SOLUTION: a(n) stands for sequence a(1)=2, a(2)=3, {{{a(n+2)=a(n+1)+1/ln(a(n))}}} Prove that the limit when n->∞, {{{a(n)ln(a(n))/n}}} exists and evaluate it

Question 1177273: a(n) stands for sequence
a(1)=2, a(2)=3,
Prove that the limit
when n->∞,
exists and evaluate it
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Let $a(1) = 2$ and $a(2) = 3$. We have the recurrence relation $a(n+2) = a(n+1) + \frac{1}{\log(a(n))}$.
First, we observe that $a(n)$ is an increasing sequence. Since $a(1) = 2 < 3 = a(2)$, we can show by induction that $a(n) < a(n+1)$ for all $n$.
Assume $a(k) < a(k+1)$ for all $k \le n$. Then $a(n+1) > a(n)$, so $\log(a(n+1)) > \log(a(n))$, and $\frac{1}{\log(a(n))} > 0$.
Thus, $a(n+2) = a(n+1) + \frac{1}{\log(a(n))} > a(n+1)$. Therefore, $a(n)$ is an increasing sequence.
Since $a(n)$ is increasing, $\log(a(n))$ is also increasing.
Also, we can observe that $a(n) \to \infty$ as $n \to \infty$. If $a(n)$ were bounded, it would converge to some limit $L$. Then $L = L + \frac{1}{\log(L)}$, which implies $\frac{1}{\log(L)} = 0$, a contradiction. Thus, $a(n) \to \infty$.
Let's consider the differences:
$a(n+2) - a(n+1) = \frac{1}{\log(a(n))}$
Summing from $n=1$ to $N-2$, we get:
$\sum_{n=1}^{N-2} (a(n+2) - a(n+1)) = \sum_{n=1}^{N-2} \frac{1}{\log(a(n))}$
The left-hand side is a telescoping sum:
$a(N) - a(2) = \sum_{n=1}^{N-2} \frac{1}{\log(a(n))}$
$a(N) = 3 + \sum_{n=1}^{N-2} \frac{1}{\log(a(n))}$
Since $a(n) \to \infty$, $\log(a(n)) \to \infty$. Thus, $\frac{1}{\log(a(n))} \to 0$.
The sum $\sum_{n=1}^{N-2} \frac{1}{\log(a(n))}$ grows slowly.
Let's consider $a(n+1) - a(n-1) = \frac{1}{\log(a(n-2))}$.
We have $a(n+1) \approx a(n)$, so $a(n+1) - a(n) \approx 0$.
Since $a(n) \to \infty$, $\log(a(n)) \to \infty$, so $\frac{1}{\log(a(n))} \to 0$.
We have $a(n+2) - a(n+1) = \frac{1}{\log(a(n))}$.
Summing from $n=1$ to $N-1$, we get $a(N+1) - a(2) = \sum_{n=1}^{N-1} \frac{1}{\log(a(n))}$.
Since $a(n)$ is increasing, we have $\frac{1}{\log(a(n))}$ decreasing.
Thus, $\sum_{n=1}^{N-1} \frac{1}{\log(a(n))} \approx \int_1^{N-1} \frac{1}{\log(a(x))} dx$.
Since $a(n+2) - a(n+1) = \frac{1}{\log(a(n))}$, we expect $a(n) \approx cn$ for some $c$.
Then $\log(a(n)) \approx \log(cn) = \log(c) + \log(n)$.
Thus, $a(N) \approx \sum_{n=1}^{N-2} \frac{1}{\log(cn)}$.
Let $a(n) \sim cn$. Then $cn+2c-cn-c = \frac{1}{\log(cn)}$.
$c = \frac{1}{\log(cn)}$.
$c \log(cn) = 1$.
$c (\log(c) + \log(n)) = 1$.
$c \log(n) \approx 1$.
$c \approx \frac{1}{\log(n)}$.
Then $a(n) \approx n/\log(n)$.
Let $a(n) \approx c n$. Then $c(n+2)-c(n+1)=\frac1{\log(cn)}$
$c \approx \frac1{\log(cn)}$. Thus $c \log(cn) \approx 1$.
$c(\log c + \log n) \approx 1$, so $c \log n \approx 1$. Thus $c\approx 1/\log n$.
$a(n) \approx n/\log n$.
Then $\frac{a(n) \log(a(n))}{n} \approx \frac{(n/\log n) \log(n/\log n)}{n} = \frac{\log n - \log\log n}{\log n} \to 1$.
Final Answer: The final answer is $\boxed{1}$

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