Problem: Using Cauchy criteron, examine the sequence for convergence: $\sum_{k=1}^{n}\frac{sin(2k+3)}{k(k+2)}$.
I need to check that $\forall$ $\epsilon>0$ $\exists N$: $\forall n \ge N$, $\forall p$ $\in$ $\mathbb{N}$ |$x_{n+p}$ $-$ $x_n$| < $\epsilon$.
$\lvert$$x_{n+p}$ $-$ $x_n$$\rvert$ is basically $\sum_{k=n+1}^{n+p} \frac{sin(2k+3)}{k(k+2)}|$. Using triangle inequality:
|$\sum_{k=n+1}^{n+p} \frac{sin(2k+3)}{k(k+2)}|$ $\le$ $\sum_{k=n+1}^{n+p} |\frac{sin(2k+3)}{k(k+2)}|$.
Sine is bounded function and its values do not exceed $1$. Also, decreasing the denominator of a fraction, will increase the value of the fraction. Therefore, if in all these fractions the denominators are replaced by the denominator of the very first fraction, we will increase the sum.
$\sum_{k=n+1}^{n+p} |\frac{sin(2k+3)}{k(k+2)}|$ $\le$ $\frac{1}{(n+1)(n+3)}+\underbrace{...}_\text{$p$ times}+\frac{1}{(n+1)(n+3)}$ $=$ $\frac{p}{(n^2+4n+3)}$.
Let $p=n^3$. Then, $\frac{p}{(n^2+4n+3)}=\frac{n^3}{n^2+4n+3}$, and this sequence diverges. Thus, the Cauchy criterion fails and the sequence diverges.
However, the correct answer is that the sequence converges. Where is the mistake in my chain of reasoning?
The mistake is that you have given a divergent upper bound for the difference of two terms of the sequence, but that doesn't prove that the differences themselves diverge. A divergent lower bound proves divergence, and a convergent upper bound proves convergence, but the converses are not valid.
Cartoon example: suppose I put $a_n = \sum_{k=1}^n 2^{-k}$. I note the true fact that $|a_{n+p}-a_n| \le p$, which is a divergent upper bound. Does that prove that $\{a_n\}$ diverges?