Hello,

I am trying to find the limit of this sequence:

$\displaystyle \sqrt[n]{\frac{k^{n}+l^{n}}{k^{2n}+l^{2n}}}, k,l>0$

but I cannot factorize the fraction at all.

Any idea is very much appreciated!

Thank you all!

- Jul 30th 2011, 02:30 PMMelsicannot find limit of a sequence because of a non factoring fraction
Hello,

I am trying to find the limit of this sequence:

$\displaystyle \sqrt[n]{\frac{k^{n}+l^{n}}{k^{2n}+l^{2n}}}, k,l>0$

but I cannot factorize the fraction at all.

Any idea is very much appreciated!

Thank you all! - Jul 30th 2011, 02:47 PMPlatoRe: cannot find limit of a sequence because of a non factoring fraction
- Jul 30th 2011, 03:03 PMMelsiRe: cannot find limit of a sequence because of a non factoring fraction
Hello,

no it is the very first time I see this theorem, I will check it out tomorrow (02:00 at midnight now) and come back with a full solution!

Very kind of you!

Thank you very much! - Jul 30th 2011, 03:17 PMPlatoRe: cannot find limit of a sequence because of a non factoring fraction
Here is the proof of that theorem.

$\displaystyle b = \sqrt[n]{{b^n }} \leqslant{\color{blue} \sqrt[n]{{a^n + b^n }}} \leqslant \sqrt[n]{{b^n + b^n }} = b\sqrt[n]{2} \to b$ - Jul 31st 2011, 12:13 AMMelsiSolved!
The proof... that's a good one, cool! Now it can be solved..

the initial formula of the general term could be written like this:

$\displaystyle \sqrt[n]{\frac{k^{n}+l^{n}}{k^{2n}+l^{2n}}}=\frac{(k^{n}+l ^{n})^{1/n}}{(k^{2n}+l^{2n})^{1/n}}$

and the limit like this:

$\displaystyle \lim_{n \to \infty}{a_{n}=\lim_{n\to \infty }\frac{(k^{n}+l^{n})^{1/n}}{(k^{2n}+l^{2n})^{1/n}}=\frac{\lim_{n\to \infty }(k^{n}+l^{n})^{1/n}}{\lim_{n\to \infty }(k^{2n}+l^{2n})^{1/n}}$

if we apply the theorem for the numerator we have:

$\displaystyle l\leq \lim_{n\to \infty }(k^{n}+l^{n})^{1/n}\leq \lim_{n\to \infty }(l+\sqrt[n]{2})\Leftrightarrow \lim_{n\to \infty }(k^{n}+l^{n})^{1/n} = l$

and for the denominator it goes like this:

$\displaystyle l^{2}\leq \lim_{n\to \infty }(k^{2n}+l^{2n})^{1/n}\leq \lim_{n\to \infty }(l^{2}+\sqrt[n]{2})\Leftrightarrow \lim_{n\to \infty }(k^{2n}+l^{2n})^{1/n} = l^{2}$

Finally having found those two limits we can write:

$\displaystyle \lim_{n \to \infty}a_{n}=\frac{l}{l^{2}}=\frac{1}{l}$

and all these supposing that $\displaystyle a\leq b$ and $\displaystyle a,b> 0$ this means that if $\displaystyle b\leq a$ it will go the other way around, the limit will be

$\displaystyle \frac{1}{k}$

Thank you once again, for your valuable help!(Hi)