How can I show the quotient of the $k$th partial sums of $\sum\limits_{n=1}^{k} n$ and $\sum\limits_{n=1}^{k} n^2$ is $\frac{3}{2k+1}$? I've found using pen and paper that any trivial case of the sum of a sequence of integers from $1$ to $k$ divided by the sum of the squares of these integers  is equal to $\frac{3}{2k+1}$
for example,
$$
\begin{split}
\frac{1+2+3}{1^{2}+2^{2}+3^{2}} &= \frac{3}{7}\\
\frac{1+2+3\ +\ 4}{1^{2}+2^{2}+3^{2}+4^{2}} &= \frac{3}{9}\\
\frac{1+2+3\ +\ 4\ +5}{1^{2}+2^{2}+3^{2}+4^{2}+5^{2}} &= \frac{3}{11}\\
\frac{1+2+3\ +\ 4\ +5+6}{1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}} &= \frac{3}{13}
\end{split}
$$
Using a computer I can verify this is true for very large values of $k$ and so I intuitively assume it is true for all values of $k$ but how do I prove this?

I've come up with the following equation to represent this as the quotient of two partial sums but don't know how to get from the left side of the equation to the right.

$$\frac{\displaystyle \sum_{n=1}^k n}{\displaystyle \sum_{n=1}^k n^2} = \frac{3}{2k+1}$$

As we know or we can easily prove, $\sum_{n=1}^kn = \dfrac{k(k+1)}{2}$
Now expanding, $(n-1)^3 = n^3 - 3n^2 + 3n - 1$
$n^3 - (n-1)^3 = 3n^2 - 3n + 1$
$ \sum_{n=1}^k n^3 - (n-1)^3 = 3\sum_{n=1}^kn^2 - 3\sum_{n=1}^kn + \sum_{n=1}^k1$
$ k^3 = 3\sum_{n=1}^kn^2 - 3\dfrac{k(k+1)}{2} + k$
$ 3\sum_{n=1}^kn^2 = k^3 + 3\dfrac{k(k+1)}{2} - k$
$ \sum_{n=1}^kn^2 = \dfrac{k(k+1)(2k+1)}{6}$
