Limes problem no. 2

[mathwannabe]

Hello everybody

Here's the problem:

Calculate the limit of the sequence:

\(\displaystyle \lim_{n \to \infty}\dfrac{1+\sqrt{3}+\sqrt{5}+...+\sqrt{2n-1}}{1+\sqrt{7}+\sqrt{13}+...+\sqrt{6n+1}}\)

I started off like this:

\(\displaystyle \lim_{n \to \infty}\dfrac{n}{1+\sqrt{3n^2+4n}}\)

Then I tried to rationalize like this:

\(\displaystyle \lim_{n \to \infty}\dfrac{n(1-\sqrt{3n^2-4n})}{1-3n^2+4n}\)

I have no idea how to proceed... or even if what I have already done is correct. Limits are giving me a lot of trouble. Please help.

 

[MarkFL]

I would use the Stolz-Cesàro theorem (the discrete analog to L'Hôpital's rule) to write the limit as:

\(\displaystyle \lim_{n\to\infty}\dfrac{\sqrt{2n+3}}{\sqrt{6n+7}}=\sqrt{\lim_{n\to\infty}\dfrac{2n+3}{6n+7}}\)

Can you proceed from here?

https://en.wikipedia.org/wiki/Stolz–Cesàro_theorem

 

[lookagain]

Hello everybody

Here's the problem:

Calculate the limit of the sequence:

\(\displaystyle \lim_{n \to \infty}\dfrac{1+\sqrt{3}+\sqrt{5}+...+\sqrt{2n-1}}{1+\sqrt{7}+\sqrt{13}+...+\sqrt{6n+1}}\)

Two things I noticed is 1) that this is not a sequence, and 2) the first number in the
denominator should be \(\displaystyle \sqrt{7} \) and not a "1," because it begins with n = 1,
and 6n + 1 = 7 when n = 1.






I started off like this:

\(\displaystyle \lim_{n \to \infty}\dfrac{n}{1+\sqrt{3n^2+4n}}\)

Then I tried to rationalize like this:

\(\displaystyle \lim_{n \to \infty}\dfrac{n(1-\sqrt{3n^2-4n})}{1-3n^2+4n}\)

I have no idea how to proceed... or even if what I have already done is correct. Limits are giving me a lot of trouble. Please help.

One version of the sequence is

\(\displaystyle \dfrac{1}{\sqrt{7}}, \ \ \dfrac{\sqrt{3}}{\sqrt{13}}, \ \ \dfrac{\sqrt{5}}{\sqrt{19}}, ... , \dfrac{\sqrt{2n - 1}}{\sqrt{6n + 1}} \)



The limit of the sequence can be expressed as

\(\displaystyle \displaystyle\lim_{n \to \infty} \ \dfrac{\sqrt{2n - 1}}{\sqrt{6n + 1}}.\)


Then you can factor out a \(\displaystyle \sqrt{n}\) from the numerator and the denominator.


And continue...

 

[Deleted member 4993]

One version of the sequence is

\(\displaystyle \dfrac{1}{\sqrt{7}}, \ \ \dfrac{\sqrt{3}}{\sqrt{13}}, \ \ \dfrac{\sqrt{5}}{\sqrt{19}}, ... , \dfrac{\sqrt{2n - 1}}{\sqrt{6n + 1}} \)



The limit of the sequence i can be expressed as

\(\displaystyle \displaystyle\lim_{n \to \infty} \ \dfrac{\sqrt{2n - 1}}{\sqrt{6n + 1}}.\)


Then you can factor out a \(\displaystyle \sqrt{n}\) from the numerator and the denominator.


And continue...

The sequence could also be:

\(\displaystyle \dfrac{1}{\sqrt{1}}, \ \ \dfrac{\sqrt{3}}{\sqrt{7}}, \ \ \dfrac{\sqrt{5}}{\sqrt{13}}, ... , \dfrac{\sqrt{2n - 1}}{\sqrt{6n - 5}} \)