I am trying to prove this theorem about sequences.
Define the sequence $s_n = t_{n+1} - t_n$, which is monotone by assumption, where $(t_n)$ is a bounded sequence of real numbers. Prove that $(t_n)$ is convergent.
I'm not sure if this fact could not be proved from the definition of convergence, so it may require an application of the fact that any bounded, monotonic sequence converges (an increasing sequence to its supremum, a decreasing sequence to its infimum). Since $(t_n)$ is bounded, it suffices to show that it is monotonic. This is where I am struggling.
This is my updated attempt after reading some of the responses.
Since $(t_n)$ is bounded, $|t_n| \leq M$ for any $n$. First, we prove that $(s_n)$ is bounded. By applying the triangle inequality, \begin{align*} |s_n| = |t_{n+1} - t_n| \leq |t_{n+1}| + |-t_n| \leq |t_{n+1}| + |t_n| \leq 2M. \end{align*} Thus, $(s_n)$ is monotone and bounded and hence converges. It is, likewise, a Cauchy sequence. To show that $(t_n)$ converges, we show (for some finite $t$): \begin{align*} \forall \epsilon > 0, \; \exists N, \; \forall n > N, \; |t_n - t| < \epsilon. \end{align*}
Here, I am stuck, because I would need a "guess" for $t$ in order to prove this. It's clear to me that if $s$ exists, it has to equal $0$ since $t_{n+1}, t_n \to 0$. I cannot directly assert this, though, as to do so would be to assert that the limit of $(t_n)$ exists.
Any help on this would be greatly appreciated.
You have some wrong inequalities. Here is a correct proof: Since $s_n$ is monotonic and bounded it has a finite limit. Call it $s$. By Cesaro's Theorem $\frac {s_1+s_2+\cdots+s_n} n \to s$. This says $\frac {t_{n+1}-t_1} n \to s$. But boundedness of $(t_n)$ implies that this limit is $0$. Hence $s=0$. But $(s_n)$ is increasing , so $s_n \leq 0$ for all $n$. This means $t_{n+1} \leq t_n$ for all $n$. Now $(t_n)$ is bounded and monotonic, hence convergent.