Show that a ratio test is inconclusive for a given series, then determine whether the series converges/diverges?

Question

4. Consider the series $\sum a_n$ where $$a_n=\begin{cases}n/2^n&n\text{ odd}\\1/2^n&n\text{ even}\end{cases}$$

• a. Show the Ratio Test is inconclusive
• b. Use the Root Test to determine whether the series is convergent or divergent.

Not sure how to do this..

Answer 1

Hints:

The ratio test gives

$$\frac{a_{n+1}}{a_n}=\begin{cases}\cfrac{\frac{n+1}{2^{n+1}}}{\frac1{2^n}}=\frac{n+1}2\,,&n\;\text{ is even}\\{}\\\cfrac{\frac1{2^{n+1}}}{\frac n{2^n}}=\frac1{2n}\,,&n\;\text{ is odd}\end{cases}$$

and thus the limit doesn't exist so...

For the $\;n\,-$ th root test:

$$\sqrt[n]{a_n}=\begin{cases}\cfrac{\sqrt[n]n}2\,,&n\;\text{ is odd}\\{}\\\;\;\;\cfrac12\,,&n\;\text{ is even}\end{cases}\;\xrightarrow[n\to\infty]{}\ldots$$

and thus the limit exists and...

Answer 2

The series is a combination of geometric series and Gabriel's Staircase: $$\sum_{k=0}^\infty a_k = \sum_{k=0}^\infty a_{2k}+\sum_{k=0}^\infty a_{2k+1} = \sum_{k=0}^\infty (\frac{1}{2})^{2k} + \sum_{k=0}^\infty (2k+1)(\frac{1}{2})^{2k+1}$$ We split again, rearrange and combine again: $$= \sum_{k=0}^\infty (\frac{1}{2})^{2k} + \sum_{k=0}^\infty k (\frac{1}{4})^{k} + \sum_{k=0}^\infty (\frac{1}{2})^{2k+1} = \sum_{k=0}^\infty (\frac{1}{2})^{k} + \sum_{k=0}^\infty k (\frac{1}{4})^{k}$$ Now both series converge and we get as the combined value: $$= \frac{1}{1-\frac{1}{2}} + \frac{1}{4} \frac{1}{(1-\frac{1}{4})^2} = 2 + \frac{1}{4} \frac{16}{9} = \frac{22}{9}$$