Convergent and Divergent Series

In summary, the question asks for what value of k, where k is an integer greater than 1, will both series: sigma n=1 to infinity ((-1)^(kn))/n and sigma n=1 to infinity (k/4)^n converge. The answer is 2. The method used to determine this answer was the Alternating Series test, which states that alternating series with terms that decrease in absolute value will converge. It was determined that (-1)^(2n)/n, which is equivalent to 1/n, converges and since (-1)^(kn)/n is similar to this series, it will also converge.
  • #1
tasveerk
24
0

Homework Statement


For what integer k, k > 1, will both sigma n=1 to infinity ((-1)^(kn))/n and sigma n=1 to infinity (k/4)^n converge?
A) 6 B) 5 C)4 D)3 E)2

Homework Equations





The Attempt at a Solution


I tried to use the ratio test and after some simplifying I got (-1)^k (n/n+1) for the first one and k/4 for the second one. I understand that for the second one any number smaller than 4 will work, but I am unsure of the first one. The answer that is listed is 2.
 
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  • #2
Well, you know that any number smaller than 4 will work. Now I'll ask you, what is the first series for k even? How about k odd?
 
  • #3
Thanks for the quick reply. Is there some type of rule that states if the ratio is negative 1, the series will converge? I thought there were absolute value bars around the ratio meaning that -1 and 1 would be the same thing.
 
  • #4
Actually, when you have an alternating series like (-1)n/n, you can use the Alternating Series test, which is far more efficient than a ratio test. As for (-1)2n/n = 1/n, well that's a series you should recognize.
 
  • #5
I now see that it is a harmonic series, but how did you conclude that (-1)2n/n = 1/n?
 
  • #6
Well, n is a natural number. And for every natural number n, 2n is even. And -1 to an even power is 1.
 
  • #7
I see what you mean now. I was confusing my results from the ratio test with what you were saying. Just to clarify one last thing, (-1)/n is a convergent series, correct?
 
  • #8
Yes it is, as you can verify with the Alternating Series test.
 
  • #9
It all makes sense now. Thanks
 

What is the difference between a convergent and divergent series?

A convergent series is one in which the sum of its terms approaches a finite value as the number of terms increases. In contrast, a divergent series is one in which the sum of its terms does not approach a finite value as the number of terms increases.

How can you determine if a series is convergent or divergent?

One way to determine if a series is convergent or divergent is to use the convergence tests, such as the ratio test, integral test, or comparison test. These tests involve checking the behavior of the terms of the series to determine if they approach a finite value or not.

What is the importance of knowing if a series is convergent or divergent?

Knowing if a series is convergent or divergent is important in many areas of mathematics and science. It allows for the calculation of limits, the approximation of values, and the evaluation of integrals. In addition, the convergence or divergence of a series can provide insights into the behavior and properties of the underlying function.

Can a series be both convergent and divergent?

No, a series cannot be both convergent and divergent. A series can either have a finite sum (convergent) or have an infinite sum (divergent). It cannot have both at the same time.

What are some real-life applications of convergent and divergent series?

Convergent and divergent series have various applications in fields such as physics, engineering, and economics. For example, in physics, they are used to calculate the total energy of a system. In engineering, they are used to approximate complex systems and in economics, they are used to model the growth of populations or investments.

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