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Seeking some help with the following proof. I have given it an attempt but not sure if my logic is correct, or if I have left out valuable information.

If we let $(a_n)$ be a sequence, and suppose $\lim |\frac{a_{n+1}}{a_n}|=L$ exists. I need to prove that the $\lim|a_n|^{\frac{1}{n}} = L$

So basically we assume that the ratio test is equal to $L$ and now we need to prove that the root test is also equal to $L$

My solution attempt,

If we define the sequence $(b_n)$ by $b_1=a_1$ and $b_n=\frac{a_n}{a_{n-1}}$ for $n>=2$. Since $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}= L$ we have $\lim_{n \to \infty}b_n=L$. With $a_n=b_1b_2...b_n$.

Then $\lim_{n \to \infty} |a_n|^{\frac{1}{n}}=\lim_{n \to \infty} |b_1b_2...b_n|^{\frac{1}{n}} = L$.

If this proof is correct what can we say about the relationship between the ratio test and the root test?

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    It is correct provided you have a justification of the claim that $$b_n \to L \implies (b_1\cdots b_n)^{1/n} \to L.$$ – 2017-02-09
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    @Umberto P do I need to add in this justification? and if so, is that the whole justification? – 2017-02-09
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    That depends whether you consider that a known result, or not. – 2017-02-09
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    okay then my answer must not have enough detail could you submit a answer? – 2017-02-09
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    Your question was about the correctness of the proof. I stand by my first comment. Perhaps someone else will submit an answer. – 2017-02-09
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    The step in question is definitely the crux of the entire proof, so you will need to justify that. What you have done is come up with a convenient notational transformation (considering $b_n$ instead of $a_n$ is nice), but you haven't proved anything at all. http://math.stackexchange.com/questions/287932/convergence-of-ratio-test-implies-convergence-of-the-root-test Somebody asking the same question here, this a great answer. You can transform the notation to think in $b_n$s http://math.stackexchange.com/questions/287932/convergence-of-ratio-test-implies-convergence-of-the-root-test – 2017-02-10
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    @BadamBaplan thanks for that link, that provides me with a answer with great explanation – 2017-02-10

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(a version of) Cesaro's lemma asserts that, given any convergent sequence $(x_n)$ of real numbers :

$$\lim_{n\to\infty}\frac 1n\sum_{k=1}^nx_k=\lim_{n\to\infty}x_n$$

and this conclusion also holds whenever the sequence $(x_n)$ has limit $\pm\infty$.

You may apply this result to the sequence $(\ln(b_n))_{n\ge1}$ in the context of your proof (provided that $b_n>0$ holds for all $n$).

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    I have not used this lemma before, could you comment on my proof? – 2017-02-09