I need to show that the series of general term $$\tanh \frac{1}{n}+ \ln \frac{n^2-n}{n^2+1}$$
converges.
I was thinking to use an equivalence as $n \rightarrow \infty$
We know that $ \tanh \frac{1}{n}= \frac{1}{n} - \frac{1}{6n^3}+ o(\frac{1}{n^4})\sim \frac{1}{n}$
and $\ln\frac{n^2-n}{n^2+1}=\ln \frac{n^2(1-\frac{1}{n})}{n^2(1+\frac{1}{n^2})}= \ln\frac{1-\frac{1}{n}}{1+\frac{1}{n^2}}= \ln (1-\frac{1}{n})- \ln(1+\frac{1}{n^2})= -\frac{1}{n}-\frac{1}{n^2} \sim -\frac{1}{n} $
which when we add both gives me zero (therefore my equivalence is incorrect). Can someone explain to me where is my mistake (I should be getting and equivalence egal to $\frac{-3}{2n^2}$
Thank you in advance
You should keep the error term in your computations. In particular in the taylor approximation of $\ln$ you should keep the term $\frac 1 {n^2}$. $$ \tanh \frac 1 n + \ln\frac{n^2-n}{n^2+1}= \left(\frac 1 n + o(\frac 1 {n^2})\right) + \left(-\frac 1 n -\frac 3{2n^2} + o(\frac 1 {n^2})\right) = -\frac{3}{2n^2} + o(\frac 1 {n^2}). $$
In general a series of the form $$ \sum_n \frac{c}{n^\alpha} + o(\frac{1}{n^\alpha}) $$ converges if $\alpha>1$ and diverges if $\alpha\le 1$.