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In Principles of Mathematical Analysis Rudin states that:

Theorem 3.33: Given $\sum a_n$, put $\alpha = \lim_{n \to \infty}\sqrt[n]{\lvert a_n \rvert}$. Then $$$$ ... $$$$ (c) if $\alpha = 1$, the test gives no information

For proving (c) he states:

We consider the series $$\Sigma\frac{1}{n}, \Sigma\frac{1}{n^2}$$ For each of these series $\alpha = 1$, but the first diverges, the second converges.

However, he does not provide any evidence as to why $\alpha = 1$ for any one of those series. I've tried to prove it myself by copying the proof for $\lim_{n \to \infty} \sqrt[n]{n} = 1$, which is presented in his book but I had no success. Can anyone help?

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    $\begingroup$ Definition of $\alpha$ is wrong. $(\frac 1 n)^{\frac 1 n}=\frac 1 {n^{1/n}}$ and $(\frac 1 {n^{2}})^{\frac 1 n}=(\frac 1 {n^{1/n}})^{2}$ $\endgroup$ Commented Mar 16 at 11:40
  • $\begingroup$ @geetha290krm you are right it's a typo, I would correct it. $\endgroup$
    – john
    Commented Mar 16 at 11:42

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Say $a_n=\dfrac{1}{n}$. Then, $$\sqrt[n]{|a_n|}=\dfrac{1}{\sqrt[n]{n}}$$

As it is known that $$\displaystyle\lim_{n\to\infty}\sqrt[n]{n}=1\neq 0,$$

limit algebra tells us that $$\displaystyle\lim_{n\to\infty}\sqrt[n]{|a_n|}=\displaystyle\lim_{n\to\infty}\dfrac{1}{\sqrt[n]{n}}=\dfrac{1}{1}=1$$

Analogously, if $b_n=\dfrac{1}{n^2}$, we have $$\sqrt[n]{|b_n|}=\dfrac{1}{\sqrt[n]{n^2}}=\dfrac{1}{\sqrt[n]{n}}\cdot \dfrac{1}{\sqrt[n]{n}}$$

As we know that $$\displaystyle\lim_{n\to\infty}\dfrac{1}{\sqrt[n]{n}}=1,$$ it follows immediately that $$\displaystyle\lim_{n\to\infty}\sqrt[n]{|b_n|}=\displaystyle\lim_{n\to\infty}\dfrac{1}{\sqrt[n]{n^2}}=\displaystyle\lim_{n\to\infty}\dfrac{1}{\sqrt[n]{n}}\cdot \dfrac{1}{\sqrt[n]{n}}=1\cdot 1=1.$$

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  • $\begingroup$ I'm sure you are right as far as your calculations, however, this book does not have anything such as limit algebra. For example, it is not proven that $lim_{n \to \infty}{(a_n * b_n)} = (lim_{n \to \infty}{a_n}) * (lim_{n \to \infty}{b_n})$. $\endgroup$
    – john
    Commented Mar 16 at 12:13
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    $\begingroup$ @Hamed go to page 49 $\endgroup$ Commented Mar 16 at 12:16
  • $\begingroup$ you are right. Thanks. $\endgroup$
    – john
    Commented Mar 16 at 12:25

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