Converging And Diverging Series

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

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.

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

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

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.

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