Converging And Diverging Series

1. What does this series do: ∞ ∑ 8/ (3n²+ 2n+1) n=1

Converge

Diverge

Cannot Be Determines

Correct Answer: Converge.
1. Since the nth term divergence test does not work, the easiest way to do this is by the Direct Comparison Test:
8/(n^2)> 8/(3n^2+2n+1)
2. P-Series:
8/n^2 and 2>1 so 8/n^2 converges
3. Since 8/n^2 converges so does the given series since it is smaller

Wrong Answers:
Diverge. There are not too many ways to come up with this conclusion. However a common mistake is that when you use any type of comparison test, you must make sure you are comparing a series to the same degree. For example in this problem if you compared the series to 8/n rather than 8/(n^2) you would have come that 8/n diverges. And even though that is true you may also make the mistake of saying the the given series also diverges, although you cannot compare those two series because the greater diverges, so the smaller could converge or diverge. However both of those mistake are fairly common.

Cannot Be Determined. You may conclude that it cannot be determine because there are restrictions on the direct comparison test (as I just described above). For example, if some series "a" diverges, and a>b then you could not conclude that series "b" also diverges. However, that rule does not apply for converging series, but the rule is in fact the opposite. It is a common mistake to mix up those rules for comparing converging and diverging series.

Explanation

Correct Answer: Converge.
1. Since the nth term divergence test does not work, the easiest way to do this is by the Direct Comparison Test:
8/(n^2)> 8/(3n^2+2n+1)
2. P-Series:
8/n^2 and 2>1 so 8/n^2 converges
3. Since 8/n^2 converges so does the given series since it is smaller

Wrong Answers:
Diverge. There are not too many ways to come up with this conclusion. However a common mistake is that when you use any type of comparison test, you must make sure you are comparing a series to the same degree. For example in this problem if you compared the series to 8/n rather than 8/(n^2) you would have come that 8/n diverges. And even though that is true you may also make the mistake of saying the the given series also diverges, although you cannot compare those two series because the greater diverges, so the smaller could converge or diverge. However both of those mistake are fairly common.

Cannot Be Determined. You may conclude that it cannot be determine because there are restrictions on the direct comparison test (as I just described above). For example, if some series "a" diverges, and a>b then you could not conclude that series "b" also diverges. However, that rule does not apply for converging series, but the rule is in fact the opposite. It is a common mistake to mix up those rules for comparing converging and diverging series.

2. Which of the following series Converge: ∞ ∞ ∞ 1. ∑ 5/n 2. ∑ 5/(n⁵⁰) 3. ∑ 5/(n^1/50) n=1 n=1 n=1

1 only

2 only

2 and 3 only

1 and 3 only

Correct Answer: 2 Only.
According to the P-Series Test, if you have
∞
∑ #/(n^p) and p is less than or equal to 1, the series diverges, and if p is greater than 1 it conv.
n=1
50>1 thus the series converges
Wrong Answers:
A. if this series is evaluated using a p-series test, then remember if p is less than or equal to one then the series divergres. People often forget what happens when p=1.

B. It cannot be option 2 either according to the p-series test as I just described above, since 1/50
Since both options 1 an 3 are incorrect, the only choice left is the one that contains only 2.

Explanation

Correct Answer: 2 Only.
According to the P-Series Test, if you have
∞
∑ #/(n^p) and p is less than or equal to 1, the series diverges, and if p is greater than 1 it conv.
n=1
50>1 thus the series converges
Wrong Answers:
A. if this series is evaluated using a p-series test, then remember if p is less than or equal to one then the series divergres. People often forget what happens when p=1.

B. It cannot be option 2 either according to the p-series test as I just described above, since 1/50
Since both options 1 an 3 are incorrect, the only choice left is the one that contains only 2.

3. Does the nth root of ((7/n) +8))^n:

Diverge

Cannot Be Determined

Converge

Correct Answer:Diverges.
1. Use the Root Test.
lim (-7/n) +8 =0+8=8
n->∞
8>1 so the series diverges.

Wrong Answers:
B. If someone thought the lim as n-> infinite of -7/n=-7 then they would get 1, and according to the root test if L=1 then the test is inconclusive.
C. some one may have mixed this up with the p-series test where if n>1, then the series converges. Another mistake that may be made is that the sequence of numbers gets smaller as n gets bigger, so that may lead to the faulty conclusion that the series converges, however the partial sum of the series always increases because of the 8's presence.

