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Given the series $$\frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \cdots,$$

I don't know how the author gets

$$\liminf_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \left(\frac{2}{3}\right)^n = 0$$

$$\liminf_{n\to\infty}a_n^{1/n} = \lim_{n\to\infty}\left(\frac{1}{3^n}\right)^{1/2n} = \frac{1}{\sqrt{3}}$$

$$\limsup_{n\to\infty}a_n^{1/n} = \lim_{n\to\infty}\left(\frac{1}{2^n}\right)^{1/2n} = \frac{1}{\sqrt{2}}$$

$$\limsup_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{1}{2}\left(\frac{3}{2}\right)^n = +\infty$$

And more generally, the book I'm using doesn't give a much precise definition on how to evaluate $\limsup$/$\liminf$.

Michael Burr
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AnalyticHarmony
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2 Answers2

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In general, if you have a sequence $\{b_1,b_2,\dots\}$, then $$ \limsup b_i $$ is defined to as follows: For each $i$, let $B_i=\{b_i,b_{i+1},\dots\}$. Then, $$ \limsup b_i=\lim_{i\rightarrow\infty}\left(\sup B_i\right). $$ In other words, you take the sup's of each of the tails of the sequence and take the limit of those sup's. In your case (for the first and last equalities), $b_i=\frac{a_{i+1}}{a_i}$. $\liminf$ is defined similarly.

In your case (for the first and last equalities), \begin{align} b_1&=\frac{2}{3}&b_2&=\frac{3}{2^2}\\ b_3&=\frac{2^2}{3^2}&b_4&=\frac{3^2}{2^3} \end{align} so you can start to see the pattern:

  • When the index on $b$ is odd, $$ b_{2i-1}=\frac{a_{2i}}{a_{2i-1}}=\frac{2^i}{3^i}=\left(\frac{2}{3}\right)^i. $$

  • When the index on $b$ is even, $$ b_{2i}=\frac{a_{2i+1}}{a_{2i}}=\frac{3^i}{2^{i+1}}=\frac{1}{2}\left(\frac{3}{2}\right)^i. $$

For a fixed $k$, the inf of the elements in $B_k$ is $0$ because $B_k$ contains higher and higher powers of $\frac{2}{3}$ (for all odd indices $i$ with $i>k$). On the other hand, the supremum of $B_k$ is $\infty$, because $B_k$ contains $\frac{1}{2}$ times higher and higher powers of $\frac{3}{2}$ (for even powers of $i$ with $i>k$).

I'm leaving the second and third cases as an exercise.

Divide1918
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Michael Burr
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  • Higher and higher powers of $\frac{2}{3}$? And what about the terms of the form $\frac{3^m}{2^n}$? – AnalyticHarmony Jun 01 '17 at 09:07
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    I put a few more details into the calculation, for example, illustrating the values of $b$, explicitly. Also, you don't get general terms of the form $\frac{3^m}{2^n}$, only terms of the form $\frac{3^m}{2^{m+1}}$. – Michael Burr Jun 01 '17 at 10:23
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The series' general term is

$$a_n=\begin{cases}\cfrac1{2^{(n+1)/2}}\,,&\text{odd}\;n\\{}\\ \cfrac1{3^{n/2}}\,,&\text{even}\;n\end{cases}$$

In this case, we have the quotients:

$$\frac{a_{n+1}}{a_n}=\begin{cases}\cfrac{\frac1{2^{(n+1)/2}}}{\frac1{3^{n/2}}}=\cfrac1{\sqrt2}\left(\cfrac32\right)^{n/2}\,,\,&\text{even}\;n\\{}\\ \cfrac{\frac1{3^{n/2}}}{\frac1{2^{(n+1)/2}}}=\sqrt2\,\left(\cfrac23\right)^{n/2}\,,\,&\text{odd}\;n\end{cases}$$

and thus we get

$$\limsup_{n\to\infty}\frac{a_{n+1}}{a_n}=\sqrt2\,\cdot\,1=\sqrt2>1$$

and thus the series diverges. The same result you get with the $\;n\,-$ th root test.

DonAntonio
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