nature of a series 
Let $\epsilon>0$ and $\lambda>0.$ Discuss, according to the values of $\epsilon$, the nature of the series  $$\sum_{n \geq 3} \lambda^{\left\lfloor{\epsilon\frac{\ln(n)}{\ln(\ln(n))}}\right\rfloor+1}/(\left\lfloor{\epsilon\frac{\ln(n)}{\ln(\ln(n))}}\right\rfloor+1)!$$
  They gave a hint: if $\epsilon>1,$ the series converges, if $\epsilon<1$ the series diverges.  

I found, using Stirling formula that $$\lambda^{\left \lfloor{\epsilon\frac{\ln(n)}{\ln(\ln(n))}}\right\rfloor+1}/(\left\lfloor{\epsilon\frac{\ln(n)}{\ln(\ln(n))}}\right\rfloor+1)! \sim (e\lambda)^{\left \lfloor{\epsilon\frac{\ln(n)}{\ln(\ln(n))}}\right \rfloor}/\sqrt{2 \pi \epsilon\frac{\ln(n)}{\ln(\ln(n))}}(\left\lfloor{\epsilon\frac{\ln(n)}{\ln(\ln(n))}}\right \rfloor+1)^{\left\lfloor{\epsilon\frac{\ln(n)}{\ln(\ln(n))}}\right\rfloor+1}$$
but I don't if this will help. 
 A: Take $\alpha=(\epsilon+1)/2,u_n=\left \lfloor{\frac{\epsilon\ln(n)}{\ln(\ln(n))}}\right \rfloor.$
As you mentioned $n^{\alpha}\frac{\lambda^{u_n+1}}{(u_n+1)!}\sim_{+\infty} n^{\alpha}(e\lambda)^{u_n}(2\pi\epsilon\frac{\ln(n)}{\ln(\ln(n))})^{-1/2}(u_n+1)^{-u_n-1}$
We have $$n^{\alpha}(e\lambda)^{u_n}(\frac{\ln(n)}{\ln(\ln(n))})^{-1/2}(u_n+1)^{-u_n-1}=\exp(\alpha\ln(n)+u_n(1+\ln(\lambda))-\frac{1}{2}\ln(\frac{\ln(n)}{\ln(\ln(n))})-(u_n+1)\ln(u_n+1))$$
we have $$\lim_n\frac{u_n}{\ln(n)}=0,\lim_n\frac{1}{\ln(n)}\ln(\frac{\ln(n)}{\ln(\ln(n))})=0,$$$$\frac{1}{\ln(n)}(u_n+1)\ln(u_n+1)\sim_{+\infty}\frac{1}{\ln(n)}\frac{\epsilon\ln(n)}{\ln(\ln(n))}\ln(\frac{\ln(n)}{\ln(\ln(n))})\sim_{+\infty}\epsilon,$$
if $\epsilon>1,$ then $\alpha>1,\lim_n\alpha\ln(n)+u_n(1+\ln(\lambda))-\frac{1}{2}\ln(\frac{\ln(n)}{\ln(\ln(n))})-(u_n+1)\ln(u_n+1)=\lim_n\ln(n)(\frac{\epsilon+1}{2}-\epsilon)=-\infty,$ 
which means, in this case that $\lim_n n^\alpha\lambda^{u_n+1}/(u_n+1)!=0,$ and then the series converges.
if $\epsilon<1,$ then $\alpha<1$ in this case that $\lim_n n^\alpha\lambda^{u_n+1}/(u_n+1)!=+\infty,$ and then the series diverges. 
