What would be the limit What would be the limit of:
$$\lim_{n \to \infty}\frac{1^2+3^2+...+(2n-1)^2}{2^2+4^2+...+(2n)^2}$$
Knowing that $2n-1$ is an arithmetic sequence the sum of it would be $\frac{n(a_{1}+a_{n})}{2}$, same goes as for $2n$
Using this fact i am able to evaluate it, $$\lim_{N \to \infty}\frac{\sum_{n=1}^{N}(2n-1)^2}{\sum_{n=1}^{N}(2n)^2}=1$$
Is this the general idea i should be going with, will it work for similar series, or is there any other way to solve it.
 A: The correct way to evaluate this limit is to recall that $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6},$$ hence the numerator is $$\begin{align*} \sum_{k=1}^n (2k-1)^2 &= \sum_{k=1}^n (2k-1)^2 + (2k)^2 - \sum_{k=1}^n (2k)^2 \\ &= \sum_{k=1}^{2n} k^2 - \sum_{k=1}^n 4k^2 \\ &= \sum_{k=1}^{2n} k^2 - 4 \sum_{k=1}^n k^2 \\ &= \frac{2n(2n+1)(4n+1)}{6} - 4 \frac{n(n+1)(2n+1)}{6}. \end{align*}$$  But notice the denominator is the second term of the above expression, so we can get $$\frac{n(2n+1)(4n+1)}{2n(n+1)(2n+1)} - 1 = \frac{4n+1}{2(n+1)} - 1,$$ the limit of which as $n \to \infty$ is simply $1$.
A: We have
$\displaystyle \begin{aligned}\dfrac{{\displaystyle \sum_{k = 1}^{n}(2k - 1)^{2}}}{{\displaystyle \sum_{k = 1}^{n}(2k)^{2}}} &= \dfrac{{\displaystyle \sum_{k = 1}^{n}(2k - 1)^{2}}}{{\displaystyle \sum_{k = 1}^{n}(2k)^{2}}} + 1 - 1\\
&= \dfrac{{\displaystyle \sum_{k = 1}^{2n}k^{2}}}{{\displaystyle \sum_{k = 1}^{n}(2k)^{2}}} - 1\\
&= \dfrac{\dfrac{2n(2n + 1)(4n + 1)}{6}}{\dfrac{4n(n + 1)(2n + 1)}{6}} - 1\\
&= \frac{4n + 1}{2(n + 1)} - 1\\
&= \dfrac{4 + \dfrac{1}{n}}{2\left(1 + \dfrac{1}{n}\right)} - 1\\
&\to 2 - 1 = 1 \text{ as }n \to \infty\end{aligned}$
A: Use the fact that $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ and $\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$. This simplifies things.
