Solution-verification of $\lim_{n\to\infty}\left(\frac{n^3}{1 \cdot 3+3\cdot5+\cdots+(2n-1)(2n+1)}\right)$

Question

I solved this limit and got the solution $\frac{3}{4}$. I tryed checking on WolframAlpha, but when generating the presentation of the expression it shows $\lim{n\to n}$ instead of $\lim{n \to \infty}$ and tells me that the limit diverges. So I'm unsure whether it diverges because of the misinterpretation of the problem or the limit truly is diverging.

EDIT: Fellow colleague Acheca provided the solution in the comments to the WolframAlpha problem, here it is, which btw answers my question with a yes.

$$\lim_{n\to\infty}\left(\frac{n^3}{1 \cdot 3+3\cdot5+\cdots+(2n-1)(2n+1)}\right)$$

I applied the Stolz-Cesaro theorem and eventually (after the initial steps) got

$$\lim_{n \to \infty}\frac{(n+1)^3-n^3}{1 \cdot 3+3\cdot5+\cdots+(2n-1)(2n+1)-(1 \cdot 3+3\cdot5+\cdots+(2n-1)(2n+1)+(2n+1)(2n+3))}=\lim_{n \to \infty}\frac{3n^2+3n+1}{(2n+1)(2n+3)}=\frac{3}{4}$$

Is the result correct ?

Thanks in advance

P.S. Should I delete these kinds of questions if the answer to them is a simple yes, since they don't provide much information and may not be of great use to anyone except me ?

Answer 1

You answer is correct but Cesaro-Stolz isn't really necessary since

• $\sum_{k=1}^n(2k-1)(2k+1) = \sum_{k=1}^n(4k^2-1) = 4\underbrace{\sum_{k=1}^nk^2}_{=\frac{n(n+1)(2n+1)}{6}} -n$
• $\Rightarrow \sum_{k=1}^n(2k-1)(2k+1) \stackrel{n\to\infty}{\sim}\frac 43n^3$

Hence, $\displaystyle \lim_{n\to\infty}\frac{n^3}{\sum_{k=1}^n(2k-1)(2k+1)}=\frac 34$.