Evaluating $\lim_{n\rightarrow \infty} \frac{n\sqrt{\ln n}}{(n+1)\sqrt{\ln(n+1)}}$ I'm supposed to look at
$$\lim_{n\rightarrow \infty} \frac{n\sqrt{\ln n}}{(n+1)\sqrt{\ln(n+1)}}$$
which is of course the result of a ratio test. While Wolfram Alpha tells me the limit is $1$, I don't know how to simplify this expression any further to come to the same conclusion.
Any suggestions on how to proceed would be appreciated. 
 A: We have
$$\dfrac{n}{n+1} \dfrac{\sqrt{\ln(n)}}{\sqrt{\ln(n+1)}} = \dfrac{n}{n+1} \dfrac{\sqrt{\ln(n)}}{\sqrt{\ln(n) + \ln(1+1/n)}} = \dfrac1{1+1/n} \dfrac1{\sqrt{1 + \dfrac{\ln(1+1/n)}{\ln(n)}}}$$
Now recall that
$$\lim_{n \to \infty} a_n b_n = \lim_{n \to \infty} a_n \cdot \lim_{n \to \infty} b_n$$
whenever $\lim_{n \to \infty} a_n$ and $\lim_{n \to \infty} b_n$ exist. Use this to compute what you want.
A: $$
\lim_{n \to \infty} \frac{n}{n+1} \cdot \lim_{n \to \infty}\sqrt{\frac{\log n }{\log n + \log(1+\frac{1}{n})}}=1
$$
A: Note that $\frac n {n + 1} \to 1$ as $n \to \infty$, so it suffices to show that $$\sqrt{\frac {\ln n}{\ln (n + 1)}} \to 1$$
In fact, by continuity of the square root function, it's sufficient to check that
$$\frac {\ln n}{\ln (n + 1)} \to 1$$
To do this, apply L'Hospital's rule to the function $f(x) = \frac{\ln x}{\ln (x + 1)}$; this gives
$$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{1/x}{1/(x + 1)} = \lim_{x \to \infty} \left(1 + \frac 1 x\right) = 1$$
as desired.
A: HINT: You can write
$$\frac{n\sqrt{\ln n}}{(n+1)\sqrt{\ln(n+1}}=\frac{n}{n+1}\cdot\sqrt{\frac{\ln n}{\ln(n+1)}}$$
and then compute the limits as $n\to\infty$ of the two factors on the righthand side. The first is clear, and for the second you really need only deal with
$$\lim_{n\to\infty}\frac{\ln n}{\ln(n+1)}=\lim_{x\to\infty}\frac{\ln x}{\ln(x+1)}\;,$$
since the square root function is continuous.
