---
title: "Convergence of 1+x/2+2!x²/3²+..."
subject: "Mathematics"
chapter: 6
key: "semester_2/mathematics:convergence-1-x-2-2"
source: https://www.bctengineer.com/q/semester_2/mathematics/convergence-1-x-2-2
site: BCT Engineering
---

# Convergence of 1+x/2+2!x²/3²+...

Mathematics · Chapter 6: Infinite Series

- Asked in: 2074 Bhadra, 5 marks; 2070 Bhadra, 5 marks (asked 2 times)
- Video: [D'alembert ratio test examples | problem 2](https://www.youtube.com/watch?v=L66XRBTcBL8) — Mathematics Analysis
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/convergence-1-x-2-2
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-6

## Question

Test convergent or divergent of the series 1 + x/2 + (2!/3²)x² + (3!/4³)x³ + ... ∞

## Short answer

Converges for x < e, diverges for x ≥ e. Ratio test limit is x/e; at x = e, terms do not approach zero.

## Answer

The series 1 + x/2 + (2!/3²)x² + (3!/4³)x³ + … has general term, starting from n = 0 with u₀ = 1,

uₙ = n! xⁿ / (n + 1)ⁿ.

#### Ratio test
```text
uₙ₊₁/uₙ = [(n+1)! xⁿ⁺¹/(n+2)ⁿ⁺¹] · [(n+1)ⁿ/(n! xⁿ)]
        = x (n+1)ⁿ⁺¹/(n+2)ⁿ⁺¹
        = x [(n+1)/(n+2)]ⁿ⁺¹
        = x / (1 + 1/(n+1))ⁿ⁺¹
        → x/e
```
(because (1 + 1/m)ᵐ → e.)

- x < e: converges.
- x > e: diverges.

#### The case x = e
uₙ₊₁/uₙ = e / (1 + 1/(n+1))ⁿ⁺¹. Since (1 + 1/m)ᵐ < e for every m, this ratio is greater than 1, so the terms increase and cannot tend to 0. The series **diverges**.

**Answer: the series converges for x < e (x > 0 assumed) and diverges for x ≥ e.**

## Check yourself

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

1. What is the general term of 1 + x/2 + (2!/3²)x² + (3!/4³)x³ + ...?
   A. n! xⁿ/(n+1)ⁿ
   B. n! xⁿ/(n+1)!
   C. (n+1)! xⁿ/nⁿ
   D. n xⁿ/(n+1)²
   Answer: A. The pattern n!/(n+1)ⁿ gives 1, 1/2, 2/9, 6/64, ..., matching the series. (Key point: General term)

2. The ratio simplifies to x[(n+1)/(n+2)]^(n+1) = x/(1 + 1/(n+1))^(n+1). Its limit is?
   A. x
   B. x/e
   C. xe
   D. 0
   Answer: B. Since (1 + 1/m)^m → e, the ratio tends to x/e. (Key point: Limit used in ratio test)

3. From the limit x/e, for which x does the series converge by the ratio test?
   A. x < 1
   B. x > e
   C. x < e
   D. all x
   Answer: C. x/e < 1 means x < e; it diverges for x > e. (Key point: Convergence condition)

4. Why does the series diverge at x = e?
   A. The ratio is less than 1
   B. (1 + 1/m)^m < e, so the ratio exceeds 1 and terms increase, not tending to 0
   C. The series is alternating
   D. It is a p-series with p ≤ 1
   Answer: B. The ratio e/(1+1/(n+1))^(n+1) > 1, so terms increase and cannot tend to 0. (Key point: Case x = e)

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