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I want to determine the limit of the following sequence

$$x_n=\frac{\frac{2}{1}+\frac{3^2}{2}+\frac{4^3}{3^2}+...+\frac{(n+1)^n}{n^{n-1}}}{n^2}$$

It is clear that the sequence $x_n$ is a sequence of the form $z_n=\frac{u_n}{v_n}$, with this try to apply Stolz's criterion by finding the limit of $$\frac{u_{n+1}-u_n}{v_{n+1}-v_n}$$

However, I was not able to reach the specific limit. I would be very grateful if someone could help me.

jimjim
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Wrloord
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3 Answers3

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After applying Stolz-Cesàro we get

$$ \lim_{n\to\infty}\frac{\sum_{k=1}^n\frac{(k+1)^k}{k^{k-1}}}{n^2} = \lim_{n\to\infty}\frac{\frac{(n+1)^n}{n^{n-1}}}{n^2-(n-1)^2} = \lim_{n\to\infty}\frac{(\frac{n+1}{n})^{n}}{2-1/n} = e/2 $$

jjagmath
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This directly follows from Stolz–Cesàro theorem

\begin{align*} &\lim_{n\to\infty}\frac{\sum_{k=1}^n\frac{(k+1)^k}{k^{k-1}}}{n^2}\\ =&\lim_{n\to\infty}\frac{\frac{(n+1)^n}{n^{n-1}}}{n^2-(n-1)^2}\\ =&\lim_{n\to\infty}\frac{\left(\frac{n+1}{n}\right)^{n}}{2-\frac 1n}\\ =&\frac e2 \end{align*}

Sayan Dutta
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  • Any explanation for the downvote? Anything wrong with my answer? – Sayan Dutta Jan 09 '22 at 16:51
  • May be the downvote is because your answer and mine are the same, but mine come out a few minutes before yours, so yours is redundant. (I didn't downvote) – jjagmath Jan 09 '22 at 18:45
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we have : $$\frac{(n + 1)^n}{n^{n - 1}} = \frac{n^n}{n^{n - 1}} \left(1 + \frac{1}{n}\right)^n = n \left(1 + \frac{1}{n}\right)^n \sim n e$$ then : $$\sum_{k = 1}^n \frac{(k + 1)^k}{k^{k - 1}} \sim \sum_{k = 1}^n k e = e \frac{n (n + 1)}{2}$$ So ; $$\frac{\sum_{k = 1}^n \dfrac{(n + 1)^n}{n^{n - 1}}}{n^2} \sim e \frac{n (n + 1)}{2 n^2} \to \frac{e}{2}$$

Essaidi
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  • Your answer suggests that from a asymptotic equivalence $a_n \sim b_n$ it follows that $\sum_{k=1}^n a_k \sim \sum_{k=1}^n b_k$ which is false in general. – jjagmath Jan 09 '22 at 15:26
  • It's true when the series diverges and non negative which is the case. – Essaidi Jan 09 '22 at 15:28
  • I know, it's called Stolz-Cesàro theorem, look at the other two answers. The point is that there's no mention of it in your proof. – jjagmath Jan 09 '22 at 15:31
  • I don't know why my answer is considered as not useful !!! – Essaidi Jan 09 '22 at 16:50