Math Problem Statement

a1=1,(an+1)^3×(an)^2=2,an>0で求められる数列{an}の極限を求めよ

Solution

Given the problem:

a1=1,(an+1)3×(an)2=2,an>0a_1 = 1, \quad (a_{n+1})^3 \times (a_n)^2 = 2, \quad a_n > 0

We are asked to find the limit of the sequence {an}\{a_n\} as nn \to \infty.

Solution:

  1. Given Recurrence Relation: (an+1)3×(an)2=2(a_{n+1})^3 \times (a_n)^2 = 2 We can rewrite this as: (an+1)3=2(an)2(a_{n+1})^3 = \frac{2}{(a_n)^2} Taking the cube root of both sides: an+1=(2(an)2)13=213an23a_{n+1} = \left(\frac{2}{(a_n)^2}\right)^{\frac{1}{3}} = \frac{2^{\frac{1}{3}}}{a_n^{\frac{2}{3}}}

  2. Assuming the sequence converges to a limit LL, we can set anLa_n \approx L as nn \to \infty: L=213L23L = \frac{2^{\frac{1}{3}}}{L^{\frac{2}{3}}} To find LL, multiply both sides by L23L^{\frac{2}{3}}: L53=213L^{\frac{5}{3}} = 2^{\frac{1}{3}} Raise both sides to the power of 35\frac{3}{5}: L=215L = 2^{\frac{1}{5}}

Conclusion:

The limit of the sequence {an}\{a_n\} as nn \to \infty is L=215L = 2^{\frac{1}{5}}.


Would you like more details or have any other questions?

Related Questions:

  1. How do you prove the convergence of the sequence {an}\{a_n\}?
  2. What happens to the sequence if we change the initial condition a1a_1?
  3. How does the recurrence relation change if the powers of an+1a_{n+1} and ana_n are different?
  4. Can this method be used to find the limit of other types of recursive sequences?
  5. What are the implications of an>0a_n > 0 in the behavior of the sequence?

Tip:

When dealing with recursive sequences, assume the sequence converges and set up the equation involving the limit to solve for it. This is often the simplest way to find the limit of the sequence.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Sequences and Series
Recurrence Relations
Limits

Formulas

Limit formula involving recursive sequences

Theorems

-

Suitable Grade Level

Advanced High School