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!
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!
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!
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!