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I working to prove the following statement:

If $\lim \limits_{k \to \infty} \lvert \frac{a_k}{a_{k+1}}\rvert$ exists then the power series $\sum_{k=0}^\infty a_kx^k$ has the convergence radius $$R= \lim \limits_{k \to \infty} \lvert \frac{a_k}{a_{k+1}}\rvert$$

So the proof in my book is making me really confused. I understand that it starts by using the ratio test:

$$y=\lim \limits_{k \to \infty}\lvert\frac{a_{k+1}x^{k+1}}{a_{k}x^k}\rvert = \frac{\rvert x \lvert}{1}\lim \limits_{k \to \infty}\lvert\frac{a_{k+1}}{a_{k}}\rvert = \frac{\lvert x\rvert}{R}$$

I understand that I can use the ratio test in order to see where the series is converging. So the text book is plugging in the power series in the ratio test. I see that $\lvert \frac{x^{k+1}}{x^k}\rvert=\lvert x \rvert$ but what I don't see is how: $$\lim \limits_{k \to \infty}\lvert\frac{a_{k+1}}{a_{k}}\rvert = \frac{1}{R}$$

Then it states that if $\frac{\lvert x \lvert}{R}<1$ the series is absolutely convergent and if $\frac{\lvert x \lvert}{R}>1$ the series is divergent.

Any clues on why the limit equals $\frac{1}{R}$ would be greatly appreciated.

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  • $\begingroup$ Note: $\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right|=\frac{1}{\lim_\limits{k\to\infty} \left|\frac{a_{k}}{a_{k+1}}\right|}=\frac 1R.$ $\endgroup$
    – farruhota
    Commented May 12, 2018 at 9:28

2 Answers 2

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$$R= \lim \limits_{k \to \infty} \lvert \frac{a_k}{a_{k+1}}\rvert \implies $$

$$ \lim \limits_{k \to \infty} \lvert \frac{a_{k+1}}{a_k}\rvert = \frac {1}{R}$$

The rest is just applying the ratio test.

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The limit is $R$ because the author of your textbook defined $R$ as$$\lim_{k\to\infty}\left|\frac{a_k}{a_{k+1}}\right|.$$

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