# Evidence for Theorem 3.33 part (c) Principles of Mathematical Analysis (Baby Rudin)

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?

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

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 $$\displaystyle\lim_{n\to\infty}\sqrt[n]{|b_n|}=\displaystyle\lim_{n\to\infty}\dfrac{1}{\sqrt[n]{n^2}}=\displaystyle\lim_{n\to\infty}\dfrac{1}{\sqrt[n]{n}}\cdot \dfrac{1}{\sqrt[n]{n}}=1\cdot 1=1.$$

• 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 $lim_{n \to \infty}{(a_n * b_n)} = (lim_{n \to \infty}{a_n}) * (lim_{n \to \infty}{b_n})$.
– john
Commented Mar 16 at 12:13
• @Hamed go to page 49 Commented Mar 16 at 12:16
• you are right. Thanks.
– john
Commented Mar 16 at 12:25