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Question about the integral test

by salazar888
Tags: integral, test
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salazar888
#1
Jul24-11, 06:45 AM
P: 12
1. The problem statement, all variables and given/known data

We have to determine whether [itex]\sum 1/n^2 + 4[/itex]
is convergente or divergent
2. Relevant equations

I'm trying to work the problem through trigonometric substitution. I was wondering if I could just determine that by the P-series test, the function 1/n^2 will always be larger than the other one, since p is greater than 1 in this case, both are convergent.
3. The attempt at a solution
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tiny-tim
#2
Jul24-11, 07:55 AM
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hi salazar888! welcome to pf!
Quote Quote by salazar888 View Post
I was wondering if I could just determine that by the P-series test, the function 1/n^2 will always be larger than the other one, since p is greater than 1 in this case, both are convergent
yes, always 0 < 1/(n2+4) < 1/n2,

the latter sum converges (from the p-series test), so so must the former (from the direct comparison test)
Ray Vickson
#3
Jul24-11, 09:45 AM
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Obviously, since the series [itex] \sum \frac{1}{n^2} [/itex] converges, the sum you wrote, [itex] 4 + \sum \frac{1}{n^2}[/itex] converges also.

RGV

salazar888
#4
Jul24-11, 09:47 AM
P: 12
Question about the integral test

The series 1/n does not converge. It's the harmonic series. You read it wrong. I get what you're saying though.
Ray Vickson
#5
Jul24-11, 10:22 AM
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I saw the error and edited it immediately.

RGV
HallsofIvy
#6
Jul24-11, 10:24 AM
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Quote Quote by Ray Vickson View Post
Obviously, since the series [itex] \sum \frac{1}{n^2} [/itex] converges, the sum you wrote, [itex] 4 + \sum \frac{1}{n^2}[/itex] converges also.

RGV
But the series
[tex]\sum\left(\frac{1}{n^2}+ 4\right)[/tex]
does NOT converge!
Ray Vickson
#7
Jul24-11, 10:53 AM
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I agree, but that is not what he wrote. We all know he meant sum 1/(n^2 + 4), but he wrote sum (1/n^2) + 4, which is very different according to standard math expression padding rules. Since he was using 'tex' anyway, he should have been able to enter "{n^2+4}" as the second argument of the '\frac' command.

RGV
salazar888
#8
Jul24-11, 11:08 AM
P: 12
Yes it was my fault. I've only been on the forum for a couple of days. Thanks for the help guys. I will improve at typing the commands.


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