How can we evaluate the following limit? 
How can this problem be solved?
  $$
\lim_{(n,r) \rightarrow (\infty, \infty)} \frac{\prod\limits_{k=1}^{r} \left( \sum\limits_{i=1}^{n} i^{2k-1} \right)}{n^{r+1} \prod\limits_{k=1}^{r-1} \left( \sum\limits_{i=1}^{n} i^{2k} \right)}
$$ 

 A: The sum of the $k$ powers of the first $n$ natural numbers is a polynomial of degree $k+1$ with leading coefficient $\frac{1}{k+1}$, that is,
$$
\sum_{i=1}^n i^k
=
\frac{1}{k+1}n^{k+1} + p_k(n)
$$
where $p_k(n) \in O(n^k)$ is a polynomial of degree (at most) $k$.
It follows that
\begin{align}
\frac{
  \prod\limits_{k=1}^{r}
  \left(\sum\limits_{i=1}^{n} i^{2k-1} \right)
}{
  n^{r+1}
  \prod\limits_{k=1}^{r-1}
  \left(\sum\limits_{i=1}^{n} i^{2k} \right)
}
= {} &
\frac{
  \prod\limits_{k=1}^{r}
  \left(\frac{1}{2k}n^{2k} + O(n^{2k-1}) \right)
}{
  n^{r+1}
  \prod\limits_{k=1}^{r-1}
  \left(\frac{1}{2k+1}n^{2k+1} + O(n^{2k}) \right)
}
\end{align}
When passing to the limit for $n\to\infty$ only the highest order powers are relevant, so that this simplifies to
\begin{align}
\lim_{(n,r)\to(\infty,\infty)}
\frac{
  \prod\limits_{k=1}^{r}
  \left(\sum\limits_{i=1}^{n} i^{2k-1} \right)
}{
  n^{r+1}
  \prod\limits_{k=1}^{r-1}
  \left(\sum\limits_{i=1}^{n} i^{2k} \right)
}
= {} &
\lim_{(n,r)\to(\infty,\infty)}
\frac{
  \prod\limits_{k=1}^{r}
  \frac{1}{2k}n^{2k}
}{
  n^{r+1}
  \prod\limits_{k=1}^{r-1}
  \frac{1}{2k+1}n^{2k+1}
}
\\
= {} &
\lim_{(n,r)\to(\infty,\infty)}
\frac{
  \frac{1}{2^r r!}n^{r(r+1)}
}{
  n^{r+1}
  \frac{2^{r-1}(r-1)!}{(2r-1)!}
  n^{r^2-1}
}
\\
= {} &
\lim_{(n,r)\to(\infty,\infty)}
\frac{1}{2^{2r-1}}
\binom{2r-1}{r}
\end{align}
Above, I used the following equalities:


*

*$\quad\displaystyle
\prod_{k=1}^r n^{2k}
=
n^{\sum_{k=1}^{r} 2k}
=
n^{r(r+1)}
$

*similarly, $\quad\displaystyle
\prod_{k=1}^{r-1} n^{2k+1}
=
n^{\sum_{k=1}^{r-1} (2k+1)}
=
n^{r^2-1}
$

*and $\quad\displaystyle
\prod_{k=1}^{r-1}(2k+1)
=
\frac{(2r-1)!}{2^{r-1}(r-1)!}
$


I'll leave the evaluation of the limit to you now... :)
