Series - The Basics — Question 4

*Question 4 Consider $\displaystyle\sum_{n=1}^{\infty}\ln\left(\frac{n+1}{n}\right)$.

1. Rewrite the $N$th partial sum as the logarithm of a product.

2. Simplify that product and determine whether the series converges.

3. Explain how the term test and the growth of the partial sums illustrate different facts.

*Question 4 – Solution

For positive factors, $\ln u+\ln v=\ln(uv)$. Therefore $$s_N=\ln\prod_{n=1}^N\frac{n+1}{n} =\ln\left(\frac21\frac32\frac43\cdots\frac{N+1}{N}\right) =\ln(N+1).$$

Every interior factor cancels: the numerator $2\cdot3\cdots N$ cancels the same factors in the denominator, leaving $N+1$. This is multiplicative telescoping.

Since $s_N=\ln(N+1)\to\infty$, the sequence of partial sums is unbounded and the series diverges to $+\infty$.

The individual terms do satisfy $$\ln\left(\frac{n+1}{n}\right)=\ln(1+1/n)\to0,$$ which is necessary but not sufficient for convergence. The logarithm grows slowly, explaining why a short numerical table can look nearly stable even though the partial sums are unbounded.