Question

For which values of x is the sequence convergent

X-2, (X-2)³/16, (X-2)²/4

Comments

The sequence is out of order.

Answer 1

The denominators in the given sequence can be expressed in powers of 4:

4⁰, 4², 4¹ but the numerators are out of sequence. So assuming that the terms are placed in order and continuing the sequence, we have:

(X-2)/4⁰, (X-2)²/4¹, (X-2)³/4², (X-2)⁴/4³, (X-2)⁵/4⁴, ..., (X-2)ⁿ⁺¹/4ⁿ, ...

This is a geometric progression with common ratio r=(X-2)/4, a=(X-2)/4⁰.

The sum of a GP is S_n (sum to nth term) and is given by:

S_n=a(1-rⁿ)/(1-r). When n→∞, rⁿ→0 if the series converges, that is, S_∞=a/(1-r)→k where k is a fixed constant. So (X-2)ⁿ/4ⁿ→0. If X-2<4, then we have a series of fractions each of which is less than 1 and is decreasing. So X<6. For example, if X=5 then r=(X-2)/4=¾. S_∞=¾/¼=3.