Indefinite Convergence

[NaN-Gram]

I wanted to confirm that I had used the proper technique to show that the expression in part c converges. I believe that I also had the right expression of the series in parts a and b, but if they aren't correct, I could use some help finding the actual expression.

 

[Romsek]

Your answers don't inspire confidence in your grasp on the subject.

First off the \(\displaystyle \sum\) symbol refers to a series, not a sequence. So right off your first two answers are incorrect.

You had the right idea for (b) but expressed it incorrectly.
The correct way of expressing it is [MATH]b_n = -3+n[/MATH]
For part (a) you had the right idea about getting the sign to alternate, but you got the power of -1 incorrect, and you completely
ignored the rest of the denominator. Try again

For part (c) I'd use the ratio test, find [MATH]L = \left |\dfrac{a_{n+1}}{a_n}\right |[/MATH]
If [MATH]L < 1[/MATH] then the sequence converges.

I leave that to you to work out. The answer you gave is incorrect.

 

[Deleted member 4993]

I wanted to confirm that I had used the proper technique to show that the expression in part c converges. I believe that I also had the right expression of the series in parts a and b, but if they aren't correct, I could use some help finding the actual expression.
View attachment 17251

Hints for part (c):

\(\displaystyle \frac{1- 5n^4}{n^4 \ + \ 8n^3} \ = \ \frac{n^4(1/n^4 - 5)}{n^4(1 + 8/n)}\) ........ continue....

 

[Steven G]

In 1a do you see that your denominator will be 1 or -1? Yes, you want the signs to alternate but you also want the denominator to have this pattern--- 1,4,9,16,25. Have you seen this list of numbers before?

 

[Steven G]

Back to 1a.
Did you even try plugging in 0, then 1, then 2, then 3,...for n?

I can see how you can't see the pattern but to come up with a wrong answer I do not see how that happens as you can check to see if you are correct.

When n=0, using your solution will yield a neg sign while the first number is positive. What about getting the 4 and the 9...
You have to be tougher than this. You can do it!