power series radius of convergence 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?
 A: 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?
A: 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.
A: Use the comparison test on $\sum |a_kx^k|$.
