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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.

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    $a_n-[a_n]$ converges to -1.2012-03-30

2 Answers 2

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  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 $a_k:=2\cos\left(\frac{2\pi}3k\right)$.
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    @Davide Giraudo OK, but $\lim \limits_{n\to \infty}{a_n+b_n}=L$ and $\lim \limits_{n\to \infty}-a_n=-A$ how can we know anything about $\lim \limits_{n\to \infty}b_n=?$2012-03-31
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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) $1,-1,0,1,-1,0,\dots$

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    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?2012-03-31