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I have a function $f(n)$ that is defined when $n$ is a non-negative integer. $f(n)$ is always a real number such that $1<f(n)<2$. Because of how this function is definied, I cannot prove the limit directly, but I was able to prove that $$\lim_{n\to\infty}[f(n)-f(n-1)]=0$$It makes sense to me that this might prove the existence of $\lim_{n\to\infty}f(n)$, but I do not know for sure if that is true. Is there some way to prove that this is true? And if so, is the fact that $f(n)$ is bounded between 1 and 2 necessary to the proof?

Kirk Fox
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    Modify a periodic function like $\sin x$ to conclude that this does not imply the existence of the limit. –  Apr 13 '18 at 05:05
  • Do you want to consider $\lim_{n\to \infty} f(n)$? You said that the function is defined on non-negative integer, not whole $\mathbb{R}$. So this seems more reasonable. – Seewoo Lee Apr 13 '18 at 05:11
  • I apologize, I did not realize the use of x implies all real numbers. I will change this. – Kirk Fox Apr 13 '18 at 05:16
  • Would it be helpful to add more context relating to the problem I am dealing with? – Kirk Fox Apr 13 '18 at 05:20
  • If you can bound the decay by a function that vanishes sufficiently fast, then it has a limit. For example, if $|f(n)-f(n-1)|\leq A/n^2$ for all $n \in {1, 2, 3, ...}$ and for some constant $A>0$, then it has a limit. More generally, it has a limit if $\sum_{n=2}^{\infty} |f(n)-f(n-1)|<\infty$. – Michael Apr 13 '18 at 05:54

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HINT: No. Let us think about a condition $0\leq f(n)\leq1$ (we can move and scale the values). Then we start:

up: $f(1)=0$, $f(2)=1$,

down: $f(3)=1/2$, $f(4)=0$,

up: $f(5)=1/4$, $f(6)=2/4$, $f(7)=3/4$, $f(8)=4/4=1$,

down: $f(9)=7/8$, $f(10)=6/8$, $f(11)=5/8$, $\ldots$. Can you see the pattern?

Przemysław Scherwentke
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Nope. Example:

$\sin(2\pi x)$.

marty cohen
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