MHB Understanding Infinite Sequences: Difference of 1 & n/(n+1)

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The discussion focuses on the limit of the sequence defined by \( a_n = \frac{n}{n+1} \) as \( n \) approaches infinity. It highlights that the difference between 1 and \( a_n \), expressed as \( 1 - \frac{n}{n+1} = \frac{1}{n+1} \), approaches 0 as \( n \) increases. This demonstrates that the sequence converges to 1, allowing the difference to be made arbitrarily small with sufficiently large \( n \). The limit notation \( \lim_{n \to \infty} \frac{n}{n+1} = 1 \) encapsulates this concept. Understanding this limit is crucial for grasping the behavior of infinite sequences.
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My textbook reads :

The graph of $$a_n=\frac{n}{n+1}$$ are approaching 1 as n becomes large . In fact the difference
$$1-\frac{n}{n+1}=\frac{1}{n+1}$$ can be made as small as we like by taking n sufficently large. We indicate this by writing $$\lim_{n \to \infty} \frac{n}{n+1}=1$$

I don't understand where they pull $$\frac{1}{n+1}$$ from and what difference they refer to?
 
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They refer to the difference
$$1-\frac{n}{n+1}$$
If $n \to \infty$ we have $\frac{1}{n+1} \to 0$ and thus $1-\frac{n}{n+1} \to 0$. In other words, the difference can be made as small as you like by taking $n$ sufficently large.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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