Convergence of $\sum_{n=0}^{+\infty} \frac{\log(n+4n^\alpha)-\log(n)}{2n}$ I have to determine for which values of $\alpha$ the series 
$\sum\limits_{n=0}^{\infty} \frac{\log(n+4n^\alpha)-\log(n)}{2n}$
converges.
First of all I simplified the expression to
$\sum\limits_{n=0}^{\infty} \frac{\log(1+4n^{\alpha-1})}{2n}$
Since the terms are definitely positive I used the ratio test, so I studied the limit 
$\lim\limits_{n \to +\infty} \frac{n\log\left(1+4(n+1)^{\alpha-1}\right)}{(n+1) \log(1+4n^{\alpha -1 })}$
and I found out it equals to zero for any $\alpha$, so it should converge for any $\alpha$. However the solution of the problem is $\alpha<1$. What am I getting wrong? 
 A: Solution of the problem.
Denote $a_n=\dfrac{\log(n+4n^{\alpha})-\log(n)}{2n}=\dfrac{\log(1+4n^{\alpha-1})}{2n}$.
Note that for all $n\in\mathbb{N}$ we have $a_n>0$. Now, consider two cases:
Case 1. $\alpha\geq 1$.
In this case we have $\log(1+4n^{\alpha-1})\geq\log 5$ for $n\in\mathbb{N}$, so $a_n\geq \dfrac{\log 5}{2n}$. Hence, series $\sum\limits_{n=1}^{\infty}a_n$ diverges because harmonic series $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ also diverges.
Case 2. $\alpha<1$. 
In this case $n^{\alpha-1}\rightarrow 0$ when $n\rightarrow 0$, so from the equivalence $\log(1+t)\sim t$ when $t\rightarrow 0$ we obtain
$$
a_n=\frac{\log(1+4n^{\alpha-1})}{2n}\sim\frac{4n^{\alpha-1}}{2n}=\frac{2}{n^{2-\alpha}},~ n\rightarrow \infty.
$$
Note that $2-\alpha>1$, so series $\sum\limits_{n=1}^{\infty}a_n$ converges because series $\sum\limits_{n=1}^{\infty}\dfrac{1}{n^p}$ converges whenever $p>1$.
Summing up two cases we get that series $\sum\limits_{n=1}^{\infty}a_n$ converges iff $\alpha<1$.
Remark. Notice that $\frac{a_{n+1}}{a_n}\rightarrow 1$ when $n\rightarrow\infty$, so I think that there is a mistake in your calculations.
