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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?

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  • $\begingroup$ You are missing a log when re-writing the limit. Also at the beginning you use $< n(n^2)$ which can lead to an overestimate in the limit, i.e., the conclusion is that if the quantity in the parentheses tends to anything than it is at most $1$. $\endgroup$
    – Gary
    Nov 16, 2023 at 3:33
  • $\begingroup$ You applied the theorem incorrectly. The sum $1\cdot n + 2 \cdot (n-1) + ... + n \cdot 1$ is not of the form $\sum_{k=1}^n f(k)$, since each term depends on $k$ and on $n$. So your can't apply Cesaro-Stolz theorem. $\endgroup$
    – jjagmath
    Nov 16, 2023 at 3:36
  • $\begingroup$ The term inside parentheses tends to $0$, not $1$. $\endgroup$
    – jjagmath
    Nov 16, 2023 at 3:44
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    $\begingroup$ @Gary I have used the well known result for limits of the form $\lim_{n \rightarrow \infty} (f(x))^g(x) = e^{(f(x)-1)g(x)}$ where $f(x) \rightarrow 1$ and $g(x) \rightarrow \infty$ as $x \rightarrow \infty$ $\endgroup$ Nov 16, 2023 at 3:54
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    $\begingroup$ Ok, that part makes sense now. Still, the application of the Cesaro-Stolz theorem is incorrect. $\endgroup$
    – jjagmath
    Nov 16, 2023 at 3:54

3 Answers 3

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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.

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As a note:

The limit, by direct expansion, can be seen as: \begin{align} L &= \lim_{n \to \infty} \left( 1 + \frac{\sum_{k=1}^{n} k (n+1-k)}{\sum_{k=1}^{n} k^3} \right)^n \\ &= \lim_{n \to \infty} \left( 1 + \frac{\binom{n+2}{3}}{\binom{n+1}{2}^2} \right)^{n} \\ &= \lim_{n \to \infty} \left(1 + \frac{2}{3} \, \frac{n+2}{n \, (n+1)} \right)^{n} \\ &= \lim_{n \to \infty} \left( e^{2/3} + \left(\frac{2}{3}\right)^2 \, \frac{(6 - e^{2/3})}{n} + \mathcal{O}\left(\frac{1}{n^2}\right) \right) \\ &= e^{2/3}. \end{align}

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$\sum\limits_{k=1}^n k(n-k+1) =\sum\limits_{k=1}^n kn -\sum\limits_{k=1}^n k^2 +\sum\limits_{k=1}^n k =(n-1)\frac{n (n+1)}{2}-$ $\frac{(n)(n+1)(2n+1)}{6}$ since $\sum\limits_{k=1}^n k^3 =\frac{n^2 (n+1)^2}{4}$

$$M_n:= \frac{\frac{n (n^2-1)}{2}-\frac{(n)(n+1)(2n+1)}{6}}{\frac{n^2 (n+1)^2}{4}} = \frac{\frac{n+1}{6} (3n^2-3n -2n^2-n)}{\frac{n^2 (n+1)^2}{4}}=\frac{2(n+1) (n^2-4n )}{3{n^2 (n+1)^2}} $$

$$\lim_{n \to \infty} M_n =0 $$

$$\lim_{n \to \infty} L_n = (1+M_n)^n = \lim_{n \to \infty} (1+M_n)^{\frac{M_n}{M_n}n}=\lim_{n \to \infty}e^{{n}{M_n}} $$

$$\text{and } \ \lim\limits_{n\to \infty } {n}{M_n} =\frac{2n(n+1) (n^2-4n )}{3{n^2 (n+1)^2}} =\frac{2}{3}$$ so $$\text{so} \ \lim\limits_{n \to \infty} L_n = e^{\frac{2}{3}}$$

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