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Let $(a_n)$, $(b_n)$ be bounded sequences. Prove that if $a_n + b_n \rightarrow c$, then $c\le \limsup \,a_n + \liminf\, b_n$.

I have tried a proof. I don't know if it is correct.

Proof: since $(a_n)$ and $(b_n)$ are bounded sequences, it holds that $ \forall \frac{\epsilon}{2} > 0$, there is $n_0 \in \mathbb N$ such that $n>n_0 \Rightarrow$ $\liminf a_n - \frac{\epsilon}{2} < a_n < \limsup \, a_n + \frac{\epsilon}{2} $ ( 1) and
$\liminf\, b_n \,- \frac{\epsilon}{2} < b_n < \limsup \, b_n + \frac{\epsilon}{2}$ (2)

By adding (1) and (2):

$\liminf \, a_n + \liminf \, b_n - \epsilon < a_n + b_n < \limsup \, a_n + \limsup \, b_n + \epsilon > 0$ (3)

But $( a_n + b_n) \rightarrow c$. So, $\forall \epsilon >0$, there is $n_1 \in \mathbb N$, such that $n>n_1 \Rightarrow c - \epsilon < a_n + b_n < c + \epsilon$ (4)
Now take $n=\max\{n_0, n_1\}$. So, in consequence of (3) and (4) it is possible to write: $c= \liminf \, a_n + \liminf \, b_n$. But $\liminf \, a_n < \limsup \, a_n$. Then, $c< \limsup a_n + \liminf \, b_n$. I'd like to know your opinion.

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Your argument falls apart in the last paragraph. It is not necessarily true that $c=\liminf\limits_{n\rightarrow\infty} a_n +\liminf\limits_{n\rightarrow\infty} b_n$. Take for example $a_n=(-1)^n$ and $b_n=-a_n$. Here, $c=0$ while the sum of the limit inferiors is $-2$.

Try this though:

Let $(b_{n_k})_{k=1}^\infty$ be a subsequence of $(b_n)_{n=1}^\infty$ converging to $\liminf\limits_{n\rightarrow\infty} b_n$. Then $\lim\limits_{k\rightarrow\infty}(a_{n_k}+b_{n_k})=c$. But since both $(b_{n_k})_{k=1}^\infty$ and $(a_{n_k}+b_{n_k})_{k=1}^\infty$ are convergent, so is $(a_{n_k})_{k=1}^\infty$. Moreover, $\lim\limits_{k\rightarrow\infty} a_{n_k}$ is at most $\limsup\limits_{n\rightarrow\infty} a_n$.

The desired result follows from the above.

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  • $\begingroup$ Thank you, David. Your approach is somewhat different, and very interesting. $\endgroup$ – Walter r Apr 28 '13 at 19:11
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If $\epsilon>0$, then $b_n<\liminf b_n+\epsilon$ for infinitely many $n$ and $a_n<\limsup a_n+\epsilon$ for almost all $n$, hence $a_n+b_n<\limsup a_n+\liminf b_n+2\epsilon$ for infinitely many $n$. This implies $c\le \limsup a_n+\liminf b_n+2\epsilon$ and, since $\epsilon$ was arbitray, $c\le \limsup a_n+\liminf b_n$.

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  • $\begingroup$ Good afternoon Hagen von Eitzen. Your answer helped me to find out my mistake. $\endgroup$ – Walter r Apr 28 '13 at 19:12
  • $\begingroup$ @Walterr That's good night actuzally, but nice to hear :) $\endgroup$ – Hagen von Eitzen Apr 28 '13 at 19:17

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