• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

Math Homework Help Please

man that was so last semester...wish i could help you, but i just crammed for the test 🙁 there is a table in the book somewhere that has all of the conditions for convergence/divergence, and IIRC if you knew that inside and out the whole part of calculus involving series was way easier.
 
ok, its coming back...

its an alternating series because of the (-1)^n. the first term will be +, the second -, third +, etc.
the condition to find convergence of an alternating series for Sum from n=1 to infinity of [(-1)^(n-1)*(a_n)] is 0 < a_(n+1) </= a_n and the limit as n->infinity of a_n = 0. does that help any?

edit:

it is convergent because of this:

1/{(4^n)*n+1} is </= 1/{(4^n)*n} and since the limit of 1/{(4^n)*n} as n->infinity is = to 0, the alternating harmonic series must also be equal to 0. if the limit of the sum is = 0, then the series converges. its called the alternating series test.

edit2:

let it be known that im not entirely sure this is right, so if someone else sees a problem with my response, please correct it for MrCodeDudes sake.
 
Originally posted by: MrCodeDude
I'm looking for the sum of the sequence though.

I know the chart you're referencing, it's ln (x) = (-1)^n * x^n
Click to expand...

oh my bad. umm let me get my book and try to figure it out.
 
Originally posted by: Bigsm00th
did they not specify a number of terms to solve for? or do they want the sum of the whole series?
Click to expand...
Sum of the whole series.

If they just wanted a certain number of terms, it's just plug and chug 🙂
 
Originally posted by: MrCodeDude
Originally posted by: Bigsm00th
did they not specify a number of terms to solve for? or do they want the sum of the whole series?
Click to expand...
Sum of the whole series.

If they just wanted a certain number of terms, it's just plug and chug 🙂
Click to expand...

yeah yeah gimme a break...i told you i just learned this for the test and thats all 🙂
 
The answer is ln(5/4).

Note that 1-x+x^2-x^3+x^4 - x^5 + ... = 1/(1+x) by the infinite geomtric series for -1<x<1.
Taking integrals of both sides, we get:

x-x^2/2+x^3/3-x^4/4+x^5/5-x^6/x + ... = ln(1+x)
Note that when x = 0, left hand side = right hand side so we took care of the constant when we do indefinite integrals.

Your sum is precisely the case when x=1/4.

i.e. your sum is ln(1+1/4) = ln(5/4).
 
Originally posted by: chuckywang
The answer is ln(5/4).

Note that 1-x+x^2-x^3+x^4 - x^5 + ... = 1/(1+x) by the infinite geomtric series for -1<x<1.
Taking integrals of both sides, we get:

x-x^2/2+x^3/3-x^4/4+x^5/5-x^6/x + ... = ln(1+x)
Note that when x = 0, left hand side = right hand side so we took care of the constant when we do indefinite integrals.

Your sum is precisely the case when x=1/4.

i.e. your sum is ln(1+1/4) = ln(5/4).
Click to expand...
Okay, I understand where you got the 1/(1+x) sequence and how you integrated it to get the ln(x) sequence.

However, how do you know that x = 1/4?
 
Originally posted by: MrCodeDude
Originally posted by: chuckywang
The answer is ln(5/4).

Note that 1-x+x^2-x^3+x^4 - x^5 + ... = 1/(1+x) by the infinite geomtric series for -1<x<1.
Taking integrals of both sides, we get:

x-x^2/2+x^3/3-x^4/4+x^5/5-x^6/x + ... = ln(1+x)
Note that when x = 0, left hand side = right hand side so we took care of the constant when we do indefinite integrals.

Your sum is precisely the case when x=1/4.

i.e. your sum is ln(1+1/4) = ln(5/4).
Click to expand...
Okay, I understand where you got the 1/(1+x) sequence and how you integrated it to get the ln(x) sequence.

