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

Question

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.

Answer 1

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 $\sum_{k=1}^n \log(\frac{k+1}{k})=\log(n+1)\to\infty$.

Answer 2

Why not to write that $$\sum_{n=1}^{p} \log \left(\frac{n+1}{n}\right)=\log\Bigg[\prod_{n=1}^{p} \frac{n+1}{n}\Bigg]=\log\Bigg[\frac{ \prod_{n=1}^{p}(n+1)} {\prod_{n=1}^{p}n }\Bigg]=\log(p+1)$$