# simple sequence proof

I'm very unsure of myself with proofs so I wanted to know if my reasoning here is correct and if this proof is valid:

Given a sequence ${a_n}$ which converges and a sequence ${b_n }$ that is bounded such that for every n the following is true, $b_{{n+1}}≤b_n+(a_{{n+1}}-a_n)$. Prove that $b_n$ converges.

Proof: Since $a_n$ converges we know that it is also bounded and monotonic, therefore: $|a_{{n+1}}-a_n |≤0$ for all $n$. (Otherwise $a_n$ would be unbounded and that’s contrary to the definition of a converging sequence)

If $b_n$ indeed converges, I would expect the following inequality to be true: $|b_{{n+1}}-b_n |≤0$ (The difference between consecutive terms must be decreasing and $0$ as $n$ reaches infinity)

Now I’m also given the inequality:

$b_{{n+1}}≤b_n+(a_{{n+1}}-a_n)$

$b_{{n+1}}-b_n<a_{{n+1}}-a_n$

$|b_{{n+1}}-b_n |≤|a_{{n+1}}-a_n |$

$|b_{{n+1}}-b_n |≤|a_{{n+1}}-a_n |≤0$ (according to my earlier statement about the definition of ${a_n}$.)

$|b_{{n+1}}-b_n |≤0$

According to the last inequality, the difference between the consecutive terms of $b_n$ are decreasing and eventually become zero as $n$ reaches infinity. Therefore $b_n$ converges. ∎

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Bounded sequences need not be monotonic. Consider the sequence $1,0,\frac{1}{2},0.\frac{1}{3},0,\frac{1}{4}, \cdots$. – user21436 Feb 4 '12 at 15:54
And, your proof is flawed in many other ways! – user21436 Feb 4 '12 at 15:55
yes but all converging sequences are monotonic, no? – nofe Feb 4 '12 at 15:56
could you detail why? That's why I've posted it on here, because I know it's flawed ... – nofe Feb 4 '12 at 15:58
your ability to accept criticism will make you great, good luck. – Arjang Feb 4 '12 at 17:14

Your intuition that that difference between two terms must decrease and eventually vanish is right upto intuition.

Definition of Convergence

First of all, a sequence $\{x_n\}$ is said to converge to $x$ if, to each $\epsilon>0$, there exists $N$ such that whenever $n>N$, $|x_n-x|< \epsilon$.

What is the intuitive meaning?

After a finite stage, the terms of the sequence are as close as we please, to the limit.

The sequences satisfying your intuition are called the Cauchy Sequences. They are more intricate and special objects on any space.

Your intuition confirms to the definition of convergence on real line and other spaces called "complete spaces". That is, Cauchy Sequences are convergent on real line and other complete spaces.

Using this definition, reflect on why the inequalities you tell us must be wrong. If you don't make much progress, ping me here and I'll be glad to tell you what it is!

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First, I must point out that if $b_n \to 0$ as $n \to \infty$ doesn't suffice to show the convergence of a series. It is necessary but not sufficient. You'd also need to show that for every $\epsilon >0$ $$|b_n -b| < \epsilon \text{ ; for } N \geq n$$

If $\left\{a_n\right\}$ converges, then $a_n \to 0$ as $n \to \infty$, thus $|a_{n+1}-a_{n}| < \varepsilon$ and $|a_n-a| < \epsilon$

What you need to prove if that a sequence is bounded and it's general term tends to zero, then it converges by the least upper bound property (i.e. every set of real numbers that has an upper bound has a LUB)

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