---
title: "Convergence of 1/2+(2/3)x+(3/4)²x²+(4/5)³x³+..."
subject: "Mathematics"
chapter: 6
key: "semester_2/mathematics:convergence-1-2-2-3"
source: https://www.bctengineer.com/q/semester_2/mathematics/convergence-1-2-2-3
site: BCT Engineering
---

# Convergence of 1/2+(2/3)x+(3/4)²x²+(4/5)³x³+...

Mathematics · Chapter 6: Infinite Series

- Asked in: 2067 Mangsir, 5 marks
- Video: [Alternating Series, Types of Convergence, and the Ratio Test](https://www.youtube.com/watch?v=-Md4CEW0-zM) — Professor Dave Explains
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/convergence-1-2-2-3
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-6

## Question

Test the convergence of the series 1/2 + (2/3)x + (3/4)²x² + (4/5)³x³ + ...

## Short answer

Root test: (u_n)^{1/n} → x; converges for 0 < x < 1. At x=1, nth term → e^{-1} ≠0, so diverges. Converges only for 0 < x <1.

## Answer

The series 1/2 + (2/3)x + (3/4)²x² + (4/5)³x³ + … has general term, for n = 1, 2, 3, …,

uₙ = ((n + 1)/(n + 2))ⁿ xⁿ,

after the first term 1/2. Leaving out one term does not change convergence.

#### Cauchy's root test
```text
(uₙ)^(1/n) = ((n+1)/(n+2)) · x = (1 − 1/(n+2)) x → x
```
- x < 1: **converges**.
- x > 1: **diverges**.

#### The case x = 1
uₙ = (1 − 1/(n + 2))ⁿ = ((n+1)/(n+2))ⁿ → e⁻¹ ≠ 0 as n → ∞. Since uₙ does not tend to zero, the series **diverges**.

**Answer: the series converges for 0 < x < 1 and diverges for x ≥ 1.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which test suits uₙ = ((n+1)/(n+2))ⁿ xⁿ, and why?
   A. Cauchy's root test, since uₙ is an nth power
   B. Alternating series test
   C. Integral test
   D. Comparison with 1/n
   Answer: A. Taking the nth root removes the power n, leaving ((n+1)/(n+2)) x. (Key point: Choice of test)

2. What does (uₙ)^(1/n) tend to, and what follows?
   A. x; converges for x < 1, diverges for x > 1
   B. x/e; converges for x < e
   C. 1; always converges
   D. 0; converges for all x
   Answer: A. (1 − 1/(n+2)) x → x, so the series converges for x < 1 and diverges for x > 1. (Key point: Root test limit)

3. At x = 1, uₙ = ((n+1)/(n+2))ⁿ. Why does the series diverge?
   A. uₙ → 1, not 0
   B. uₙ → e⁻¹ ≠ 0, so the terms do not tend to zero
   C. It is the harmonic series
   D. The root test shows root > 1
   Answer: B. uₙ → e⁻¹ ≠ 0, so the series diverges; it converges only for 0 < x < 1. (Key point: Case x = 1)

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