Series - The Basics — Question 4

Question 4 – Solution

Step 1: Work with a finite partial sum.

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

Step 2: Simplify the product.

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

Step 3: Classify the series.

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

Step 4: Interpret the nth-term condition.

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.