Covergence tests

[Guest]

hi i have a couple of questions to ask:
?
1) How do you figure out ? [5 (1/2)^k- 3 (1/7)^k+1]?
k=1

2) what can i compare ?
? cos^2n/2^n to to perform the limit comparison test?

3)Last question, how would i figure out ? (n= 2 to infinity) n/n^3-1?

 

[skeeter]

hi i have a couple of questions to ask:
?
1) How do you figure out ? [5 (1/2)^k- 3 (1/7)^k+1]?
k=1

2) what can i compare ?
? cos^2n/2^n to to perform the limit comparison test?

3)Last question, how would i figure out ? (n= 2 to infinity) n/n^3-1?

#1 ... your post is vague do you mean

\(\displaystyle \L \sum_{k=1}^{\infty} 5\left(\frac{1}{2}\right)^k - 3\left(\frac{1}{7}\right)^{k+1}\) ???

if so, just split it into two series that are both geometric with |r| < 1 ... you know how to calculate the sum of such a series, correct?

#2 ... I would do a direct comparison with 1/2ⁿ ... 0 < cos²n < 1 for all n.

#3 ... figure out what? ... the sum or convergence/divergence?

 

[Guest]

I'm sorry for the vagueness of question #1, but yes I do know how to solve for the sum of a series.

Also, for question number 3, I wanted to know how to figure out to figure out how to see if the limit was convergent

Thank you so much for your help

 

[skeeter]

Also, for question number 3, I wanted to know how to figure out to figure out how to see if the limit was convergent

you mean if the series is convergent? ... try a limit comparison test with 1/n².

 

[galactus]

The sum of the series is as such:

Use a geometric series.

\(\displaystyle \L\\5\sum_{k=1}^{\infty}\frac{1}{2^{k}}=5(\frac{\frac{1}{2}}{1-\frac{1}{2}})=5\)

\(\displaystyle \L\\3\sum_{k=1}^{\infty}\frac{1}{7^{k+1}}=\frac{3}{7}\sum_{k=1}^{\infty}\frac{1}{7^{k}}=\frac{3}{7}(\frac{\frac{1}{7}}{1-\frac{1}{7}})=\frac{1}{14}\)

\(\displaystyle 5-\frac{1}{14}=\frac{69}{14}\)

I hope I don't get chided for showing you the entire thing, but it's difficult to describe without showing. Some series are relatively easy to find the sum. Others, not so easy.