I know this limit can be evaluated with application of elementary methods, but having recently learnt the Cesaro-Stolz theorem, I tried applying the theorem on this limit.
We can see that $1\cdot n + 2 \cdot (n-1) + … + n \cdot 1 < n(n^2)$ and $1^3 + 2^3 + … + n^3 = \frac{n^2(n-1)^2}{4}$ thus the term inside parantheses tends to 1 as $n \rightarrow \infty$. Hence the limit can be written as
$$e^{\lim_{n \rightarrow \infty} n\left(\frac{1\cdot n + 2 \cdot (n-1) + … + n \cdot 1}{1^3 + 2^3 + … + n^3}\right) } $$
To evaluate it, we apply Cesaro-Stolz theorem (since the sequence in denominator is monotonically increasing), obtaining
$$ \lim_{n \rightarrow \infty} n\left(\frac{1\cdot n + 2 \cdot (n-1) + … + n \cdot 1}{1^3 + 2^3 + … + n^3}\right) = \lim_{n \rightarrow \infty} \frac{(n+1)\cdot 0 – n^2}{(n+1)^3 – n^3} = \lim_{n \rightarrow \infty} \frac{-n^2}{3n^2 – 3n + 1} = \frac{-1}{3} $$
Which gives the (wrong) value of limit as
$$ L = e^{-\frac{1}{3}} $$
The correct value is known to be $e^{\frac{2}{3}}$ which can be evaluated through other methods. Have I applied the theorem correctly?
Best Answer
As pointed out by other users, you computed the limit incorrectly. Note that
\begin{align*} \lim_{n \to \infty} n \cdot \frac{1\cdot n + 2 \cdot (n-1) + \cdots + n \cdot 1}{1^3 + 2^3 + \cdots + n^3} &= \lim_{n \to \infty} \cdot \frac{\sum_{k=1}^{n} \frac{k}{n}\left(1-\frac{k-1}{n}\right) \frac{1}{n}}{\sum_{k=1}^{n} \left(\frac{k}{n}\right)^3 \cdot \frac{1}{n}} \\ &= \frac{\int_{0}^{1} x(1-x) \, \mathrm{d}x}{\int_{0}^{1} x^3 \, \mathrm{d}x} \\ &= \frac{1/6}{1/4} = \frac{2}{3}. \end{align*}
Note. It is not impossible to apply Stolz–Cesaro theorem here, but it requires a bit nasty computation. Indeed, set
$$ A_n = n (1 \cdot n + 2 \cdot (n-1) + \cdots + n \cdot 1), \qquad B_n = 1^3 + 2^3 + \cdots + n^3. $$
Writing $A_n$ in summation notation, we get $A_n = n \sum_{k=1}^{n} k (n+1 - k)$. Hence,
\begin{align*} A_n - A_{n-1} &= n^2 + \sum_{k=1}^{n-1} [nk(n+1-k) - (n-1)k(n-k)] \\ &= n^2 + \sum_{k=1}^{n-1} (2nk - k^2). \end{align*}
Using this, we find
$$ \frac{A_n - A_{n-1}}{B_n - B_{n-1}} = \frac{1}{n} + \sum_{k=1}^{n-1} \left( \frac{2k}{n} - \frac{k^2}{n^2} \right) \, \frac{1}{n} \to \int_{0}^{1} (2x-x^2) \, \mathrm{d}x = \frac{2}{3} $$
and therefore $A_n/B_n \to \frac{2}{3}$ by Stolz–Cesaro theorem.