$|a_{n+1}-a_n|\le \frac{\ln(n+1)}{n^2}$ Prove that $a_n$ is convergent

Question

Given a sequence $\{a_n\}$ such that for all $n$:

$$|a_{n+1}-a_n|\le \frac{\ln(n+1)}{n^2}$$

Prove that $a_n$ is a convergent sequence. I was thinking of using Cauchy but I can't quite put my finger on it.

Please any help.

Answer 1

Once you know that $\displaystyle \sum_{n=1}^\infty \frac{\ln (n+1)}{n^2}$ is convergent, the comparison test tells you that $\displaystyle \sum_{n=1}^\infty |a_{n+1} - a_n|$ is convergent, and the absolute convergence test tells you that $\displaystyle \sum_{n=1}^\infty (a_{n+1} - a_n)$ is convergent. Write $$ a_k = a_1 + \sum_{n=1}^{k-1}(a_{n+1} - a_n)$$ to conclude via the usual limit law that $\{a_k\}$ converges and $$ a_k \to a_1 + \sum_{n=1}^\infty (a_{n+1} - a_n), \quad k \to \infty.$$

Answer 2

It is a Cauchy sequence since for every $p\ge 1$

$$\begin{align}|a_{n+p}-a_n| &\le \sum_{j=1}^{p}|a_{n+j}-a_{n+j-1}| \le\sum_{j=1}^{p} \frac{\ln(n+j)}{(n+j-1)^2}\\ &\le \sum_{j=1}^{p} \frac{\ln(n+j)}{n^2} =\frac{\ln(n)}{n^2}\sum_{j=1}^{p}\ln(1+\frac{j}{n}) \\&\le C\frac{\ln(n)}{n}\to 0\end{align} $$

Indeed $\frac{1}{n}\sum_{j=1}^{p}\ln(1+\frac{j}{n})$ is a bounded sequence since by Riemann sum, we have $$ \frac{1}{n}\sum_{j=1}^{p}\ln(1+\frac{j}{n}) \le\frac{1}{n}\sum_{j=1}^{\max(p,n)}\ln(1+\frac{j}{n}) \to\int_0^1\ln(1+x)dx$$

Answer 3

Since $\log\left(\frac{n+1}{n+2}\right)\ge-\frac1{n+1}$, we have $$ \begin{align} \frac{\log(n+1)}{n-1}-\frac{\log(n+2)}n &=\frac{\log(n+1)+n\log\left(\frac{n+1}{n+2}\right)}{n(n-1)}\\ &\ge\frac{\log(n+1)}{n^2}-\frac1{(n+1)(n-1)} \end{align} $$ Therefore, $$ \begin{align} &\sum_{n=1}^m\frac{\log(n+1)}{n^2}\\ &=\log(2)+\sum_{n=2}^m\frac{\log(n+1)}{n^2}\\ &\le\log(2)+\sum_{n=2}^m\left(\frac{\log(n+1)}{n-1}-\frac{\log(n+2)}n\right)+\sum_{n=2}^m\frac1{(n+1)(n-1)}\\ &=\log(2)+\log(3)-\frac{\log(m+2)}m+\frac12\sum_{n=2}^m\left(\frac1{n-1}-\frac1{n+1}\right)\\ &=\log(2)+\log(3)-\frac{\log(m+2)}m+\frac12\left(1+\frac12-\frac1m-\frac1{m+1}\right)\\ \end{align} $$ which is bounded above by $\log(6)+\frac34$.

Since $\sum\limits_{n=1}^m\frac{\log(n+1)}{n^2}$ is an increasing sequence bouned above, it converges.

Since $\sum\limits_{n=1}^\infty|a_{n+1}-a_n|$ converges, the sequence is Cauchy and therefore converges.