Question:
Does there exist a positive sequence $\{a_{n}\}$ such $$\lim_{n\to\infty}\dfrac{n(a_{n+1}-a_{n})+1}{a_{n}}=0?$$
If it exists, can you make an example? if not, why not?
My try: we consider this sequence $$a_{n}=\dfrac{1}{n}$$ then the $$\lim_{n\to\infty}\dfrac{n(a_{n+1}-a_{n})+1}{a_{n}}=\lim_{n\to\infty}\dfrac{\dfrac{n}{n(n+1)}+1}{\dfrac{1}{n}}\to+\infty$$ But I can't take a example such condition? Thank you
Its better to rewrite your expression as: $$\lim_{n\to\infty}\dfrac{n \Delta a_{n}+1}{a_{n}}$$ where $\Delta$ is the difference operator. The difference operator is the discrete analogue of continuous differentiation. If you consider the continuous version of your limit: $$\lim_{x\to\infty}\dfrac{x f'(x) + 1 }{f(x)}$$ then it is clear that you have to look at functions that grow much slower than their values. The identity function $f(x) = x$ grows actually too fast. The next function to consider is $f(x) = \mathit{ln}(x)$: $$\lim_{x\to\infty}\dfrac{x \mathit{ln}'(x) + 1 }{\mathit{ln}(x)} = \lim_{x\to\infty}\dfrac{2}{\mathit{ln}(x)} = 0$$ All of the above carries to the discrete case --- where the analogue of the logarithmic function is the harmonic sequence $H_n = \sum_{k=1}^n \frac1k$.
There is also a more systematic way to guess the logarithmic function. Let us assume that $\lim_{x\to \infty} f(x) = \infty$. Then: $$\lim_{x\to\infty}\dfrac{x f'(x) + 1 }{f(x)} = \lim_{x\to\infty}\dfrac{x f'(x)}{f(x)}$$ The function $\frac{f'(x)}{f(x)}$ may be rewritten as $\mathit{ln}(f(x))'$. Therefore, our task is to find a function $f$ such that: $$\lim_{x\to\infty}x\mathit{ln}(f(x))' = \lim_{x\to\infty}\dfrac{\mathit{ln}(f(x))'}{\mathit{ln}(x)'} = 0$$ Now the heuristic goes as follows: under some reasonable conditions if the above limit exists, then there exists limit: $$\lim_{x\to\infty}\dfrac{\mathit{ln}(f(x))}{\mathit{ln}(x)} = 0$$ and is also equal to the same value. Substituting $e^y$ for $x$ yields: $$\lim_{y\to\infty}\dfrac{\mathit{ln}(f(e^y))}{y} = 0$$ and the choice $f(x) = \mathit{ln}(x)$ seems most natural (of course, one still needs to check that the logarithmic function really works!).
If we take $a_n=H_n$ to be the $n^\text{th}$ Harmonic number, then $$ \begin{align} \frac{n(H_{n+1}-H_n)+1}{H_n} &=\frac{\frac{n}{n+1}+1}{H_n}\\ &\le\frac2{H_n}\tag{1} \end{align} $$ and since the Harmonic Series diverges, the right side of $(1)$ tends to $0$.
If you chose a constant sequence $a_n = C$, your limit is equal to $\frac1C$: $$\lim_{n\to\infty}\frac{n(a_{n+1} - a_n) + 1}{a_n} = \lim_{n\to\infty}\frac{n(C-C) + 1}{C} = \frac1C$$ So if you chose an "almost" cauchy-sequence with unbounded limit, it will work. An example of this is $$a_n = \log n$$ Yielding $$\lim_{n\to\infty} \frac{n(\log(n+1) - \log(n)) + 1}{\log(n)} = 0$$ Which you can prove using l'Hospital, for example.
Take $a_n=\ln(n)$:
$$\lim_{n\to\infty}\frac {n\cdot\ln\left(\frac{n+1}{n}\right)+1}{\ln n}$$
Can you take it from here?
EDIT: the intuition behind taking a logarithmic function is that you need a slowly growing function $f$ such that the difference between the values of $f$ for consecutive integers is small, so it will cancel the linear factor, but as $n$ goes to infinity $f(n)$ goes to infinity as well.
Notice that taking $a_n =\sqrt{n}$ will yield limit of a constant, so we do need a "slower" growingfunction.
