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captcouch
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Hello! I was working some practice problems for a Calc II quiz for Friday on the alternating series test for convergence or divergence of a series. I ran into a problem when I was working the following series, trying to determine whether it was convergent or divergent:
∞
\sum (-1)n\frac{3n-1}{2n+1}
n=1
Alternating Series Test:
For an alternating series
∞
\sum (-1)n-1bn
n=1
to converge, it must satisfy the following conditions:
1. bn+1 ≤ bn for all n in the series
2. lim bn = 0
n\rightarrow∞
Test for Divergence of a Series:
If the limit as n approaches infinity of a series does not exist or does not equal 0, the series is divergent.
I first began by analyzing the limit as n approaches infinity for the series. By dividing the coefficients of the highest exponential power of the top and bottom (in this case, 1), I found the limit of the series to be \frac{3}{2}. It did not satisfy the second condition of the alternating series test, and as such, I sought to double-check with the Test for Divergence of a Series.
(-1)n's limit does not exist, as it is always alternating back and forth between -1 and 1.
\frac{3n-1}{2n+1}'s limit, as mentioned above, I found to be \frac{3}{2}.
From this information, I concluded that the limit did not exist, and that the series was divergent.
I decided to check the answer from the back of the textbook, and it said it was convergent! I ran through it again a couple of times and got the same result: divergent. I'm not sure what I did wrong here - could I please have some insight from someone else to shed some light on the situation? Thank you very much!
Homework Statement
∞
\sum (-1)n\frac{3n-1}{2n+1}
n=1
Homework Equations
Alternating Series Test:
For an alternating series
∞
\sum (-1)n-1bn
n=1
to converge, it must satisfy the following conditions:
1. bn+1 ≤ bn for all n in the series
2. lim bn = 0
n\rightarrow∞
Test for Divergence of a Series:
If the limit as n approaches infinity of a series does not exist or does not equal 0, the series is divergent.
The Attempt at a Solution
I first began by analyzing the limit as n approaches infinity for the series. By dividing the coefficients of the highest exponential power of the top and bottom (in this case, 1), I found the limit of the series to be \frac{3}{2}. It did not satisfy the second condition of the alternating series test, and as such, I sought to double-check with the Test for Divergence of a Series.
(-1)n's limit does not exist, as it is always alternating back and forth between -1 and 1.
\frac{3n-1}{2n+1}'s limit, as mentioned above, I found to be \frac{3}{2}.
From this information, I concluded that the limit did not exist, and that the series was divergent.
I decided to check the answer from the back of the textbook, and it said it was convergent! I ran through it again a couple of times and got the same result: divergent. I'm not sure what I did wrong here - could I please have some insight from someone else to shed some light on the situation? Thank you very much!
Last edited:
