# Does the series $\sum_{n=1}^{\infty} \log(\frac{n+1}{n})$ converge?

I was working on the convergence or divergence of this series $$\sum_{n=1}^{\infty} \log(\frac{n+1}{n})$$. I would like if someone gives me some feedback on my answer and on my understanding of the limit comparison test:

Consider the sequence $$b_n = \log(\frac{n+1}{n}) = \log(n+1)-\log(n)$$. Since log(x) is a monotonically increasing concave function, then $$b_n>b_{n+1}$$, for all $$n\in\mathbb{N}$$. Thus $$b_n$$ is monotonically decreasing sequence. Moreover $$\log(n+1)>\log(n)$$, so $$\log(n+1)- \log(n)>0$$, for all $$n$$. So $$(b_n)$$ is bounded by 0.

Every monotonically decreasing sequence converges to its greatest lower bound. For $$b_n$$ it is equal to zero. Moreover, there certainly exist a sequence $$(a_n)$$ of positive terms such that its limit is some real number distinct from zero, and $$\sum_{n=1}^{\infty} a_n$$ converges. Hence because $$\lim_{n\to\infty}\frac{b_n}{a_n} = 0$$ and $$\sum_{n=1}^{\infty} a_n$$ converges, then $$\sum_{n=1}^{\infty} b_n$$ must converge.

What doesn't fit in my argument is that, it will be implying that the series of any sequence of non-negative terms that converges to zero, must converge. But certainly this is not true since $$\sum_{n=1}^{\infty} \frac{1}{n}$$ diverges. But i don't see where I am blundering.

I'll appreciate any feedback. Moreover, if someone may give an argument without appealing to the concave part.

• Consider the sum of $n$ terms of the $b_i$ sequence. This is a simple expression. What happens to it as $n\to\infty$? Commented Feb 16, 2022 at 3:21
• why not just look at it as a telescopic series? i.e $\lim_{k\to\infty}\sum_{n=1}^{k}\log\left(\frac{n+1}{n}\right)=\log\left(2\right)-\log\left(1\right)+\log\left(3\right)-\log\left(2\right)+\log\left(4\right)-\log\left(3\right)+\dots\log\left(k+1\right)-\log\left(k\right)$ which cancels out nicely to $\log\left(k+1\right)-\log\left(1\right)$ it seems that the partial sum does not converge. Commented Feb 16, 2022 at 3:21
• What is this sequence of positive terms? Are you taking them from the series itself, or somewhere else? How can the limit of this sequence be non-zero, but $\sum a_n$ converges? There are many details missing, which makes this confusing. Commented Feb 16, 2022 at 3:26
• You are correct. That's my blunder Commented Feb 16, 2022 at 3:29

Your argument that $$b_n\to 0$$ can be made much simpler: $$\frac{n+1}{n}\to 1$$ and since $$\log$$ is continuous so $$\log(\frac{n+1}{n})\to \log 1 = 0$$. However this won't help as your proof breaks down after that: you say that there is some convergent $$\sum a_n$$ such that $$a_n\to L\ne 0$$ but this is not possible as if $$\sum a_n$$ converges then $$a_n\to 0$$.
We know that $$\log x \sim x-1$$ for $$x\to 1$$ (as follows from the Taylor series expansion of $$\log x$$). Since $$\frac{k+1}{k}\to 1$$ this means that your series $$\sum b_n$$ behaves roughly like $$\sum (\frac{k+1}{k}-1)=\sum \frac{1}{k}$$ and hence diverges. To prove this rigorously use the comparison test with $$\sum_{n=1}^{\infty} \frac{1}{n}$$.
Another way to prove divergence is by writing the partial sums of the series and noticing that by telescoping all elements cancel and you are left with $$\sum_{k=1}^n \log(\frac{k+1}{k})=\log(n+1)\to\infty$$.
Why not to write that $$\sum_{n=1}^{p} \log \left(\frac{n+1}{n}\right)=\log\Bigg[\prod_{n=1}^{p} \frac{n+1}{n}\Bigg]=\log\Bigg[\frac{ \prod_{n=1}^{p}(n+1)} {\prod_{n=1}^{p}n }\Bigg]=\log(p+1)$$