However, how do you know that x = 1/4?
Click to expand...

if you expand the original series into the form x-x^2/2+x^3/3... you'll see that x is 1/4

eg if you expand it you have (1/4) - (1/4)^2/2 + (1/4)^3/3...
 
Originally posted by: dighn
Originally posted by: MrCodeDude
Originally posted by: chuckywang
The answer is ln(5/4).

Note that 1-x+x^2-x^3+x^4 - x^5 + ... = 1/(1+x) by the infinite geomtric series for -1<x<1.
Taking integrals of both sides, we get:

x-x^2/2+x^3/3-x^4/4+x^5/5-x^6/x + ... = ln(1+x)
Note that when x = 0, left hand side = right hand side so we took care of the constant when we do indefinite integrals.

Your sum is precisely the case when x=1/4.

i.e. your sum is ln(1+1/4) = ln(5/4).
Click to expand...
Okay, I understand where you got the 1/(1+x) sequence and how you integrated it to get the ln(x) sequence.

However, how do you know that x = 1/4?
Click to expand...

if you expand the original series into the form x-x^2/2+x^3/3... you'll see that x is 1/4

eg if you expand it you have (1/4) - (1/4)^2/2 + (1/4)^3/3...
Click to expand...
There is no X in the original sequence.

 
Originally posted by: MrCodeDude
Originally posted by: dighn
Originally posted by: MrCodeDude
Originally posted by: chuckywang
The answer is ln(5/4).

Note that 1-x+x^2-x^3+x^4 - x^5 + ... = 1/(1+x) by the infinite geomtric series for -1<x<1.
Taking integrals of both sides, we get:

x-x^2/2+x^3/3-x^4/4+x^5/5-x^6/x + ... = ln(1+x)
Note that when x = 0, left hand side = right hand side so we took care of the constant when we do indefinite integrals.

Your sum is precisely the case when x=1/4.

i.e. your sum is ln(1+1/4) = ln(5/4).
Click to expand...
Okay, I understand where you got the 1/(1+x) sequence and how you integrated it to get the ln(x) sequence.

However, how do you know that x = 1/4?
Click to expand...

if you expand the original series into the form x-x^2/2+x^3/3... you'll see that x is 1/4

eg if you expand it you have (1/4) - (1/4)^2/2 + (1/4)^3/3...
Click to expand...
There is no X in the original sequence.
Click to expand...

1) In my expansion of ln(1+x), plug in x=1/4.
2) Compare what you get with your sequence that you're trying to find the sum of.
3) Notice they are one and the same.
4) ...
5) Profit!
 
Originally posted by: chuckywang
Originally posted by: MrCodeDude
Originally posted by: dighn
Originally posted by: MrCodeDude
Originally posted by: chuckywang
The answer is ln(5/4).

Note that 1-x+x^2-x^3+x^4 - x^5 + ... = 1/(1+x) by the infinite geomtric series for -1<x<1.
Taking integrals of both sides, we get:

x-x^2/2+x^3/3-x^4/4+x^5/5-x^6/x + ... = ln(1+x)
Note that when x = 0, left hand side = right hand side so we took care of the constant when we do indefinite integrals.

Your sum is precisely the case when x=1/4.

i.e. your sum is ln(1+1/4) = ln(5/4).
Click to expand...
Okay, I understand where you got the 1/(1+x) sequence and how you integrated it to get the ln(x) sequence.

However, how do you know that x = 1/4?
Click to expand...

if you expand the original series into the form x-x^2/2+x^3/3... you'll see that x is 1/4

eg if you expand it you have (1/4) - (1/4)^2/2 + (1/4)^3/3...
Click to expand...
There is no X in the original sequence.
Click to expand...

1) In my expansion of ln(1+x), plug in x=1/4.
2) Compare what you get with your sequence that you're trying to find the sum of.
3) Notice they are one and the same.
4) ...
5) Profit!
Click to expand...
How did you find out that x = 1/4 in the first place though?
 
You must log in or register to reply here.

TRENDING THREADS

Back
Top