Prove if $\limsup$ $s_n = +\infty $ and $k > 0$ then $\limsup (ks_n) = +\infty$. Prove if $\limsup$ $s_n = +\infty $ and $k > 0$ then $\limsup (ks_n) = +\infty$.
Hi, I'm preparing for Analysis and found this exercise in a Book. The proof simply uses a theorem which was established earlier, one of the limit properties that I was familiar with.
But I thought we can also prove this by using the definition. That the $\limsup$ $s_n = +\infty $ when for every $M$ there exists an $s_n $ $ n > N $ such that $s_n > M $.
Could I say that because the limit of $s_n = +\infty $ I let $M_0$ be $\frac {M}{k}$ for some $M$ and therefore $s_n > M_0$ and $ks_n > M$ ? Sorry if this is very basic. I do limits and that quite okay, but I'm struggling a bit with getting comfortable with the formal proofs. Thanks a lot!  
 A: Your proof is overall right, although there are many approximations.
Firstly, you say that the limit of $s_n$ is $+\infty$. This is false since, it is only $\operatorname{lim sup}(s_n) = +\infty$. However, you can extract a subsequence $s_{\phi(n)}$ such that $\lim(s_{\phi(n)}) = +\infty$.
This is not even necessary, formally, the fact that $\operatorname{lim sup}(s_n) = +\infty$ means that for any number $M$, there exists an integer $l$ such that $s_l > M$. In order to be clearer, I will denote $s'_n$ the sequence $s'_n = ks_n$, and our goal is to show the above property for the sequence $s'$ assuming it is true for the sequence $s$. Let's fix a number $M_0$, we want to find an integer $l$ such that $s'_l>M_0$. This simplifies to $ks_l > M_0$, i.e. (since k>0), $s_l > \frac{M_0}{k}$. By the above property for the sequence $s$ and the number $\frac{M_0}{k}$, there exists a rank $l_0$ such that $s_{l_0}>\frac{M_0}{k}$, and for this rank, we have $s'_{l_0} > M_0$.
So yes, in essence your proof was correct, but you need to refer more precisely to the properties you are using, so that you can see by yoursefl that it is correct
