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This question was inspired by another question I asked on this site. For that question, I had thought that if the differences in subsequent evaluations of a function, $f(n)$, defined with a domain of only the natural numbers (including $0$) converged to $0$ as $n$ approached infinity, then this would show that $f(n)$ would converge to some value. I was wrong in this assumption. Another idea I had is that if the ratio of one evaluation to the previous converged to $1$, this would also imply existence of a limit. This seems more likely because of the ratio test, but that applies only to terms of an infinite sum and only implies convergence when $|a_n / a_{n-1}| < 1$. Is my assumption true? Can it be proven somehow by the ratio test? If not, what is a counterexample?

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    $\begingroup$ Another counterexample is $f(n)=n$. $\endgroup$ – rtybase May 3 '18 at 16:31
  • $\begingroup$ When you say that $\lim\limits_{n\to\infty} f(n)$ exists, do you mean that exists finite? $\endgroup$ – user May 3 '18 at 16:34
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Let consider as counterexample $f(n)$ such that

  • $f(n)=1$ for $n$ even
  • $f(n)=-1$ for $n$ odd
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As soon as I wrote this question, I believe I found a counterexample for it. Consider the function as partial sums of the harmonic series. $$f(n) = \sum_{k=1}^n \frac{1}{k}$$ If we consider the limit of the ratio, we get $$\lim_{n\to\infty} \left|\frac{f(n)}{f(n-1)}\right| = \lim_{n\to\infty} \frac{\sum_{k=1}^n \frac{1}{k}}{\sum_{k=1}^{n-1} \frac{1}{k}} = \lim_{n\to\infty} \frac{\frac{1}{n} + \sum_{k=1}^{n-1} \frac{1}{k}}{\sum_{k=1}^{n-1} \frac{1}{k}}$$ Separating this into two fractions gives us $$\lim_{n\to\infty} \left(1 + \frac{1}{n\sum_{k=1}^{n-1} \frac{1}{k}}\right)$$ Because the harmonic series diverges, the denominator of the second fraction approaches infinity. Even if we did not know that the series diverged, we know it is greater than zero. That multiplied by $n$ as $n$ approaches infinity would show the denominator to tend to infinity anyway. This leads us to the limit being equal to $1$. Because we already knew that $\lim\limits_{n\to\infty} f(n)$ does not exist, this leads to a contradiction and disproves the original statement.

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  • $\begingroup$ Why do you say that limit does not exist? Usually we say in these cases that $f(n) \to +\infty$. We say that a limit doesn't exist when for example $f(n)$ oscillates between two different values or other cases for which we can't define a limit (finite nor infinite) according to the definition. $\endgroup$ – user May 3 '18 at 16:42
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    $\begingroup$ @gimusi I was taught in my calculus class that saying a limit tends towards infinity was just a more descriptive statement implying that the limit does not exist. Saying the limit is equal to infinity is just an abbreviated way of saying it tends towards infinity. I believe the first answer on this question: math.stackexchange.com/questions/127689/… offers a good explanation. $\endgroup$ – Kirk Fox May 3 '18 at 16:54
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    $\begingroup$ Ok it's mainly a matter of convention and notation, I'm used to distinguish between the two cases according to the definition given. Now it is clear to me what you were asking for! Thanks $\endgroup$ – user May 3 '18 at 16:58

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