Combine an alternating GP with logarithms · source reference

Printed question

(i) Find $\sum_{r=1}^k[(-\frac12)^{r+1}+\ln(r+1)]$ in terms of $k$. Simplify your answer. [4]

(ii) Hence determine if $\sum_{r=1}^{\infty}[(-\frac12)^{r+1}+\ln(r+1)]$ exists. [1]

Printed working

(i)

$$\begin{aligned}\sum_{r=1}^k[(-1/2)^{r+1}+\ln(r+1)]&=\sum_{r=1}^k(-1/2)^{r+1}+\sum_{r=1}^k\ln(r+1)\\&=\frac{\frac14[1-(-1/2)^k]}{1-(-1/2)}+\ln[(k+1)!]\\&=\frac16[1-(-1/2)^k]+\ln[(k+1)!].\end{aligned}$$

(ii) As $k\to\infty$,

$$\lim_{k\to\infty}\frac16[1-(-1/2)^k]=\frac16,\qquad\lim_{k\to\infty}\ln[(k+1)!]=\infty.$$

Therefore the sum to infinity does not exist.

Teacher notes

$\sum_{r=1}^k(-\frac12)^{r+1}=(-\frac12)^2+(-\frac12)^3+(-\frac12)^4+\cdots$ is geometric, with first term $\frac14$, common ratio $-\frac12$.

Printed answer key

(i) $\frac16[1-(-1/2)^k]+\ln[(k+1)!]$; (ii) does not exist

Mathematical review and qualifications

(i) The GP sum is $\frac{1/4}{1+1/2}[1-(-1/2)^k]=\frac16[1-(-1/2)^k]$. The log sum is $\ln[2\cdot3\cdots(k+1)]=\ln((k+1)!)$. Add them.

(ii) As $k\to\infty$, the first part tends to $1/6$ while $\ln((k+1)!)\to\infty$. The infinite series diverges to $+\infty$, so it has no finite sum.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: SIGMA NOTATION. [2022/HCI/Prelim/II/4a] Question PDF pages 1; solution PDF pages 3, 4. The original files remain private QA records outside this native content.