Identifying Types of Converging Series

[DameLight]

Hello, I'm looking for some help with this problem for my Calculus 2 class. Since it's a summer class my professor wasn't able to explain everything fully so if you can help me that would be great : )

1. Homework Statement
Select the FIRST correct reason why the given series converges.

A. Convergent geometric series
B. Convergent p series
C. Comparison (or Limit Comparison) with a geometric or p series
D. Alternating Series Test
E. None of the above

1. Σ_n=1^∞ 6(4)ⁿ/7²ⁿ
2. Σ_n=1^∞ sin²(6n)/n²
3. Σ_n=1^∞ cos(nπ)/ln(2n)
4. Σ_n=1^∞ (−1)ⁿ/(6n+5)
5. Σ_n=1^∞ (n+1)(24)ⁿ/5²ⁿ
6. Σ_n=1^∞ (−1)ⁿ * √(n)/(n+9)

Homework Equations

Geometric Series:
Σ_n=1^∞ ar^n-1 will converge if -1 < r < 1

P Series:
Σ_n=1^∞ 1/n^p converges if p > 1

Alternating Series Test:
1) b_n+1 </= b_n for all n and
2) lim_n->∞ b_n = 0

The Attempt at a Solution

1. A
2. B
3. E
4. D
5. C
6. D

 

[Ray Vickson]

Hello, I'm looking for some help with this problem for my Calculus 2 class. Since it's a summer class my professor wasn't able to explain everything fully so if you can help me that would be great : )

1. Homework Statement
Select the FIRST correct reason why the given series converges.

A. Convergent geometric series
B. Convergent p series
C. Comparison (or Limit Comparison) with a geometric or p series
D. Alternating Series Test
E. None of the above

1. Σ_n=1^∞ 6(4)ⁿ/7²ⁿ
2. Σ_n=1^∞ sin²(6n)/n²
3. Σ_n=1^∞ cos(nπ)/ln(2n)
4. Σ_n=1^∞ (−1)ⁿ/(6n+5)
5. Σ_n=1^∞ (n+1)(24)ⁿ/5²ⁿ
6. Σ_n=1^∞ (−1)ⁿ * √(n)/(n+9)

Homework Equations

Geometric Series:
Σ_n=1^∞ ar^n-1 will converge if -1 < r < 1

P Series:
Σ_n=1^∞ 1/n^p converges if p > 1

Alternating Series Test:
1) b_n+1 </= b_n for all n and
2) lim_n->∞ b_n = 0

The Attempt at a Solution

1. A
2. B
3. E
4. D
5. C
6. D

I get (in order 1-6) A, C, D, D, E, D.

 

[DameLight]

My professor told me that I got numbers 2 and 3 correct

 

[Ray Vickson]

My professor told me that I got numbers 2 and 3 correct

I disagree. Here are the details.
1. $\sum 6 \;4^n/7^{2n} = 6 \sum (4/49)^n$, so geometric (A).
2. In $\sum \sin^2 (6n) /n^2$ the terms are (i) not geometric; (ii) not alternating; (iii) not in the simple form $1/n^p$. However, $0 \leq \sin^2(6n)/n^2 \leq 1/n^2$, so a comparison with the series $1/n^2$ works (C).
3. $\sum \cos(n \pi)/ \ln(n)$ is an alternating series, since $\cos(n \pi) = (-1)^n$. So, D.
4. $\sum (-1)^n / (6n+5)$ is alternating, so D.
5. $\sum (n+1) \; 24^n / 5^{2n} = \sum (n+1) r^n, \;\; r = 24/25$. Here, A,B,C,D do not apply, and that leaves E.
6. $\sum (-1)^n \sqrt{n}/(n+9)$ (or $\sum (-1)^n \sqrt{ n/(n+9)}$---can't tell which you mean) is alternating in either interpretation, so D.

 

[DameLight]

The correct answer ended up being:

1. A
2. C
3. D
4. D
5. C
6. D

Thank you for your help : )

 

[Ray Vickson]

The correct answer ended up being:

1. A
2. C
3. D
4. D
5. C
6. D

Thank you for your help : )

How do you get 5 C? Did you read my detailed explanation?

 

[micromass]

How do you get 5 C? Did you read my detailed explanation?

Why wouldn't the comparison test apply in $5$? You can do $n+1\leq \alpha^n$ for some small enough $\alpha>1$.

 

[Ray Vickson]

Why wouldn't the comparison test apply in $5$? You can do $n+1\leq \alpha^n$ for some small enough $\alpha>1$.

OK. That would be an answer to my question (why?) that the OP did not answer. For all I know the OP may have used false reasoning.

 

[DameLight]

OK. That would be an answer to my question (why?) that the OP did not answer. For all I know the OP may have used false reasoning.

I don't want to assume, but that sounded a bit rude.

Still, your explanations are helpful.

I am assuming that going off of your explanation for 5 that we can compare that to a geometric function ∑_n=0^∞ a_nrⁿ and since the absolute value of r is less than 1 we can go ahead and reason that the limit would be equivalent to a/(1-r)

 

[Ray Vickson]

I don't want to assume, but that sounded a bit rude.

Still, your explanations are helpful.

I am assuming that going off of your explanation for 5 that we can compare that to a geometric function ∑_n=0^∞ a_nrⁿ and since the absolute value of r is less than 1 we can go ahead and reason that the limit would be equivalent to a/(1-r)

No: a geometric series would be $\sum r^n$ (or $\sum c r^n = c \sum r^n$ for constant $c$), but $\sum a_n r^n$ is NOT a geometric series if $a_n$ varies with $n$. However, if $a_n \to c = \text{constant}$ as $n \to \infty$, then comparison with the series $\sum c r^n$ can be made to work without too much effort (although some extra effort is needed), However, in your case you have $a_n = n+1 \to +\infty$ as $n \to \infty$, so that type of argument does not work automatically. If you thought it did, that is what what I would call "false reasoning".

However, as 'micromass' has pointed out, if you choose a small enough $\epsilon > 0$ so that $(1+\epsilon) r < 1$, then for some finite $N> 0$ we have $n+1 \leq (1+\epsilon)^n$ for all $n \geq N$, so for $n \geq N$ we have $0 \leq (n+1) r^n \leq [(1+\epsilon) r]^n$ and we can compare with the geometric series $\sum [(1+\epsilon) r]^n$. Note, however, that some such argument as this must be acknowledged and recognized in order for C to be a valid and well-founded answer. Without this realization, just saying it would, again, be false reasoning.