Convergence or divergence of $\sum_{n=2}^{\infty}\frac{\ln \frac{n+1}{n}}{\ln \frac{n-1}{n}}$

Question

I need to study convergence or divergence of this series $\sum_{n=2}^{\infty}\frac{\ln \frac{n+1}{n}}{\ln \frac{n-1}{n}}$.

All the terms are negative hence $\frac{n-1}{n}<1$ which implies that $\ln (\frac{n-1}{n})<0$. The other hand, my idea was trying to calculate the $\lim a_n=\lim \frac{\ln \frac{n+1}{n}}{\ln \frac{n-1}{n}}$ but this is indefinided term. By L'hopital rule

$$\frac{\frac{n}{n+1}(\frac{n-(n+1)}{n^2})}{\frac{n}{n-1}(\frac{n-(n-1)}{n^2})}\rightarrow -1$$ when $n\rightarrow \infty$.

Therefore, the series diverge because for all series that converge implies that the general term $a_n$ goes to $0$.

Am I ok?

Thank you

Answer 1

$\ln(\cdot)$ is an increasing and strictly concave function. By Jensen's inequality, for $n \geq 2$, $$ \ln\left(\frac{n+1}{n}\right) = \ln\left(1 + \frac{1}{n}\right) = \ln \left( 1\cdot \frac{n-1}{n} + 2\cdot \frac{1}{n}\right) > \frac{n-1}{n}\ln(1) + \frac{1}{n}\ln(2) = \frac{\ln (2)}{n}. $$ Also, $$ 0 > \ln\left(\frac{n-1}{n}\right) = \ln \left(1 - \frac{1}{n}\right) \geq \ln\left( 1 - \frac{1}{2}\right) = - \ln(2), \text{ for all } n \geq 2. $$ Therefore, the ratio $$ \left|\frac{\ln\frac{n+1}{n}}{\ln\frac{n-1}{n}}\right| > \frac{\frac{\ln (2)}{n}}{\ln(2)} = \frac{1}{n}. $$ Since $\sum \frac{1}{n}$ is divergent starting at any $n$, and the terms in your sum are consistently negative, your sum will also diverge.

Answer 2

While constructing the number $e$ the following inequalities are proved on the way $\left (1+{1\over n}\right )^{n}<e<\left (1+{1\over n}\right )^{n+1}.$ Properties, the derivatives in particular, of the exponential and logarithm functions are based on this construction.

The second inequality gives $\log{n+1\over n}=\log\left ({1+{1\over n}}\right)\ge {1\over n+1}.$ The first inequality applied to $n-1$ gives $\left |\log{n-1\over n}\right |=\log\left (1+{1\over n-1}\right )\le {1\over n-1}.$ Summarizing we get $${\log{n+1\over n}\over |\log{n-1\over n}|} \ge {{1\over n+1}\over {1\over n-1}}=1-{2\over n+1}\to 1$$