235k views
0 votes
Find all the values of x such that the given series would converge n =1 .∑ (3x-8)^n/n^2

1 Answer

4 votes

Answer: the series converges for x in the interval (7/3, 3).

Explanation:

The given series is a summation of the form: ∑(n = 1 to ∞) ((3x - 8)^n / n^2)

This series converges when the absolute value of the ratio, |(3x - 8)|, is less than 1, as it resembles a power series with a common ratio (3x - 8).

To determine the values of x for which the series converges, we need to solve the inequality:

|3x - 8| < 1

This inequality can be broken down into two separate inequalities:

-1 < 3x - 8 < 1

Adding 8 to all parts of the inequality, we get:

7 < 3x < 9

Dividing all parts by 3:

7/3 < x < 3

So, the series converges for x in the interval (7/3, 3).

User Dannybrown
by
8.0k points