Apply the absolute root test · source reference

Printed question

The root test is a convergence test for $\sum_{r=0}^{\infty}a_r$. It converges when $\lim_{n\to\infty}\sqrt[n]{|a_n|}<1$ and diverges when that limit exceeds 1; when it is 1 the test is inconclusive. Use it to explain why $\sum_{r=0}^{\infty}(\frac{-3r^3+rx}{5r^3+7})^r$ converges for all values of $x$. [3]

Printed working

$$\begin{aligned}\lim_{n\to\infty}\sqrt[n]{|a_n|}&=\lim_{n\to\infty}\sqrt[n]{\left|\left(\frac{-3n^3+nx}{5n^3+7}\right)^n\right|}\\&=\lim_{n\to\infty}\left|\frac{-3n^3+nx}{5n^3+7}\right|\\&=\lim_{n\to\infty}\left|\frac{n^3(-3+x/n^2)}{n^3(5+7/n^3)}\right|\\&=\lim_{n\to\infty}\left|\frac{-3+x/n^2}{5+7/n^3}\right|=\left|-\frac35\right|=\frac35<1.\end{aligned}$$

As $n\to\infty$, $x/n^2\to0$ and $7/n^3\to0$. By the root test, the series converges for all $x$.

Printed answer key

No separate short-answer entry is printed for this question; the worked solution is retained above.

Mathematical review and qualifications

For $n\ge1$, $\sqrt[n]{|a_n|}=|(-3n^3+nx)/(5n^3+7)|=|(-3+x/n^2)/(5+7/n^3)|\to3/5<1$ for every fixed real $x$. The root test therefore proves absolute convergence. The $r=0$ expression uses the usual series convention that a zeroth power contributes 1; choosing any finite value for that single term cannot affect convergence.

Keep the absolute value in the root test: the underlying ratio tends to −3/5, but the root-test limit is positive 3/5.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: SIGMA NOTATION. [2023/ACJC/Prelim/II/4b] Question PDF pages 2; solution PDF pages 10. The original files remain private QA records outside this native content.