I am trying to prove the root test for sequence convergence with $\limsup$.
I know exactly what to do but I think I misunderstood a simple detail when I learned about $\limsup$ and $\liminf$. That if $(a_n)$ is a sequence then we can find $N$ that for all $n>N$ the sequence is bounded with the $\limsup+\epsilon$.
If I knew how to prove it (in case it was true), that would help with my solution.
A hint would really make a difference.
I tried to search a lot but unfortunately I couldn't find something that answers my question.
Thank you!
Claim: If $$ \limsup a_n = A $$ and $\epsilon>0$, then there exists $N\in\mathbb{N}$ such that for all $n>N$, $$ a_n < A+\epsilon $$ Proof: I'm assuming your definition of $\limsup$ is the first one from Wikipedia. If your class uses the other one, you can modify this proof accordingly. $$ \limsup_{n\rightarrow\infty} a_n = \lim_{n\rightarrow\infty}\left(\sup_{m\ge n} a_n\right) $$ If this converges to $A$, by definition of a limit, there exists $N$ such that for $n>N$, $$ \left|\sup_{m\ge n} a_n - A\right| <\epsilon $$ and hence \begin{eqnarray} \sup_{m\ge n} a_n - A &<& \epsilon\\ \sup_{m\ge n} a_n &<&A +\epsilon \end{eqnarray} hence $a_n < A+\epsilon$ for all $n>N$.