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Denote: $\liminf a_n = a$ and $\limsup b_n = B$. Prove: $\liminf(a_n + b_n) \le a+B$.

The proof:

Let $\varepsilon > 0$.
By definition of infimum, there's a subsequence $a_{n_k}$ such that $a_{n_k} \le a +\frac{\varepsilon}{2}$

By definition of supremum, there's a subsequence $b_{n_k}$ such that $b_{n_k} \le B +\frac{\varepsilon}{2}$

Hence, (for sufficiently large $n_k$):
$$(a_{n_k} + b_{n_k}) \le a + \frac{\varepsilon}{2} + B + \frac{\varepsilon}{2} = a+ B + \varepsilon$$

Therefore, $$\liminf (a_{n_k} + b_{n_k}) \le a+ B$$

And that's implying:

$$\color{Green}{\liminf (a_n + b_n) \le a+ B}$$

I'd be glad to get an explanation for the "Green" claim.

Is it because $a_{n_k} + b_{n_k}$ is a subsequence of $a_n + b_n$?

Elimination
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1 Answers1

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Proof:

Let $\varepsilon>0$. Then some $k$ exists with $n>k\Rightarrow b_{n}\leq B+\varepsilon$.

Then $a_{n}+b_{n}\leq a_{n}+B+\varepsilon$ for $n>k$ and consequently $\liminf_{n\rightarrow\infty}\left(a_{n}+b_{n}\right)\leq a+B+\varepsilon$.

This is true for any $\varepsilon>0$ and this allows the conclusion that $\liminf_{n\rightarrow\infty}\left(a_{n}+b_{n}\right)\leq a+B$.

drhab
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  • I understand your approach. But given $\liminf(a_{n_k} + b_{n_k}) < a + B$. How to conclude $\liminf(a_n + b_n) < a + B$? – Elimination Aug 25 '14 at 18:12
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    If $(c_{n_k})$ is a subsequence of $(c_n)$ then $\liminf c_n\leq \liminf c_{n_k}$. For every $d<\liminf c_n$ you have $c_{n_k}>d$ for $k$ large enough so $\liminf c_{n_k}\geq d$. Consequently $\liminf c_{n_k}\geq \liminf c_n$. – drhab Aug 25 '14 at 18:21