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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!

Dark Malthorp
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FAF
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  • If $\limsup a_n = A$, then for any $\epsilon>0$, there exists $N$ such that for all $n>N$ you have $ a_n < A+\epsilon $. Is this the detail you miunderstood? – Dark Malthorp May 03 '20 at 21:02
  • Yes. To be more specific, I don't know how to prove it (if it was true). – FAF May 03 '20 at 21:22

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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 $n>N$.

Dark Malthorp
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