---
title: "Convergence of x/1.2 + x²/2.3 + x³/3.4"
subject: "Mathematics"
chapter: 6
key: "semester_2/mathematics:convergence-x-1-2-x2"
source: https://www.bctengineer.com/q/semester_2/mathematics/convergence-x-1-2-x2
site: BCT Engineering
---

# Convergence of x/1.2 + x²/2.3 + x³/3.4

Mathematics · Chapter 6: Infinite Series

- Asked in: 2073 Bhadra, 5 marks; 2071 Bhadra, 5 marks; 2069 Bhadra, 5 marks (asked 3 times)
- Video: [Determine if the Series is Convergent or Divergent: x/(1*2) + x²/2*3 + x³/3*4 + x⁴/4*5 + ...](https://www.youtube.com/watch?v=3M-cplWCo4Y) — PW Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/convergence-x-1-2-x2
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-6

## Question

Test the convergence of the series x/1.2 + x²/2.3 + x³/3.4 + ..., x > 0.

## Short answer

Converges for 0 < x ≤ 1, diverges for x > 1. Ratio test gives limit x; at x = 1, series telescopes to1 and converges.

## Answer

The series x/(1·2) + x²/(2·3) + x³/(3·4) + … has general term

uₙ = xⁿ / (n(n + 1)), n = 1, 2, 3, …

#### Ratio test
```text
uₙ₊₁/uₙ = x · n(n+1) / ((n+1)(n+2)) = x · n/(n+2) → x
```
- If x < 1, the series **converges**.
- If x > 1, the series **diverges**.

#### The case x = 1
uₙ = 1/(n(n + 1)) = 1/n − 1/(n + 1). Compare with Σ 1/n²: limit of uₙ/(1/n²) = n²/(n(n+1)) → 1 (finite and non-zero), and Σ 1/n² converges, so the series **converges** (its sum is 1, by telescoping).

**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. What is the general term of x/1.2 + x²/2.3 + x³/3.4 + ...?
   A. xⁿ/(n(n+1))
   B. xⁿ/(n+1)²
   C. xⁿ/(n(n−1))
   D. xⁿ/n²
   Answer: A. The denominators are 1·2, 2·3, 3·4, i.e. n(n+1). (Key point: General term)

2. What does the ratio uₙ₊₁/uₙ tend to, and what does this say?
   A. x; converges for x < 1 and diverges for x > 1
   B. 1; inconclusive for all x
   C. x; converges for x > 1
   D. 0; converges for all x
   Answer: A. uₙ₊₁/uₙ = x·n/(n+2) → x, so the series converges if x < 1 and diverges if x > 1. (Key point: Ratio test result)

3. How is the case x = 1 settled?
   A. Ratio test, since the limit is 1
   B. Divergence, because terms do not tend to 0
   C. Limit comparison with Σ 1/n², so it converges
   D. Limit comparison with Σ 1/n, so it diverges
   Answer: C. uₙ = 1/(n(n+1)) and uₙ/(1/n²) → 1, with Σ 1/n² convergent, so the series converges at x = 1 (0 < x ≤ 1). (Key point: Case x = 1)

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