Sum of series of logarithms

Question

It’s from a STEP paper. How can I show this summation converges to give the answer required?

$$1(\ln2-\ln1)+2(\ln3-\ln2)+\dots+(n-1)(\ln n-\ln(n-1))$$ which you can rearrange to give $n\ln n-\ln n!$ as required.

Answer 1

Knowing your logarithmic laws, the expression is $$\ln\left(\frac{2^1}{1^1}\cdot\frac{3^2}{2^2}\cdot\frac{4^3}{3^3}\cdots\frac{n^{n-1}}{{(n-1)^{n-1}}}\right)=\ln\left(\frac{n^{n-1}}{1\cdot2\cdot3\cdots(n-1)}\right)=\ln\left(\frac{n^n}{n!}\right)$$ where we multiplied the numerator and denominator by $n$ as the last step.

This expression, again using logarithmic laws, is $n\ln(n)-\ln(n!)$ as required.

Answer 2

Let $$ S = 1(\ln2−\ln1)+2(\ln3−\ln2)+⋯+(n−1)(\ln n−\ln(n−1)) $$

Then, $$\begin{align*} S &= -\ln1 + (2-1)\cdot(-\ln2)+\dots+\big((n-1)-(n-2)\big)\cdot(-\ln (n-1)) + -\ln n + n\ln n \\ &=-\sum_{i=1}^n \ln i + n\ln n\\ &=-\ln(\prod_{i=1}^n i ) + n\ln n\\ &=-\ln(n!)+n\ln n \end{align*}$$