Let $\{x_n\}$ be sequence of real numbers such that $\lim_{n\to\infty}({x_{n+1}}-{x_n})=5$. Then find $\lim_{n\to\infty}\frac{x_n}{n \log n}$.

Question

I am unable to solve this problem. I tried using the definition of the limit of sequences. I took an example by considering a sequence $x_n= 5n$. Then I got that, the limit was zero.

$x_{n+1}-x_n=(5n+5)-5n=5$. Then,

$$\lim_{n\to\infty}\frac{x_n}{n \log n}=\lim_{n\to\infty}\frac{5n}{n \log n}=\lim_{n\to\infty}\frac{5}{\log n}=0$$.

But I am unable to solve it in general. Please help me. Thank you so much!!!

Answer 1

Just apply the Stolz-Cesaro theorem, as $n\log(n)$ is strictly monotone and approaches $+\infty$, and $$\lim_{n\rightarrow\infty} \frac{x_{n+1}-x_n}{(n+1)\log(n+1)-n\log(n)}=\lim_{n\rightarrow\infty}\frac{x_{n+1}-x_n}{n\log(\frac{n+1}{n})+\log(n+1)}=0$$

because $x_{n+1}-x_n$ converges hence bounded while $n\log(\frac{n+1}{n})+\log(n+1)\ge n$ approaches $\infty$.

Answer 2

As mentioned in the comments, there is an $N$ such that for $n \ge N$, $4 < x_{n+1} - x_n < 6$. Then, for all $n\ge N$,

$$x_N + 4(n-N) < x_n = x_N + \sum_{k=N}^{n-1} x_{k+1} - x_k < x_N + 6(n-N)$$

And now we divide by $n \log n$ and use the Sandwich theorem, so the sought limit is $0$.

Answer 3

Every convergent sequence is bounded. Thus $|x_{n+1}-x_n|\le C$ for some $C> 0$ and every $n.$ WLOG we may assume that $C\ge |x_1|.$ By the triangle inequality $$|x_n|\le \sum_{k=2}^n|x_k-x_{k-1}|+|x_1|\le Cn$$ and the conclusion follows.