Why does Stolz- Cesaro fail to evaluate the limit of $\dfrac{n + n^2 + n^3 + n^4 + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$, I need to find the limit of the sequence
$\dfrac{n + n^2 + n^3 + n^4  + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$,
My strategy is to use Stolz's Cesaro theorem for this sequence.
Now, the numerator is given by :
$x_r = n^1+ n^2 +n^3 + \ldots +n^r$, so $x_{n+1} - x_{n} = n^{n+1}$
Similarly for denominator
$y_r = 1^n + 2^n + 3^n +\ldots +r^n$, so $y_{n+1}- y_{n} = (n+1)^n$
Using Stolz Cesaro, this limit is equivalent to 
$\displaystyle \lim \dfrac{n^{n+1}}{(n + 1)^n}$, which diverges to $ +\infty$,
However ans given to me is $\dfrac{e-1}{e}$, Can anyone tell where is the error in my solution ?
Thanks.
 A: Note that, as mentioned in the comments below, your computation of the ratio is incorrect. Regardless, the hypothesis of Stolz-Cesaro assumes that the limit $\lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n}$ exists. If it doesn't exist, it does not imply that the original limit does not exist.
A better way to approach is to write it as follows:
$$
\frac{n + n^2 + \cdots + n^n}{1^n + 2^n + \cdots +n^n} = \frac{n^{-(n-1)} + n^{-(n-2)} + \cdots + n^{-1} + 1}{\left(\frac{1}{n}\right)^n + \left(\frac{2}{n}\right)^n  + \cdots + \left(\frac{n-1}{n}\right)^n + 1}
$$
As $n \to \infty$, clearly the numerator $\to 1$. For the denominator, see this.
A: Divide nunberator and denominator by $n^n$. So your question consists of two limits, numerator and denominator, we'll deal with them separately.
For the numerator the limit would become $lim_{n \to \infty} 1+\frac{1}{n}+\ldots+\frac{1}{n^n} = 1*\frac{(1/n)^{n+1}-1}{(1/n)-1} = \lim_{h \to 0} \frac{h^{1+1/h} -1}{h-1} = -[e^{(1+1/h)\ln(h)} -1] = -[e^{-\infty}-1]=1$
For the denominator I'll prove a even more general limit, for any constant $k \neq 0$
$$\lim_{n to \infty}\frac{1^{kn}+2^{kn}+\ldots+n^{kn}}{n^{kn}}=\sum_{r=1}^{n} \frac{r^{kn}}{n^{kn}} = \sum_{r=0}^{n-1} \frac{(n-r)^{kn}}{n^{kn}}= \sum_{r=0}^{n-1} (1-r/n)^{kn} = \sum_{r=0}^{n-1} e^{-rk}  = \frac{1}{1-e^{-k}} = \frac{e^k}{e^k-1}$$.
Here k=1, so final answer is $\frac{1}{(e/e-1)}=\frac{e-1}{e}$
A: If you approximate the sums with the corresponding integrals it converges to 0
$$
L = \frac{\int_{1}^{n}n^x dx}{\int_{1}^{n} x^n dx} = \frac{n^{n+1} + n^n }{n^{n+1}\log n - \log n} = 0
$$
