Evaluate: $$\mathop {\lim }\limits_{n \to \infty } \frac{{(n + 1){{\log }^2}(n + 1) - n{{\log }^2}n}}{{{{\log }^2}n}}$$
Intuitively, I feel that for large $n$, ${\log}(n+1) \approx \ {\log}(n) $. So, the above limit should reduce to:
$$=\mathop {\lim }\limits_{n \to \infty } \frac{{\{ (n + 1) - n\} {{\log }^2}n}}{{{{\log }^2}n}} \ = 1$$
However, can someone please suggest how can one mathematically show this.
Thanks!
If you want to be strict, write $$\log(1+n)=\log(n)+\log(1+\frac 1n)$$ So, the numerator is $$A=(n + 1) \log^2(n + 1) - n\log^2(n)=(n+1)\left(\log(n)+\log(1+\frac 1n)\right)^2-n \log^2(n)$$ Expanding the square and grouping $$A=a^2 n+a^2+2 a n \log (n)+2 a \log (n)+\log ^2(n)$$ where $a=\log(1+\frac 1n)\approx \frac 1n $ is small when $n$ is large.
Is this what you are looking for ?
Edit
Alternatively, you could set $n=\frac 1x$ and use Taylor series around $x=0$. Doing so, the entire expression will be $$\left(1-\frac{2}{\log (x)}\right)+x \left(\frac{1}{\log ^2(x)}-\frac{1}{\log (x)}\right)+\frac{x^2}{3 \log (x)}+O\left(x^3\right)$$
I would proceed as follows: $$\frac{(n+1)\log^2(n+1)-n\log^2n}{\log^2 n} = \frac{n}{\log^2n}(\log^2(n+1)-\log^2n) + 1 \left( \frac{\log(n+1)}{\log n}\right)^2$$ Now, the second summand tends to $1$ (for example, you can easily prove this using l'Hopital), while for the first summand you use $A^2-B^2=(A+B)(A-B)$ $$\frac{n}{\log^2n}(\log^2(n+1)-\log^2n) = \frac{n}{\log^2n}(\log(n+1)+\log n)(\log(n+1)-\log n)=$$ $$=\frac{n}{\log^2n}(\log(n^2+n))(\log(1+\frac1n)) \to 0$$ since $(\log(1+\frac1n))n \to 1$ and $\frac{\log(n^2+n)}{\log^2n} \sim \frac{2 \log n}{\log^2 n} = \frac{2}{\log n} \to 0$.
So finally your limit is $0+1 = 1$.
It should be $$\lim_{n\to \infty} \left((n+1)(\frac{\ln(n+1)}{\ln n})^2 -n\right) = \lim_{n\to \infty} (n+1 - n) = 1.$$
