I'm trying to find a sequence $a_n$ with $a_n\xrightarrow{}{0}$ and $a_n>0$, but such that $\sum_{n=1}^{\infty}(-1)^n{a_n}$ does not admit limit, or its limit is oscillating. From Leibniz's rule if $a_n$ satisfies the previous hypothesis and decreases monotonically, than the series would converge. So at least $a_n$ cannot be monotonic, I was thinkig about a series like: $$ \sum_{n=1}^{\infty}(-1)^n{|sin(n)|\over{n}}, $$ the sequence above is not monotonic, is always positive and converges to 0. The series does not diverge since $0\le{|sin(n)|}\le{1}$, so or it converges or it doesn't admit limit. Now I don't know how to move on.
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@RobertoPalermo: What difference do you make between "diverges" and "does not admit a limit"? Most textbooks will define convergent=having a limit in $\mathbb R$, divergent=not convergent. – Taladris Aug 27 '23 at 22:42
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With "divergent" I mean that the series goes to $+\infty$ or $-\infty$ such as the series of $1/n$. With "does not admit limit" I mean a series like one whose general term is $(-1)^n$. In this case if I follow the order of the sum I got 0 or 1 depending when I stop. – Roberto Palermo Aug 28 '23 at 08:09
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Here is an example: Take some positive constant $x\neq1$ and the sequence
$$ a_n = 1, x, \frac12, \frac12x, \frac13, \frac13x, \frac14, \frac14x,\dots $$ The sequence converges against 0 and is positive. Adding two subsequent elements yields: $$ (-1)^{2n-1}a_{2n-1}+(-1)^{2n}a_{2n}=a_{2n}-a_{2n-1}=\frac1nx-\frac1n=(x-1)\frac1n $$ so you are basically getting the Harmonic Series.
An example of a series that does not converge to infinity is the sum over:
$$ s_n= \underbrace{2,-1}_{=1 (2 \text{ terms})}, \underbrace{\frac12,-1,\frac12,-1}_{=-1 (4 \text{ terms})}, \underbrace{\frac12,-\frac14\cdots,\frac12,-\frac14}_{=1 (8 \text{ terms})}, \underbrace{\frac18,-\frac14\cdots,\frac18,-\frac14}_{=-1 (16 \text{ terms})},\dots $$ with $a_n=|s_n|$.
The idea is to use the sum $1-1+1-1\pm\cdots$ and split the $n$-th term into $2^n$ alternating sub-summands.
If you want something that is not bounded, split the sum $1-2+3-4+5\mp\cdots$
emacs drives me nuts
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The sequence converges to 0, is positive and non-monotonic, but since the series is harmonic I think it diverges to $\infty$ if $x>1$ and $-\infty$ if $x<1$, am I right? – Roberto Palermo Aug 27 '23 at 15:10
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Yes. And you can do this "trick" with every positive series that diverges, not just the Harmonic Series. – emacs drives me nuts Aug 27 '23 at 15:17
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In this case the series admits limit and it's $+\infty$ or $-\infty$ depending on $x$ I fixed. But I'm trying to find a series such that the limit does not exist (e.g. $\sum_{n=0}^{\infty}(-1)^n$), but whose general term satisfies the hypothesis above. – Roberto Palermo Aug 27 '23 at 15:43
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I like this one. $$ a_{2n} = \frac{1}{n}, \\ a_{2n+1} = \frac{1}{n^2} $$ Of course $a_n > 0$ and $a_n \to 0$. Note $\sum a_{2n}$ diverges to $\infty$, while $\sum a_{2n+1}$ converges. So $\sum_{n=1}^\infty (-1)^n a_n$ diverges to $\infty$.
GEdgar
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