Investigate the convergence of the series $\sum_{n=1}^{+\infty}(\ln(n-1)-\ln(n)-\frac{1}{n})$ from Taylor My try:I tried to use Taylor in this task but then I have: $$(\ln(n-1)-\ln(n)-\frac{1}{n})=-n-\frac{n^{2}}{2}-\frac{n^{3}}{3}-\frac{n^{4}}{4}+o(n^{4})-(n-1)+\frac{(n-1)^{2}}{2}-\frac{(n-1)^{3}}{3}+o(n^{3})-\frac{1}{n}$$I think it is a more complicated version than the original one so I can't what I can do with it.Can you get some tips?
 A: You have$$\ln(n-1)-\ln(n) - \frac{1}{n} = \ln\left(1- \frac{1}{n} \right) - \frac{1}{n} = - \frac{1}{n}+  o\left( \frac{1}{n}\right) - \frac{1}{n} = -\frac{2}{n} + o\left( \frac{1}{n}\right) $$
So $$\ln(n-1)-\ln(n) - \frac{1}{n}  \sim -\frac{2}{n} $$
so the series diverges.
A: I'm assuming the sum is supposed to start at $n=2$ to avoid $\ln(0)$ appearing in the expression.
In that case, I assume you know that $-\sum_{n=2}^\infty\frac1n$ diverges to $-\infty$. Adding $\ln(n-1)-\ln(n)$ to each term is adding a negative amount to $-\frac1n$ and is just going to speed up the divergence to $-\infty$.

Is there is a typo and that first subtraction should be addition? If so, there is some telescoping with a partial sum:
$$
\begin{align}
\sum_{n=1}^N\left[\ln(n+1)-\ln(n)-\frac1n\right]&=\ln(N+1)-\ln(1)-\sum_{n=1}^N\frac1n\\
&=\ln(N+1)-\sum_{n=1}^N\frac1n\\
&=\ln(N+1)-\ln(N)+\ln(N)-\sum_{n=1}^N\frac1n\\
&=\ln(1+1/N)+\ln(N)-\sum_{n=1}^N\frac1n\\
&\to0-\gamma
\end{align}
$$
where $\gamma$ is the Euler-Mascheroni constant.
A: Considering $-\sum_{n=1}^{+\infty}(\ln(n-1)-\ln(n)-\frac{1}{n})$, you have 


*

*$\frac{1}{n} + \ln n -\ln (n-1) \geq \frac{1}{n}$
So, the given series diverges to $-\infty$.
Maybe you mean $\sum_{n=1}^{+\infty}(\ln(n)-\ln(n-1)-\frac{1}{n})$, which is convergent as


*

*$\ln(n)-\ln(n-1)-\frac{1}{n} \stackrel{Taylor}{=} \frac{1}{\xi_n}-\frac{1}{n}$ with $n-1 < \xi_n <n \Leftrightarrow \frac{1}{n} < \frac{1}{\xi_n} < \frac{1}{n-1}$
It follows
$$0 < \sum_{n=1}^{+\infty}(\ln(n)-\ln(n-1)-\frac{1}{n}) < \sum_{n=1}^{+\infty}(\frac{1}{n-1}-\frac{1}{n}) < \infty$$
A: Why use Taylor? The summands are less than $-1/n$ for $n>1$ and $\sum_{n=2}^{\infty} (-1/n)$ $= -\infty.$
