Limit of ${n\log n}\left(\frac{\log(n+2)}{\log (n+1)}-1\right)$ I want to determine the limit of the following sequence
$x_n={n\log n}\left(\frac{\log(n+2)}{\log (n+1)}-1\right)$
To try to do so, I took into account that
$${n\log n}\left(\frac{\log(n+2)}{\log (n+1)}-1\right)=n\log n\frac{\log\frac{n+2}{n+1}}{\log(n+1)}$$
At this point I do not know how to proceed, I understand that this limit is 1 but I do not know how to arrive at this result.
Any help?
 A: Hint: Your start is good. Now use $\frac{\log n}{\log(n+1)}\to 1$ and $n\log\frac{n+2}{n+1}=\log\left(1+\frac{1}{n+1}\right)^n$
A: This is an example for calculating with known limits.
But this is a tricky example.
While the first factor is for itself divergent the second in convert. And that is worth the trick. The quotient of the to logarithm itself converges to 1. So the second factor converges to zero. The task is multiplying a divergent factor with a convergent one.
Use your calculator for proving for example for $10^5$ that this is slowly converging altogether. So my result is $0.999984$.
How to prove that $n \log(n)$ is diverging slower than $\frac{\log(n+2)}{\log(n+1)}-1$ is converging to zero?
It is possible to calculated Taylor expansion for greater value for both factors. Each time the second factor give a value that compensates the first. For $10^5$: $0.0000108556 * 92104.4037 = 0.99985 $.
This is a product of two limits $ \infty 0$ and it is finite because the second factor converges faster to zero than the first to infinity.
The proof is via l'Hopitals rule. The third row in the section "Other indeterminate forms". $\frac{1}{n log(n)}$ the the quotient.
A: We have :
$$x_n = \dfrac{\log(n)}{\log(n + 1)} \dfrac{\log \dfrac{n + 2}{n + 1}}{\dfrac{n + 2}{n + 1} - 1} \dfrac{n}{n + 1} \to 1 \times 1 \times 1 = 1$$
because :

*

* $\dfrac{\log(n)}{\log(n + 1)} \to 1$.

* $\dfrac{\log \dfrac{n + 2}{n + 1}}{\dfrac{n + 2}{n + 1} - 1} \to 1$.

* $\dfrac{n}{n + 1} \to 1$.

