The infinite sum in question is $\sum\_{1\leq n\leq\infty} ln\left(\frac{n}{n+1}\right)$ (I'm writing my sums in general Sigma-notation) which I broke down into $\sum\_{1\leq n\leq\infty} ln(n) - \sum\_{1\leq n\leq\infty} ln(n + 1)$. Writing out the partial sum and seeing the pattern, you can prove that it obviously diverges. However, when solving telescopic series, I like to reindex because it's easier, but when I reindex, I get that it converges as shown below.

$$\sum\_{1\leq n\leq\infty} ln(n) - \sum\_{1\leq n\leq\infty} ln(n + 1)$$
$$\sum\_{1\leq n\leq\infty} ln(n) - \sum\_{1\leq n - 1\leq\infty} ln((n - 1) + 1 )$$
$$\sum\_{1\leq n\leq\infty} ln(n) - \sum\_{1 + 1\leq n\leq\infty + 1} ln(n)$$
$$\sum\_{1\leq n\leq\infty} ln(n) - \sum\_{2\leq n\leq\infty} ln(n)$$
$$\sum\_{1\leq n\leq\infty} ln(n) - \left(\sum\_{1\leq n\leq\infty} ln(n) - ln(1)\right)$$
$$\sum\_{1\leq n\leq\infty} ln(n) - \sum\_{1\leq n\leq\infty} ln(n) + 0$$
$$0 \rightarrow converges$$

I've looked over this over and over and feel as if I'm doing something dumb or missing something. Someone explained to me that in the partial sum $S\_n = ln(1) + (ln(2) - ln(2)) + (ln(3) - ln(3))... - ln(n+1)$ that because there will always be that last term which doesn't have a pair, that it will approach negative infinity and therefore diverge, but I don't get this because of the fact that there are an infinite amount of terms means that the last $ln(n + 1)$ term will always have a pair to cancel it out which is the next $ln(n)$ term in the series. I see how the limit of the partial sum diverges, but by my reasoning, it has to converge at infinity.
So there are two questions here, why does reindexing fail in this scenario and what is wrong about my thought process with the partial sum?