0
$\begingroup$

If it is known that the series $\sum_{n=1}^{\infty} a_nx^n$ is convergent at $x=4$. What can we conclude about the series $\sum_{n=1}^{\infty} a_n(-7)^n$?

A. Convergent
B. Conditionally Convergent
C. Conditionally Convergent
D. Divergent
E. May be convergent or divergent

Is my logic right? Since we are told that the series is convergent at $x=4$, then this might be a point inside the convergence interval or one of the endpoints, however, we do not know. Hence, the series $\sum_{n=1}^{\infty} a_n(-7)^n$ might be convergent or divergent (option $E$ is right).

$\endgroup$
2
  • 1
    $\begingroup$ E is correct. Your logic is essentially right; however, it would be good to illustrate that it can be convergent with an example, and divergent with an example in order to be totally sure. For divergent, consider $a_i = \frac{1}{(-7)^i}$. For convergent, consider $a_i = 0$. $\endgroup$
    – Doctor Who
    Aug 20, 2020 at 19:42
  • $\begingroup$ @user62487108 A minor point is your option B and C are identical, so I assume it's a typo. $\endgroup$
    – John Omielan
    Aug 20, 2020 at 21:15

1 Answer 1

Reset to default
3
$\begingroup$

You chose the right option, but you should explain why it may converge or diverge. For instance, both series $\sum_{n=0}^\infty\frac{x^n}{5^n}$ and $\sum_{n=0}^\infty\frac{x^n}{8^n}$ converge when $x=4$. However, the first one diverges when $x=-7$, whereas the second one converges.

$\endgroup$

You must log in to answer this question.