Test for convergence

**taurus** · May 23rd 2008, 11:20 PM

For the following summations from 1 to infinity, which tests for convergence would I use? I am confused on what to select.

1) 9/((n^2)*(sqrt(n)))

2) (-3*n*sqrt(n))/(n^2)

3) 3*(-1)^n

4) cos((n^5)+1)

5) (n^7)-10*n

Thanks

**Chris L T521** · May 23rd 2008, 11:45 PM

Originally Posted by taurus

For the following summations from 1 to infinity, which tests for convergence would I use? I am confused on what to select.

1) 9/((n^2)*(sqrt(n)))

2) (-3*n*sqrt(n))/(n^2)

3) 3*(-1)^n

4) cos((n^5)+1)

5) (n^7)-10*n

Thanks

1) $\sum_{n=1}^{\infty}\frac{9}{n^2\sqrt{n}}\implies \sum_{n=1}^{\infty}\frac{9}{n^{\frac{5}{2}}}$ . You can apply the integral test, or recognize that it's a p-series, where $\frac{5}{2}>1$ (thus, the series converges).

2) $\sum_{n=1}^{\infty}\frac{-3n\sqrt{n}}{n^2} \implies \sum_{n=1}^{\infty}-\frac{3}{n^{\frac{1}{2}}}$ . Again apply the integral test, or recognize it's a p-series, where $\frac{1}{2}<1$ (thus, the series diverges).

3) I forget how to prove it, but it diverges. (expanding we get 3-3+3-3+3-3+...)

4) $\sum_{n=1}^{\infty}\cos(n^5+1)$ . I believe you would use the limit comparison test here (?) : Thus, $\lim_{n\to{\infty}}cos(n^5) \ \text{DNE}$ . Since this doesn't exist, then the series diverges. (I think this is correct)

5) $\sum_{n=1}^{\infty}n^7-10n$ . I'm sure the integral test would work here as well. It's pretty evident that the series diverges.

Can someone verify these??

**flyingsquirrel** · May 24th 2008, 12:09 AM

Hi

Originally Posted by Chris L T521

3) I forget how to prove it, but it diverges. (expanding we get 3-3+3-3+3-3+...)

4) $\sum_{n=1}^{\infty}\cos(n^5+1)$ . I believe you would use the limit comparison test here (?) : Thus, $\lim_{n\to{\infty}}cos(n^5) \ \text{DNE}$ . Since this doesn't exist, then the series diverges. (I think this is correct)

5) $\sum_{n=1}^{\infty}n^7-10n$ . I'm sure the integral test would work here as well. It's pretty evident that the series diverges.

Can someone verify these??

For the third and the fifth series, you can use the same method as for the fourth one : if $\lim_{n\to \infty} a_n \neq 0$ than the series $\sum a_n$ diverges.

**taurus** · May 24th 2008, 12:16 AM

so for
n^7 - 10n
using the limit comparison test, does that mean i can just put a few numbers in as 'n' and if it doenst appear close to any number it means it diverges?

**flyingsquirrel** · May 24th 2008, 12:20 AM

Originally Posted by taurus

so for
n^7 - 10n
using the limit comparison test, does that mean i can just put a few numbers in as 'n' and if it doenst appear close to any number it means it diverges?

Yes, you can, for example, say that " $\lim_{n\to\infty}n^7-10n=\infty\neq 0$ hence $\sum n^7-10n$ diverges."

**taurus** · May 24th 2008, 12:23 AM

oh so if i put any number as i get bigger it has to go close to zero?

also for the integral test, after you integrate, what value to you use as upper bound?

like for example this says n^2 + 7:
Visual Calculus -

i dont quite get the step after they say the integral becomes....

EDIT: ahh alright i get it

**taurus** · May 24th 2008, 01:20 AM

i know 1/(7n+4) doesnt converge but how would i prove it?

**Moo** · May 24th 2008, 01:42 AM

Originally Posted by taurus

i know 1/(7n+4) doesnt converge but how would i prove it?

Because past a certain value of n, $\frac{1}{7n+4} \sim \frac{1}{7n}$

And $\sum \frac{1}{7n}=\frac 17 \sum \frac 1n$ , which diverges

**taurus** · May 24th 2008, 02:05 AM

where did the 4 go?

**Moo** · May 24th 2008, 02:06 AM

Originally Posted by taurus

where did the 4 go?

When n becomes very big, 4 is being very small comparing to 7n. So we can neglect him.

If you prefer : $\lim_{n \to \infty} 7n+4=\lim_{n \to \infty} 7n$

**taurus** · May 24th 2008, 02:17 AM

So for this:
(2n+5)/((n+2)^4)

what rule would i use?

Would it be like this:
> (2n+5)/((n+2)^4)
> 2n/(n+2)^4
> 2n/n^4
> 2/n^3
so p series with p > 1 therefore it converges?

**flyingsquirrel** · May 24th 2008, 02:42 AM

Originally Posted by taurus

So for this:
(2n+5)/((n+2)^4)

what rule would i use?

Would it be like this:
> (2n+5)/((n+2)^4)
> 2n/(n+2)^4
> 2n/n^4
> 2/n^3
so p series with p > 1 therefore it converges?

That's it !

**taurus** · May 24th 2008, 02:47 AM

what about
(3^n)/(4+(8^n))

i got this far:
(3^n)/(8^n)

but from there what?

**Moo** · May 24th 2008, 02:52 AM

Originally Posted by taurus

what about
(3^n)/(4+(8^n))

i got this far:
(3^n)/(8^n)

but from there what?

$\frac{3^n}{8^n}=\left(\frac 38\right)^n$

This is the general term for a geometric sequence.

$\sum_{n=0}^\infty \left(\frac 38\right)^n=\lim_{n \to \infty} \frac{1-\left(\frac 38\right)^n}{1-\frac 38}$

It converges because $\frac 38 <1 \implies \lim_{n \to \infty} \left(\frac 38\right)^n=0$

**taurus** · May 24th 2008, 02:57 AM

Originally Posted by Moo

Because past a certain value of n, $\frac{1}{7n+4} \sim \frac{1}{7n}$

And $\sum \frac{1}{7n}=\frac 17 \sum \frac 1n$ , which diverges

But doesnt my number go to 0 as n becomes larger? therefore its convergent?

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