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Construct a convergent series of positive terms with $\displaystyle\limsup_{n\to\infty} {a_{n+1}\over{a_n}}=\infty$


My thoughts:

By the theorem: Suppose $a_n\ge0$ for all $n$, and let $l=\displaystyle\limsup{\sqrt[n]{a_n}}$. If $l<1$, then $\displaystyle\sum_{n=1}^\infty a_n$ converges; and if $l>1$, then $\displaystyle\sum_{n=1}^\infty a_n$ diverges.

Thus, we know $\displaystyle\sum_{n=1}^\infty {a_{n+1}\over{a_n}}$ diverges

I guess it might works, $(x_n)_{n=1}^{+\infty}$, $x_{n+1} = x_n^2 + x_n$ for all $n\ge1$..

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4 Answers 4

up vote 10 down vote accepted

You need to get big ratios $\frac{a_{n+1}}{a_n}$, but not too often. Start with a nice convergent series, say $$\sum_{n\ge 0}2^{-n}=1+\frac12+\frac1{2^2}+\frac1{2^3}+\ldots~\;.\tag{1}$$ Of course in $(1)$ the ratios are all $\frac12$. To get a bigger ratio at $\frac{a_{n+1}}{a_n}$ for some $n$, just replace $a_n=2^{-n}$ by a smaller number. For instance, let

$$a_n=\begin{cases} \frac1{n2^n},&\text{if }n\text{ is even}\\\\ \frac1{2^n},&\text{if }n\text{ is odd}\;. \end{cases}$$

Then

$$\frac{a_{n+1}}{a_n}=\begin{cases} \frac{n}2,&\text{if }n\text{ is even}\\\\ \frac1{2(n+1)},&\text{if }n\text{ is odd}\;. \end{cases}$$

Clearly $\sum_{n\ge 0}a_n$ is still convergent, and $\limsup_n\frac{a_{n+1}}{a_n}=\infty$.

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Look at the series $$\frac{1}{4}+\frac{1}{2}+\frac{1}{4^2}+\frac{1}{2^2}+\frac{1}{4^3}+\frac{1}{2^3}+\frac{1}{4^4}+\frac{1}{2^4}+\frac{1}{4^5}+\frac{1}{2^5}+\cdots.$$ Note the unusual alternation: Sometimes the next term is a lot less than the previous one, and sometimes it is a lot bigger.

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Okay, so you made a sequence of positive terms in which the limit superior of the ratio is infinite. Does the series converge though? –  CogitoErgoCogitoSum Feb 12 '13 at 3:08
    
@CogitoErgoCogitoSum Yes. Consider the sequence where the $4$s are replaced by $2$s (making the sum larger) and use the comparison test. –  Potato Feb 12 '13 at 3:19
    
@CogitoErgoCogitoSum: Sure it converges, either by Comparison as by Potato, or by noting that partial sums of the first $2n$ terms are bounded by the sum of the corresponding partial sums of the first $n$ terms of the two interleaved series. –  André Nicolas Feb 12 '13 at 3:28
    
Its simpler than that. Replacing the 4's with 2's makes it twice a geometric series. Thanks though, guys, it didnt jump out at me. –  CogitoErgoCogitoSum Feb 12 '13 at 4:01
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I have an example:

Consider $$ a_n = \begin{cases} \frac{1}{m^2} & n = 2m - 1\\ \frac{1}{2^m} & n = 2m \end{cases}\qquad m = 1, 2, \ldots. $$

It's obvious that the series $\sum a_n$ is convergent since $\sum \frac{1}{n^2}$ and $\sum \frac{1}{2^n}$ are both convergent.

However, $$\limsup \frac{a_{n+1}}{a_n}=\lim \frac{2^n}{n^2}=\infty.$$

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Try the sequence $( \frac{1}{2})^3 + ( \frac{1}{2})^2+ ...+ ( \frac{1}{k})^3+ (\frac{1}{k})^2+ ...$.

Since $\frac{a_{2n+2}}{a_{2n+1}}=\frac{(\frac{1}{n+1})^2}{(\frac{1}{n+1})^3} = n+1$, we have $\limsup_n \frac{a_{n+1}}{a_n} = \infty$. The sequence is convergent since both $\sum \frac{1}{k^2}$ and $\sum \frac{1}{k^3}$ are convergent.

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