contradicting examples for series statements I have tried to find counter examples for the following but without any success. Would appreciate your help:


*

*$a_n \to \infty$ then $a_n$ is monotonically increasing

*$|a_n| \to a$ then $a_n$ converges

*$a_nb_n \to 0$ then $a_n \to 0$ or $b_n \to 0$

*$a_n \to K$ (while $K\neq\pm\infty$) then $\frac{a_{n+1}}{a_n}$ converges


just thought about solution for 2, if I choose unbound series like $(-1)^n$, in absolute value it will be the constant series of (1) thus it converges.
I think that for 3 I have to use also unbounded series, but still no clue.
 A: Here are some possible answers:


*

*Consider a sequence like: 1,-1,2,-2,3,-3, $\dots$ that after 1,000 becomes an increasing sequence of 1000,1001, 1002, $\dots$ that would be not monotonically increasing for all $n$. A stuttering sequence here could also work: 1,1,2,2,3,3, $\dots$ as back to back terms could be the same so the $a_{n+1}>a_n$ isn't true for this sequence case.

*Consider this sequence: 1,-1,1,-1, $\dots$ where for any term the absolute value would be 1 but the sequence merely oscillates.

*Consider $a_n$=1,0,1,0,$\dots$ and $b_n$=0,1,0,1,0,$\dots$ so that the product is zero for all the terms. yet neither $a_n$ or $b_n$ even converge.

*i) Consider the sequence of: 0,0,0,0,$\dots$ where the division by zero would be undefined rather than converging even though the sequence does go to zero which is a valid value for $K$. 
To use Justpassingby's idea one could have a sequence like: 
ii) 1,1,.5,.5,.25,.25,$\dots$ where the ratio test of $\frac{a_{n+1}}{a_n}$ would be 1,.5,1,.5,$\dots$ so the ratio doesn't converge but the initial sequence does converge to zero.
iii) For a non-zero limit of K, just adjust all the terms up a fixed value:
3,3,2.5,2.5,2.25,2.25,$\dots$ so that now the sequence converges to 2 yet the consecutive terms still oscillate back and forth given the back-to-back values. In this case, the value is depending on which subsequence you want to take:
$a_{2n-1}=2+\frac{1}{2^{n-1}}$ which would be 3, 2.5, 2.25, 2.125,  $\dots$
$a_{2n}=2+\frac{1}{2^{n-1}}$ which would be 3, 2.5, 2.25, 2.125, $\dots$
Note that the sequence of $2+\frac{1}{2^{n-1}}$ on its own wouldn't meet the criteria since it does have a monotonically descending property and so the ratio of $\frac{a_{n+1}}{a_n}$ would go to zero if this alone was the sequence. By having a copy of it, the new ratio has this oscillating with 1s for the other ratio as back-to-back terms may be equal half the time as if $n$ is odd and the first term is $a_1$ then $a_n=a_{n+1}$ while if $n$ is even $a_n>a_{n+1}$ since this is where we change the terms.
Finally a good possibility for $a_n$ without dividing its definition for odds and evens would be $a_n = 3-(1-(\frac{1}{2})^{\lceil{\frac{n}{2}}\rceil})$
