Is it true that $s_{n+1} - s_n \to 0$ in a convergent sequence? Is it correct to say that if $s_n$ is a convergent sequence, then $s_{n+1} - s_n \to 0$. It feels right but I don't know how to test it rigorously. I am asking this because I am using this to prove a few questions.
Let $t_n$ be a sequence with:
$$t_1 = 1$$
$$t_n = \sqrt{t_{n-1} +1}$$
To find if the sequence converges to something, I used $t_{n+1} - t_n \to 0$, or $t_n \to t_{n+1}$.
Thus, setting $t_n = \sqrt{t_{n-1} +1} = t_{n-1}$, (again, not sure how to justify this rigorously), but this gives a quadratic on squaring both sides, which gives $\frac{1+\sqrt{5}}{2}$ as the possible value of $t_{n-1}$ where this equality happens, and thus I infer that that is the limiting value. But this seems a really bad proof. Any advice for improvement?
 A: Yes $s_n - s_{n+1} \to 0$. Here is a short proof. Let $s$ be the limit (i.e. $s_n \to s$)
$$|s_n - s_{n+1}| = |s_n - s + s - s_{n+1}| \le |s_n - s| + |s - s_{n+1}| \to 0$$.
A: There is a result which claims that a sequence converges iff every subsequence converges to the same value. More precisely, given a real-valued sequence $(s_{n})_{n\in\mathbb{N}}$, we say that $(t_{n})_{n\in\mathbb{N}}$ is a subsequence of $(s_{n})_{n\in\mathbb{N}}$ iff there exists a strictly increasing function $f:\mathbb{N}\to\mathbb{N}$ such that $t_{n} = s_{f(n)}$. If we denote by $s$ the limit of the given sequence and define $f(n) := n + 1$, we may claim from the limits' properties that:
\begin{align*}
\lim_{n\to\infty}(t_{n} - s_{n}) = \lim_{n\to\infty}t_{n} - \lim_{n\to\infty}s_{n} = s - s = 0.
\end{align*}
Hopefully this helps!
A: For your second question, you asked if
$$t_n=\sqrt{t_{n-1}+1}$$
implies $t_n\to \frac{1+\sqrt{5}}{2}=\phi$? It does not, but you are on the right track. The result you are using implicitly here is the following: if

*

*$x_n=f(x_{n-1},x_{n-2},...,x_{n-m})$

*$\lim_{n\to\infty}x_n=L$

*$f(y_1,y_2,...,y_m)$ is continuous at $f(L,L,...,L)$
then
$$L=f(L,L,...,L)$$
In your case

*

*$f(x)=\sqrt{x+1}$

*$t_n=f(t_{n-1})=\sqrt{t_{n-1}+1}$
We have to show that $\lim_{n\to\infty}t_n=L> -1$ since $\sqrt{x+1}$ is continuous for all $x>-1$. To show that the limit exists and is positive (clearly greater than $-1$) we will show two things: it is increasing and it is bounded. This is enough to show that it converges to a positive number. Specifically, we will show $t_{n+1}-t_{n}>0$ and it is bounded above by $\phi$ and below by $1$. We will do this by two induction proofs:
Base case 1: For $n=1$
$$t_1\geq 1$$
Induction 1: Assume $t_{n-1}>1$. Then
$$t_n=\sqrt{t_{n-1}+1}>\sqrt{1+1}=\sqrt{2}>1$$
Base case 2: For $n=1$ we have
$$t_2-t_1=\sqrt{2}-1>0$$
$$t_2=\sqrt{2}<\phi$$
Induction 2: Assume the proposition holds for some $n-1$. Then
$$t_{n}=\sqrt{t_{n-1}+1}<\sqrt{2+1}=\sqrt{3}<2$$
$$t_{n+1}-t_{n}=\sqrt{t_n+1}-t_n$$
Now, by our inductive assumption and our previous induction proof we know $1\leq t_n<\phi$. What can we say about the function $\sqrt{x+1}-x$ in the interval $[1,\phi)$? Well
$$\frac{d}{dx}\left[\sqrt{x+1}-x\right]=\sqrt{x+2}-\sqrt{x+1}-1$$
For $1\leq x\leq 2$ this is
$$\sqrt{x+2}-\sqrt{x+1}-1<\sqrt{2+2}-\sqrt{1+1}-1=1-\sqrt{2}<0$$
We conclude on the interval $[1,\phi)$ that $\sqrt{x+1}-x$ is decreasing. Since
$$\sqrt{\phi+1}-\phi=0$$
we may conclude
$$\sqrt{t_n+1}-t_n>0$$
and we are done.
Having shown that $t_n\to L$ for some $L>0$, we may now finish the proof using the theorem stated above. We have
$$L=\sqrt{L+1}\Rightarrow L=\phi$$
We conclude
$$\lim_{n\to\infty}t_n=\frac{1+\sqrt{5}}{2}$$
A: Some things stated also in the comments . Deducing that $t_n\to t_{n+1}$ from $t_n-t_{n+1}\to 0$ is not valid.  Also setting $t_n=t_{n-1}$ ie that $t_n$ is constant is not correct, neither and it leads you to wrong results. You may use the definition of limit to prove that $t_n, t_{n+1}$ have the same limit, if there such one.  And to prove that such a limit exists you need to prove,  if you like, that $t_n$ is bounded and monotonous .
