# 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?

• I can be wrong, but ratio test says that if the fraction of a(n+1) / a(n) from some N is less then q (where 0 < q < 1), then the sum absolutely converges. If you plug in 1 for alpha, the limit will be 1 and we cannot say anything about convergence or divergence. If alpha > 1, limit is infinity, so the answer is alpha < 1 – Kirill Korolev May 31 at 17:47
• Stefan, I think you miss 4 in the last limit. – richrow May 31 at 18:12
• I corrected it. – Stefan Jun 1 at 7:54

## 1 Answer

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.