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Power series; 2

by arl146
Tags: power, series
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arl146
#55
Mar6-12, 11:20 PM
P: 343
because the 1/n adds more to the value than just the 1/n^2 on the bottom. but as n approaches infinity that value doesnt matter so both series are pretty much equal.
is that right or am i thinking about this wrong?
vela
#56
Mar6-12, 11:31 PM
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The 1/n makes the denominator bigger in your series, so what does that imply about the terms of the series compared to 1/n2?
arl146
#57
Mar6-12, 11:38 PM
P: 343
yea thats what i said didnt i? OHHHHHH i see what dumb thing i did. yea so the terms of my series is smaller than those of 1/n^2

sorry, i know i do overlook stuff easily like that. i just cant help it, my brain just does it no matter how hard i try

so at x=5 i proved that it is convergent??? or is there some how more that im missing?
arl146
#58
Mar8-12, 06:14 PM
P: 343
so.. did i actually prove it this time?
Mark44
#59
Mar8-12, 08:07 PM
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How about writing some mathematics instead of just describing things vaguely with words?
arl146
#60
Mar9-12, 11:16 PM
P: 343
ok well

we need [itex]\frac{n}{n^3+1}[/itex] < [itex]\frac{1}{n^2}[/itex] for the test
and that can be proven by:

[itex]\frac{n}{n^3+1}[/itex] = [itex]\frac{n(1)}{n(n^2+1/n)}[/itex] = [itex]\frac{1}{n^2+1/n}[/itex]

so now we are looking at [itex]\frac{1}{n^2+1/n}[/itex] < [itex]\frac{1}{n^2}[/itex] for my series to be convergent since [itex]\frac{1}{n^2}[/itex] converges.

looking at the denominators: n2+[itex]\frac{1}{n}[/itex] is > n2

thus making [itex]\frac{n}{n^3+1}[/itex] = [itex]\frac{1}{n^2+1/n}[/itex] < [itex]\frac{1}{n^2}[/itex] this true

is that good enough?
Mark44
#61
Mar10-12, 12:05 AM
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Looks good!
arl146
#62
Mar10-12, 05:29 PM
P: 343
ok great! so can i like write exactly what i have there for my homework?
Mark44
#63
Mar11-12, 01:28 PM
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What you wrote was fine, but a little more verbose and roundabout than necessary. This is how I would say it.
Quote Quote by arl146 View Post
ok well

we need [itex]\frac{n}{n^3+1}[/itex] < [itex]\frac{1}{n^2}[/itex] for the test
and that can be proven by:

[itex]\frac{n}{n^3+1}[/itex] = [itex]\frac{n(1)}{n(n^2+1/n)}[/itex] = [itex]\frac{1}{n^2+1/n}[/itex]
Now, since n2 + 1/n > n2, for all n >= 1,
then 1/(n2 + 1/n) < 1/n2, for all n >= 1.

Therefore, n/(n3 + 1) < 1/n2, for all n >= 1.
This shows that each term of the series in question is smaller than the corresponding term of the convergent p-series Ʃ1/n2.
Quote Quote by arl146 View Post

so now we are looking at [itex]\frac{1}{n^2+1/n}[/itex] < [itex]\frac{1}{n^2}[/itex] for my series to be convergent since [itex]\frac{1}{n^2}[/itex] converges.

looking at the denominators: n2+[itex]\frac{1}{n}[/itex] is > n2

thus making [itex]\frac{n}{n^3+1}[/itex] = [itex]\frac{1}{n^2+1/n}[/itex] < [itex]\frac{1}{n^2}[/itex] this true

is that good enough?
arl146
#64
Mar11-12, 08:38 PM
P: 343
ok awesome. that is a lot shorter. and the endpoint check for x=3 is just the same concept so i think i got that. thanks a lot guys!
arl146
#65
Mar12-12, 09:05 PM
P: 343
hmm i have another question (hopefully short) ... why do we check the endpoints for convergence? whats the purpose of doing that? do i always do that if the question asks for an interval of convergence and/or radius of convergence?

does checking the endpoints change anything if its convergent or divergent?
Mark44
#66
Mar12-12, 09:13 PM
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P: 21,397
The radius of convergence is the same whether the series converges at neither, either, or both endpoints. What does change is the interval of convergence, which could look like (a, b), (a, b], [a, b), or [a, b].

If the question asks for the interval of convergence, you need to check both endpoints.
arl146
#67
Mar12-12, 09:17 PM
P: 343
ok yea i just figured that out right as i posted that.. so if they ask just to find the radius of convergence, since that answer comes before finding the interval of convergence, i wont have to check the endpoints since i dont even have to find the interval. correct?
Mark44
#68
Mar12-12, 09:21 PM
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P: 21,397
Right


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