Apply the absolute root test

The root test is a convergence test for ∑r=0∞ar\sum_{r=0}^{\infty}a_r. It converges when lim⁡n→∞∣an∣n<1\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 ∑r=0∞(−3r3+rx5r3+7)r\sum_{r=0}^{\infty}(\frac{-3r^3+rx}{5r^3+7})^r converges for all values of xx. marks

[2023/ACJC/Prelim/II/4b]

Solution

Convergence for all x

The root test

Use the root test to explain why ∑r=0∞(−3r3+rx5r3+7)r\sum_{r=0}^{\infty}\left(\frac{-3r^3+rx}{5r^3+7}\right)^r converges for all values of xx.

Explore this idea

Hint

  1. Compute ∣an∣n\sqrt[n]{|a_n|}; the power nn cancels the root.

  2. Divide top and bottom by n3n^3 and take the limit.

  3. Compare with 11.

Answer · Convergence for all x

Root-test limit L=3/5<1L=3/5<1, so converges absolutely for every real xx.

The r=0 term uses the zeroth-power convention; any finite choice for this term leaves convergence unchanged.

Explore the idea

The root-test quantity settles at 3/5 for every x

Three very different values of xx all give a quantity that eventually settles at 35\frac35, well under the line at 11.

Term numbern = 3
Try an example
x = −50 at n = 3
1.6268
x = 0 at n = 3
0.5704
x = 50 at n = 3
0.4859
Key idea

A large xx can inflate early terms, but only the limit of the root matters, and xx is overwhelmed by the n3n^3 terms.

Ideas to try
  1. At n=1n=1, large ∣x∣|x| makes the quantity bigger than 11. Slide nn to the right and watch all three curves fall to 0.60.6.
  2. The root test only cares about the limit, not about the first few terms.
Takeaway

Roots tame powers

When a term is already an nnth power, the root test reduces to the limit of its base. A limit below 11 guarantees convergence.