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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..

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  • $\begingroup$ The way to do it is to apply the test. You insert the sequence into the test, find the quantity needed, check if it is less than, greater than or equal to one and conclude. If you are stuck while applying the test, then post whatever you have done, because when it is so clear that you should apply the test, then I am sure you would have already done so. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Dec 7 '16 at 8:01
  • $\begingroup$ @астонвіллаолофмэллбэрг I did try appying the test, separately for when n is odd and for when n is even. Was I correct in doing the test twice, one for n = odd, and one for n = even? $\endgroup$
    – gticecream8
    Dec 7 '16 at 8:07
  • $\begingroup$ Yes, you were right. The answer below will give you a better clue, though. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Dec 7 '16 at 8:12
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    $\begingroup$ Also, it is not right to use the word "harmonic" simply because you like the sound of it. $\endgroup$
    – Ivan Neretin
    Dec 7 '16 at 9:20
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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...

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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: $$= \frac{1}{1-\frac{1}{2}} + \frac{1}{4} \frac{1}{(1-\frac{1}{4})^2} = 2 + \frac{1}{4} \frac{16}{9} = \frac{22}{9}$$

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