Explanation

Correct Answer:Diverges.
1. Use the Root Test.
lim (-7/n) +8 =0+8=8
n->∞
8>1 so the series diverges.

Wrong Answers:
B. If someone thought the lim as n-> infinite of -7/n=-7 then they would get 1, and according to the root test if L=1 then the test is inconclusive.
C. some one may have mixed this up with the p-series test where if n>1, then the series converges. Another mistake that may be made is that the sequence of numbers gets smaller as n gets bigger, so that may lead to the faulty conclusion that the series converges, however the partial sum of the series always increases because of the 8's presence.

4. What does the series do: ∞ ∑ 3n³/ (5n³+9) n=1

Cannot Be Determined

Converge

Diverge

Correct Answer: Diverge.
1. Do the nth Term Divergence Test and use L'Hopitals Rule
lim 3n^3/(5n^3+9) = lim 9n^2/(15n^2) = lim 18n/(30n)= 18/30
n->∞ n->∞ n->∞

3/5 does not equal zero, so the series diverges.

Wrong Answers:
A. If someone does the nth term divergence test then he/she would get infinite/infinite. Rather than using L'Hopitals rule, one may assume that the series is inconclusive.

B. There are a couple of ways one may conclude this diverges. First someone may look at the first few terms of the series and see that it id decreasing. And while the sequence does converge, the sequence of partial sums does not. Another way to get this wrong answer is if someone does the nth term divergence test and gets 3/5, then he/she may think it is what the series converges to, or that since 3/5

Explanation

Correct Answer: Diverge.
1. Do the nth Term Divergence Test and use L'Hopitals Rule
lim 3n^3/(5n^3+9) = lim 9n^2/(15n^2) = lim 18n/(30n)= 18/30
n->∞ n->∞ n->∞

3/5 does not equal zero, so the series diverges.

Wrong Answers:
A. If someone does the nth term divergence test then he/she would get infinite/infinite. Rather than using L'Hopitals rule, one may assume that the series is inconclusive.

B. There are a couple of ways one may conclude this diverges. First someone may look at the first few terms of the series and see that it id decreasing. And while the sequence does converge, the sequence of partial sums does not. Another way to get this wrong answer is if someone does the nth term divergence test and gets 3/5, then he/she may think it is what the series converges to, or that since 3/5

5. (*Note*: This is the only question I got from the an outside source, which in this case was out class notes)Find what the series converges to, if it converges: ∞ ∑ (n²+1)^-1 n=1

3

1

Diverges

0

Correct Answer: 1.
1. Try the nth term divergence test.
lim 1/(n^2+1) = 0. Thus this may converge or diverge.
n-> ∞
2.Telescoping series:
Rewrite as partial fractions:
∞
∑ ((1/n) – 1/ (n+1)) = 1-.5+.5-.333+.333+.....
n=1
Every value after the first value cancels out so it ends up just equaling the first value which is one.

Wrong Answers:
3. Just by looking at the terms you can tell that since all the values of the series are less than 1, the sum will never reach 3. You can also test this by just adding the first 5-6 values of the series.

Diverges. There are a couple of ways one can reach this conclusion. One way is that if you write it as a partial fraction you get (1/n - 1/ (n+1)), and since 1/n diverges because it is a harmonic series, one may thing the entire expression is harmonic and diverges. Another way is that if someone does the nth term divergence test and gets "0", then they may think it diverges, if they mixed it up with another test, such as the p-series test, where when n
0. If someone did the nth term divergence test, they would get "0" and thing that is what the series converges to. However, if you get 0, that only means the series may converge or diverge

Explanation

Correct Answer: 1.
1. Try the nth term divergence test.
lim 1/(n^2+1) = 0. Thus this may converge or diverge.
n-> ∞
2.Telescoping series:
Rewrite as partial fractions:
∞
∑ ((1/n) – 1/ (n+1)) = 1-.5+.5-.333+.333+.....
n=1
Every value after the first value cancels out so it ends up just equaling the first value which is one.

Wrong Answers:
3. Just by looking at the terms you can tell that since all the values of the series are less than 1, the sum will never reach 3. You can also test this by just adding the first 5-6 values of the series.

