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## Homework Statement

Consider [itex]\sum_{1}^{\infty} a_{n}, a_{n} \neq 0[/itex]

Show that

[itex]\underline{\lim\limits_{n \rightarrow \infty}}|\frac{a_{n+1}}{a_{n}}| \leq \underline{\lim\limits_{n \rightarrow \infty}}\sqrt[n]{|a_{n}|}\leq \overline{\lim\limits_{n \rightarrow \infty}}\sqrt[n]{|a_{n}|)} \leq \overline{\lim\limits_{n \rightarrow \infty}}|\frac{a_{n+1}}{a_{n}}| [/itex]

## The Attempt at a Solution

I have just been thinking about the limit superior part of the inequality:

[itex]\overline{\lim\limits_{n \rightarrow \infty}}\sqrt[n]{|a_{n}|)} \leq \overline{\lim\limits_{n \rightarrow \infty}}|\frac{a_{n+1}}{a_{n}}| [/itex]

Ok so I know that for all n

[itex]|\frac{a_{n+1}}{a_{n}}| \leq M[/itex]

and that for all n

[itex]\sqrt[n]{|a_{n}|} \leq N[/itex]

The best way I can see to prove that [itex] N \leq M[/itex] is to do a proof by contradiction

so if N>M then for a sufficiently large p such that:

[itex]|\frac{a_{p+1}}{a_{p}}| \leq M \leq \sqrt[p]{|a_{p}|} \leq N[/itex]

I've spent a lot of time manipulating this equation, but I'm not getting to a contradiction.

Can anyone see a way to prove this or an alternative way to do the proof that works?

Another thing I tried was saying:

[itex]\sqrt[n]{|a_{n}|} \leq N [/itex]

So

[itex]a_{n} \leq N^{n}[/itex]

[itex]\sqrt[n+1]{|a_{n+1}|} \leq N [/itex]

So

[itex]a_{n+1} \leq N^{n+1}[/itex]

But I can't really do anything with that as far as I can see. Was just a thought.