$ for all $C \gt 0$. If you look at the proof of this formula for the radius of convergence (usually called the Cauchy-Hadamard theorem), you'll see that this essentially comes down to the same as the first paragraph: a comparison with a geometric series which is known to converge.
t.b.
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small question: so if the series converges absolutely -1<x<1, then shouldn't the radius of convergence just be 1? how to show that the radius is at least 1? thanks! – ninja May 16 '11 at 18:38
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@ninja: Recall the definition of the radius of convergence $\rho$. It it is the largest $\rho$ such that for all $|x| \lt \rho$ the series $\sum_{n=0}^{\infty} a_{n} x^n$ converges. So if I know that the series converges for $|x| \lt 1$, then I know that $\rho \geq 1$ by the definition of the radius of convergence. – t.b. May 16 '11 at 18:46
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1@ninja: In other words: I indicated how to show that the radius of convergence is at least one by showing that the power series converges for all $|x| \lt 1$. The radius of convergence could still be greater than one. I said nothing about $|x| \geq 1$, and anything could happen. If $a_{n} = 1$ for all $n$ then the radius of convergence is precisely one (the geometric series diverges for $x = 1$), and if $a_{n} = 0$ for $n$ large enough (or $a_{n} = \frac{1}{n!}$), for example, then the radius of convergence is infinite. – t.b. May 16 '11 at 18:48
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