Determine whether the series diverges or converges I would appreciate help in determining if the following series diverges or converges: $\sum_{n=1}^{\infty} \left(\ln (n+1)-\ln n\right)^{\ln n}$. I know of one approach, but that gets very complicated, and I thought that someone in here might know of a slightly smoother way to do this?
My approach: I have started by rewriting the expression to $e^{ln(n)ln(ln(n+1)-ln(n))}$. Then I applied the Limit Chain rule where $g_1(x)=ln(n)ln(ln(n+1)-ln(n))$ and $f_1(u)=e^{u}$ and started with solving $$(1)\lim_{x\rightarrow\infty}ln(n)ln(ln(n+1)-ln(n))=\lim_{x\rightarrow\infty}ln(n)\lim_{x\rightarrow\infty}ln(ln(n+1)-ln(n))=\infty\cdot\lim_{x\rightarrow\infty}ln(ln(n+1)-ln(n))$$
And to solve that I applied the Limit Chain rule again where $g_2(x)=ln(n+1)-ln(n)$ and $f_2(u)=ln(u)$ and started solving $$(2)\lim_{x\rightarrow\infty}ln(n+1)-ln(n)=\lim_{x\rightarrow\infty}ln(1+\frac{1}{n})$$
And here I need to apply the Limit Chain rule again where $g_3(x)=1+\frac{1}{n}$ and $f_3(u)=ln(u)$ and started solving $$(3)\lim_{x\rightarrow\infty}1+\frac{1}{n}=1.$$
And only from here I can start computing all $f(u)$, so for $f_3(u)$ we have that $\lim_{u\rightarrow 1}ln(u)=0$. For $f_2(u)$ we have that $\lim_{u\rightarrow 0}ln(u)=-\infty$ that we now can put in (1) and get $\infty(-\infty)=-\infty$. Now we can finally determine $f_1(u)=\lim_{u\rightarrow-\infty}e^u=0$ and say that the series is convergent.

Here is a simple approach.  Note that for $x>0$, $0<\log(1+x)\le x$.   Therefore, we have
$$0< \log(n+1)-\log(n)\le \frac1n$$
So, for $n>3$, $\log(n)>1.09$ and hence
$$\left( \log(n+1)-\log(n) \right)^{\log(n)}<\frac1{n^{1.09}}$$
Using the p-test, we conclude that the series of interest converges.
