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JH

Suppose that $\sum_{n = 1}^{\infty} a_n \left( a_n \not= 0 \right)$ is known to be a convergent series. Prove that $\sum_{n = 1}^{\infty} 1/a_n$ is a divergent series.

Hint: Divergence test

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Catherine R.

Missouri State University

Heather Z.

Oregon State University

Samuel H.

University of Nottingham

Boston College

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suppose that this Siri's that's given his conversion and then we'LL try to prove that this sum over here by flipping a and with one over a end is diversion. So here, let's go ahead and prove this. So let's start off by just letting everyone know we're doing a proof. So let's write proof. And then we have since this Siri's converges Bye, the diversions test. We must have that the limit of as n goes to infinity of a n equals zero diversions test says that every time a Siri's converges the limit of a in must equal zero. However, if we look at our new Siri's over here, let's not call this a M. Because Anne is already being used. So is called this and prime equals one over. And so an prime is the and term that being added in the series. If this limit goes to zero, then the limit is end goes to infinity of am this a prime? Well, this is just going to equal Well, we have one in this up. What is that denominator going to the denominators going to zero, so hear this limit will not exist. So this limit is either it may exist. It's either infinity, negative infinity or under fine. And it really depends on the ends of the ends are getting close to zero If there are on ly positive. So maybe a photograph here. So if the ends are on ly positive than one over and will be positive and very big, that will go to infinity. If aliens were negative for the similar reasoning, the limit would go to minus infinity. However, a N could be both positive and negative alternating and now case one over zero in the limit will bounce around between negative infinity and infinity and that case, the limit will be under find. However, these are on ly three cases since we have a one over zero in the limit. This is the limit of this form is not going to be a real number. So we have that the limit of a N prime is not equal to zero. So that means that this new series and prime or let's go back and actually call it what it is. One over a end is diversion. And it's divergent by what the book calls the diversion by the Test four divergence, we computed the limit of a M or an prime for our new series in the limit. This term does not go to zero. Therefore, the Siri's is divergent by the test for diversions, and that's our final answer.

JH

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Catherine R.

Missouri State University

Heather Z.

Oregon State University

Samuel H.

University of Nottingham

Boston College

Lectures

Join Bootcamp