Given $\{{a_n}\}_{n=0}^{\infty}$ and $\{{b_n}\}_{n=0}^{\infty}$

Prove or disprove:

1) if $\lim \limits_{n\to \infty}a_n=0$ then $\lim \limits_{n\to \infty}a_n-[a_n]=0$

I think (1) is correct because if $\lim \limits_{n\to \infty}a_n=0$ then by the definition of limit I can show that for each $\epsilon>0, |a_n-[a_n]|<\epsilon$

2)If $a_n$ converges and $b_n$ doesn't converge then $(a_n+b_n)$ doesn't converge.

I think (2) is correct, but I'm not sure how to start proving it - maybe I can assume that it isn't correct and then get a contradiction?

3)If $\lim \limits_{n\to \infty}\frac{a_n+a_{n+1}+a_{n+2}}{3}=0$ then $\lim \limits_{n\to \infty}a_n=0$

I have no idea about (3).

My knowledge is of simple calculus theorem(limit definition, arithmetics of limits and the Squeeze Theorem).

Thanks a lot for your time and help.

  • 2
    $\begingroup$ Negative numbers can converge to zero, right? $\endgroup$
    – GEdgar
    Mar 30 '12 at 15:42
  • $\begingroup$ @ GEdgar right. the sequence -1/n converges to zero. $\endgroup$
    – Anonymous
    Mar 30 '12 at 15:43
  • 2
    $\begingroup$ And what about $a_n - [a_n]$ for that sequence? $\endgroup$
    – GEdgar
    Mar 30 '12 at 15:44
  • 3
    $\begingroup$ The author might mean rounding to integers instead of floor/ceil... $\endgroup$ Mar 30 '12 at 15:51
  • 1
    $\begingroup$ Please try not to have multiple disparate questions in one. They might all be part of one homework, but they are essentially different. Of course, you do have to consider that posting multiple questions is bad too. $\endgroup$
    – Aryabhata
    Mar 30 '12 at 15:55
  1. The assertion is not true. Indeed, $a_n$ converges to $0$, so we have to see whether $[a_n]$ (I think it's the floor function) converges to $0$. But it doesn't need to be the case, for example with $a_n=-\frac 1n$.
  2. You can show it's true by contradiction: if $a_n+b_n$ converges, since $-a_n$ converges and the sum of two converging sequences is...
  3. The sequence $a_n=\left(\frac{1+i\sqrt 3}2\right)^n$ is such that $a_n+a_{n+1}+a_{n+2}=0$ but doesn't converge to $0$. Alternatively, a simpler example given by @robjohn is $a_k:=2\cos\left(\frac{2\pi}3k\right)$.
  • 9
    $\begingroup$ The sequence $2\cos\left(\frac{2\pi}{3}k\right)=\{-1,-1,2,-1,-1,2,\dots\}$ remains in $\mathbb{R}$. $\endgroup$
    – robjohn
    Mar 30 '12 at 16:07
  • $\begingroup$ You are right, the complex numbers make the situation more... complex. $\endgroup$ Mar 30 '12 at 16:09
  • $\begingroup$ @Davide Giraudo Could you please get in-depth a little more regarding question (2)? let's say $-a_n$ converges which assumption can I contradict? thank you very much. $\endgroup$
    – Anonymous
    Mar 30 '12 at 20:45
  • $\begingroup$ We get that $a_n+b_n-a_n$ converges so $b_n$ converges. $\endgroup$ Mar 30 '12 at 20:46
  • $\begingroup$ @Davide Giraudo hmmm... why does it imply that $a_n+b_n-a_n$ converges? and why then $b_n$ converges? $\endgroup$
    – Anonymous
    Mar 30 '12 at 21:00

1) What is $|a_n - [a_n]|$, if $|a_n - 0| < \epsilon < \frac{1}{2}$?

2) Let $a = \lim_{n \to \infty}a_n$, then $|a_n + b_n - x| \geq |b_n - x + a| - |a_n - a| \geq |b_n - x'| - \epsilon$ for $x' = x-a$ and $n$ big enough.

3) $1,-1,0,1,-1,0,\dots$

  • $\begingroup$ Or a different way of seeing the answer to 3): $a_{3k+r} = r$, where $r$ is either $0$,$1$, or $-1$. This can be generalized to odd number of terms. For even, just taking alternate $1$ and $-1$ will do. $\endgroup$
    – Aryabhata
    Mar 30 '12 at 18:56
  • $\begingroup$ Or more generally take $a_1 + \dots + a_n = 0$ with at least one summand not trivial and define $a_{nk + r} = a_r$. $\endgroup$ Mar 30 '12 at 19:42
  • $\begingroup$ @Alexander Thumm regarding question (2) what is x and why is this inequality true and how is it prove what's asked in the question? regarding question (3) the sequence makes a lot of intuitive sense, however, how formally can I show that $\lim \limits_{n\to \infty}\frac{a_n+a_{n+1}+a_{n+2}}{3}=0$? $\endgroup$
    – Anonymous
    Mar 30 '12 at 20:49
  • $\begingroup$ 2) is nothing but the reverse triangle inequality: $|p + q| \geq |p| - |q|$. Formally negating the definition of convergence, we see, that a sequence $x_n$ does not converge to $x$, iff for every $\epsilon > 0$ and every $N \geq 0$, there is a $n \geq N$ such that $|x_n - x| > \epsilon$. Now try to use the inequality to show, that $a_n + b_n$ cannot converge to any $x$, since $b_n$ does not converge to any $x'$... As for 3), what is the sum of any three consecutive numbers in the sequence? $\endgroup$ Mar 31 '12 at 11:39

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