Question about the integral test

[salazar888]

Homework Statement

We have to determine whether \sum 1/n^2 + 4
is convergente or divergent

Homework 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.

The Attempt at a Solution

 

[tiny-tim]

welcome to pf!

hi salazar888! welcome to pf!

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]

Obviously, since the series \sum \frac{1}{n^2} converges, the sum you wrote, 4 + \sum \frac{1}{n^2} converges also.

RGV

 

[salazar888]

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]

I saw the error and edited it immediately.

RGV

 

[HallsofIvy]

Obviously, since the series \sum \frac{1}{n^2} converges, the sum you wrote, 4 + \sum \frac{1}{n^2} converges also.

RGV

But the series
\sum\left(\frac{1}{n^2}+ 4\right)
does NOT converge!

 

[Ray Vickson]

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]

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.