Diverges. There are a couple of ways one can reach this conclusion. One way is that if you write it as a partial fraction you get (1/n - 1/ (n+1)), and since 1/n diverges because it is a harmonic series, one may thing the entire expression is harmonic and diverges. Another way is that if someone does the nth term divergence test and gets "0", then they may think it diverges, if they mixed it up with another test, such as the p-series test, where when n
0. If someone did the nth term divergence test, they would get "0" and thing that is what the series converges to. However, if you get 0, that only means the series may converge or diverge

6. ^∞ Does ∑ ( (-1)ⁿ⁺¹n²+1))/n^{3:
n=1}

Converge

Diverge

Conditionally Converge

Correct Answer: Conditionally Converges.
First take the absolute value of the series. Then find whether the series converges or diverges, to do this I used the Integral test since a(n) decreases as n increases.When I did this, I got it diverges. So you must next use the Conditional Convergence Test. When I did this I found that a(n) decreases to zero as n increases, so it conditionally converges.

Wrong Answers:
While Converging would be a less common incorrect answer, Diverging may be be pretty common because people may forget to check for conditional convergence.

Explanation

Correct Answer: Conditionally Converges.
First take the absolute value of the series. Then find whether the series converges or diverges, to do this I used the Integral test since a(n) decreases as n increases.When I did this, I got it diverges. So you must next use the Conditional Convergence Test. When I did this I found that a(n) decreases to zero as n increases, so it conditionally converges.

Wrong Answers:
While Converging would be a less common incorrect answer, Diverging may be be pretty common because people may forget to check for conditional convergence.

7. What is the interval of convergence for the power series: ∞ ∑ ((5n!)(x+5)ⁿ⁺⁷)/n n=1

-6≤x≤-4

Cannot Be Determined

-6≤x

-6 is less than x which is less than -4

Correct Answer:
-6 is less than x which is less than -4
1. Use the Ratio Test to Find the Interval
∑ ((5n!)(x+5)^n+7)/n= ((5(n+1)!(x+5)^n+8)/n+1 x n/((5n!)(x+5)^n+7))= x+5; -1 2. Determine whether the endpoints converge
a. Abs. Value [((5n!)(-1)^(n+7))/n]= 5n!/n. Using the Ratio Test, you get that this diverges.
b. ((5n!)(1)^(n+7))/n= 5n!/n. Using the same values from part a, you get that this also diverges.

Wrong Answers:
A. If someone forgets to check whether the endpoints converge or diverge, he/she may just assume all values (including the endpoints) converge.
B. Sometimes the factorial throws people off, and they may think that the series cannot be evaluated, but remember when you have a factorial, the first things that should come to mind is the Ratio Test.
D. If someone realizes this is an alternating series test for the -6 endpoint and he/she does the absolute value convergence test, then it ends up being it diverges. He/she must then must check for conditional convergence, and if they did this step wrong, such as if they though a(n) should be increasing as n increases (when it should really be decreasing) then they may get the the -6 conditionally converges.

Explanation

Correct Answer:
-6 is less than x which is less than -4
1. Use the Ratio Test to Find the Interval
∑ ((5n!)(x+5)^n+7)/n= ((5(n+1)!(x+5)^n+8)/n+1 x n/((5n!)(x+5)^n+7))= x+5; -1 2. Determine whether the endpoints converge
a. Abs. Value [((5n!)(-1)^(n+7))/n]= 5n!/n. Using the Ratio Test, you get that this diverges.
b. ((5n!)(1)^(n+7))/n= 5n!/n. Using the same values from part a, you get that this also diverges.

Wrong Answers:
A. If someone forgets to check whether the endpoints converge or diverge, he/she may just assume all values (including the endpoints) converge.
B. Sometimes the factorial throws people off, and they may think that the series cannot be evaluated, but remember when you have a factorial, the first things that should come to mind is the Ratio Test.
D. If someone realizes this is an alternating series test for the -6 endpoint and he/she does the absolute value convergence test, then it ends up being it diverges. He/she must then must check for conditional convergence, and if they did this step wrong, such as if they though a(n) should be increasing as n increases (when it should really be decreasing) then they may get the the -6 conditionally converges.