Calculus II - Alternating Series Test - Convergent?

[captcouch]

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:

Homework Statement

∞
\sum (-1)ⁿ\frac{3n-1}{2n+1}
n=1

Homework Equations

Alternating Series Test:
For an alternating series

∞
\sum (-1)^n-1b_n
n=1

to converge, it must satisfy the following conditions:

1. b_n+1 ≤ b_n for all n in the series
2. lim b_n = 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!

 

[RoshanBBQ]

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:

Homework Statement

∞
\sum (-1)ⁿ\frac{3n-1}{2n+1}
n=1

Homework Equations

Alternating Series Test:
For an alternating series

∞
\sum (-1)^n-1b_n
n=1

to converge, it must satisfy the following conditions:

1. b_n+1 ≤ b_n for all n in the series
2. lim b_n = 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!

I'll tell you what's wrong here: your book. That series is definitely not convergent.

 

[captcouch]

Man, you've got to be kidding me. Nobody's perfect, I guess. I'll be sure to tell my professor in the morning. Thanks for the help!

 

[HallsofIvy]

The most fundamental theorem about infinite series, typically the first proved in a textbook, is
If \lim_{n\to\infty} a_n\ne 0, then \sum a_n does NOT converge.