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Find the radius of convergence of the following power series $$\sum_{n=1}^{\infty} \frac{2^n + 1}{n} x^n.$$

Using the ratio test, I have found that the radius of convergence is $R = \frac{1}{2}$. I wasn't able to find this using the root test however. Is there a clever way of finding this with the root test?

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    $\begingroup$ If you want to be formal about the $(2^n+1)^{1/n}$, note that $2^n\lt 2^n+1\le 2\cdot 2^n$. Take the $n$-th root and Squeeze. $\endgroup$ – André Nicolas Jul 23 '14 at 17:04
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$$\lim_{n \to \infty}\sqrt[n]{\frac{2^n+1}{n}} = \lim_{n \to \infty}\sqrt[n]{\frac{2^n}{n}} =\lim_{n \to \infty}\frac{2}{ \sqrt[n]{n}}=2$$

Why can we omit the $+1$ term? Good question. ;)

Write $$2^n+1 = 2^n(1+\frac{1}{2^n})$$

As $n \to \infty$, the limit of both sides becomes equal. Therefore $$\lim_{n \to \infty} 2^n+1 = \lim_{n \to \infty} 2^n(1+\frac{1}{2^n})=\lim_{n \to \infty} 2^n\cdot \lim_{n\to \infty}(1+\frac{1}{2^n})=\lim_{n \to \infty} 2^n \cdot 1= \lim_{n \to \infty}2^n$$

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  • $\begingroup$ Ah ok, why is it that you can just omit the $+1$ ? $\endgroup$ – rehband Jul 23 '14 at 17:02
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    $\begingroup$ @rehband: Because you can factor $2^n$, the thing that is left between the parentheses goes to $1$ as $n$ goes to infinity. So, you can see that at $n \to \infty$ the two things become actually the same. It is a usual trick when you deal with limits at infinity. $\endgroup$ – math.n00b Jul 23 '14 at 17:06
  • $\begingroup$ Great, I'll remember that. Thanks a lot! $\endgroup$ – rehband Jul 23 '14 at 17:08
  • $\begingroup$ @rehband: You're welcome. $\endgroup$ – math.n00b Jul 23 '14 at 17:12

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