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Suppose that you can apply the Ratio Test to $\Sigma a_{n}$. Let $r$ be the limit of $|a_{n+1}|/|a_{n}|$. Show that $\lim\sup|a_{n}|^{1/n}=r$ as $n\rightarrow\infty$.

I know by definition of lim sup that $\forall\epsilon>0$, $\exists N_{\epsilon}$ s.t. $x_{n}<r+\epsilon$ $\forall n>N_{\epsilon}$. Not sure how to apply that here though.

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This is a special case of the result from this question: math.stackexchange.com/questions/69386/… –  Martin Sleziak Mar 21 at 9:04
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2 Answers

Let $\epsilon>0$. There is an $N$ so that for all $n\ge N$, $|a_{n+1}|\le (r+\epsilon)|a_n|$ From this it follos that

$\ \ \ |a_{N+1}|\le (r+\epsilon)|a_N|$

$\ \ \ |a_{N+2}|\le (r+\epsilon)^2|a_N|$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots$

$\ \ \ |a_{N+k}|\le (r+\epsilon)^k|a_N|$

Using the above, if $n>N$: $$ |a_n|=|a_{N+(n-N)}|\le (r+\epsilon)^{n-N}|a_N| $$ Let $A= |a_N|/(r+\epsilon)^N$. Then $$ \root n\of {|a_n| }\le A^{1/n}(r+\epsilon) $$ for all $n>N$.

Now, $A^{1/n}\rightarrow1$ as $n\rightarrow\infty$; so, $$ \limsup_{n\rightarrow\infty} \root n\of{|a_n|}\le r+\epsilon. $$

Since $\epsilon$ was arbitrary, we have

$$ \limsup_{n\rightarrow\infty} \root n\of{|a_n|}\le r. $$

Now show that $\liminf\limits_{n\rightarrow\infty} \root n\of{|a_n|}\ge r$. I'll leave that for you.



Note that the above can be modified slightly to show that $\limsup\limits_{n\rightarrow\infty}\root n\of {|a_n|}\le\limsup\limits_{n\rightarrow\infty}\root n\of {\Bigl|{a_{n+1}\over a_n}\Bigr|}$. One can also show that $\liminf\limits_{n\rightarrow\infty}\root n\of {|a_n|}\ge\liminf\limits_{n\rightarrow\infty}\root n\of {\Bigl|{a_{n+1}\over a_n}\Bigr|}$. From this, your result easily follows.

Note also please, that the result has little to do with series...

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We will prove the following:

If $ \lim|\frac{a_{n+1}}{a_n}|=l$ then it is $\lim |a_n|^{1/n} =l \quad (\spadesuit)$

To prove this we need the following two inequalities:

(a) $\limsup |a_n|^{1/n} \leq \limsup|\frac{a_{n+1}}{a_n}|$

(b) $\liminf|\frac{a_{n+1}}{a_n}| \leq \liminf |a_n|^{1/n}$

Proof of $(\spadesuit)$: From (a) and (b) we have that $\liminf|\frac{a_{n+1}}{a_n}| \leq \liminf |a_n|^{1/n} \leq\limsup |a_n|^{1/n} \leq \limsup|\frac{a_{n+1}}{a_n}| \quad (\diamond)$

Since $ \lim|\frac{a_{n+1}}{a_n}|=l$ exists we have that $ \liminf|\frac{a_{n+1}}{a_n}|= \limsup|\frac{a_{n+1}}{a_n}|= \lim|\frac{a_{n+1}}{a_n}|=l$ and then $(\diamond)$ gives:

$\liminf |a_n|^{1/n}=\limsup |a_n|^{1/n}=\lim |a_n|^{1/n}=l$. Q.E.D

The converse does not hold Here it is a counterexample:

Consider the sequance $(a_n), \quad n\in\mathbb{N}$ with $a_{2n-1}=\frac{1}{2^n}$ and $a_{2n}=\frac{1}{2^{n-1}}$

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This simply postpones the proof the OP asks for to your statements (a) and (b)... which you do not prove. –  Did Jan 23 '12 at 0:16
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