nth root of an, a_n converges to 1

the limit of $a_n$ is 1.

I need to prove/disprove that the nth root of $a_n$ (as n approaches infinity) is 1

$\sqrt[n] a_n$ -> 1

any ideas?

Solution 1:

You have from Cauchy's First limit theorem That if $a_{n}\to a $ then $\frac{1}{n}\sum_{r=1}^{n}a_{r}\to a$ . Here is a proof of this :-Proof

This also gives that if $a_{n}>0$ and $a_{n}\to a$.

$$\ln(a)=\lim_{n\to\infty}\ln(a_{n})=\frac{1}{n}\sum_{r=1}^{n}\ln(a_{n})=\ln(\sqrt[n]{\prod_{r=1}^{n}a_{r}})$$

So $\lim_{n\to\infty}\sqrt[n]{\prod_{r=1}^{n}a_{r}}= a$.

i.e. The geometric mean and arithmetic mean tend to the same limit as that of the original sequence. These are Cauchy's first and 2nd limit theorem.

Now if $y_{1}=a_{1}$ and $y_{n}=\frac{a_{n}}{a_{n-1}}$ . Then using the 2nd limit theorem we get if $$\lim_{n\to\infty}y_{n}=\lim_{n\to\infty}\frac{a_{n}}{a_{n-1}}=l$$

Then

$$\lim_{n\to\infty}\sqrt[n]{a_{n}}=\lim_{n\to\infty}\frac{a_{n}}{a_{n-1}}=l$$.

Here as $\lim_{n\to\infty}a_{n}=1$ . You have $l=1$.