Question

(a) Check whether the given series\(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\)is convergent or divergent.

(b) Check whether the given series\(\sum\limits_{n = 0}^\infty {{c_n}} {8^n}\)is convergent or divergent.

(c) Check whether the given series\(\sum\limits_{n = 0}^\infty {{c_n}} {( - 3)^n}\)is convergent or divergent.

(d) Check whether the given series\(\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} {c_n}{9^n}\)is convergent or divergent.

Short answer

(a) The series \(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\) is convergent.

(b) The series \(\sum\limits_{n = 0}^\infty {{c_n}} {8^n}\) is divergent.

(c) The series \(\sum\limits_{n = 0}^\infty {{c_n}} {( - 3)^n}\) is convergent.

(d) The series \(\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} {c_n}{9^n}\) is divergent.

Step-by-step solution

Power series theorem

(1) The sequence is convergent if the limit of the sequence exists.

(2) For a given power series\(\sum\limits_{n = 0}^\infty {{c_n}} {(x - a)^n}\), there is a positive number\(R\)such that the series converges if \(|x - a| < R\)and diverges if\(|x - a| > R\).

As given series is\(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\)

Find whether given series is convergent

(a)

As given series is\(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\)

It is given that the series\(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\)is convergent when\(x = - 4\)and divergent when\(x = 6\).

Since, the given series\(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\)is convergent when\(x = - 4\)means it converges for at least\( - 4 \le x < 4\)and divergent when\(x = 6\)means it diverges for\(x \ge 6\)and\(x < - 6\).

Therefore, by the above result the series \(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\) is convergent for \(x = 1\).

Check whether given series is convergent or divergent

(b)

As given series is\(\sum\limits_{n = 0}^\infty {{c_n}} {8^n}\).

It is given that the series\(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\)is convergent when\(x = - 4\)and divergent when\(x = 6\).

Since the given series\(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\)is convergent when\(x = - 4\)means it converges for at least\( - 4 \le x < 4\)and divergent when\(x = 6\)means it diverges for\(x \ge 6\)and\(x < - 6\).

Therefore, by the above result the series \(\sum\limits_{n = 0}^\infty {{c_n}} {8^n}\) is divergent for\(x = 8\).

Determine whether given series is convergent or divergent

(c)

As given series is\(\sum\limits_{n = 0}^\infty {{c_n}} {( - 3)^n}\).

It is given that the series\(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\)is convergent when\(x = - 4\)and divergent when\(x = 6\).

Since the given series\(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\)is convergent when\(x = - 4\)means it converges for at least\( - 4 \le x < 4\)and divergent when\(x = 6\)means it diverges for\(x \ge 6\)and\(x < - 6\).

Therefore, by the above result the series\(\mathop \sum \limits^\infty {c_n}{( - 3)^n}\)is convergent for\(x = - 3\).

Find whether given series is convergent or divergent

(d)

As given series is\(\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} {c_n}{9^n}\).

It is given that the series\(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\)is convergent when\(x = - 4\)and divergent when\(x = 6\).

Since the given series\(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\)is convergent when\(x = - 4\)means it converges for at least\( - 4 \le x < 4\)and divergent when\(x = 6\)means it diverges for\(x \ge 6\)and\(x < - 6\).

Therefore, by the above result the series\(\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} {c_n}{9^n}\)is divergent for\(x = - 9\).