Let $a_{n}$ be a sequence as $|a_{n+1}-a_{n}|\leq\frac{\ln(n+1)}{n^{2}}$ for every $n\in\mathbb{N}$. Prove that $a_{n}$ converges.

Question

Let $a_{n}$ be a sequence as $|a_{n+1}-a_{n}|\leq\frac{\ln(n+1)}{n^{2}}$ for every $n\in\mathbb{N}$.

Prove that $a_{n}$ converges.

I was given a hint: prove that $\sum_{n=1}^{\infty}\frac{\ln(n+1)}{n^{2}}$ converges.

I tried to prove it using Cauchy condensation test. I got that it converges if and only if $$\sum_{n=1}^{\infty}\frac{2^{n}\ln(2^{n}+1)}{2^{2n}} = \sum_{n=1}^{\infty}\frac{\ln(2^{n}+1)}{2^{n}}$$ converges, which isn't very helpful. (I know it converges, but I can't prove it.)

Answer 1

You can also do it using the Cauchy condensation test like you tried:

$$\frac{\ln(2^n+1)}{2^n} \le \frac{\ln(2^n+2^n)}{2^n} = \frac{\ln(2^{n+1})}{2^n} = \frac{n+1}{2^n}\ln 2$$

Now prove that for $n \ge 6$ we have $n+1 \le (\sqrt{2})^n$ so

$$\sum_{n=1}^\infty \frac{\ln(2^n+1)}{2^n} \le \sum_{n=1}^5 \frac{\ln(2^n+1)}{2^n} + (\ln 2)\sum_{n=6}^\infty \frac1{(\sqrt{2})^n} < +\infty$$ because the latter is a geometric series.

Hence $\sum_{n=1}^\infty \frac{\ln(n+1)}{n^2}$ converges so $(a_n)_n$ is a Cauchy sequence.

Answer 2

Note that $$ a_{n}=a_{0}+\sum_{k=0}^{n-1}(a_{k+1}-a_{k}). $$ Let $n\to \infty$. The sequence of partial sums on the right converge, and thus the sequence $(a_{n})$ also converges, since $$ \sum_{n=0}^\infty|a_{n+1}-a_{n}|\leq \sum_{n=0}^\infty \frac{\ln (n+1)}{n^{2}}<\infty\tag{1} $$ Indeed $\dfrac{\ln (n+1)}{\sqrt{n}}\to 0.\;$ In particular $\ln (n+1)\leq \sqrt{n}$ for sufficiently large $n$ and thus by comparison to the series $\sum n^{-3/2}$, we have the result in (1).