What is the Limit Comparison Test for Series Convergence?

[rcmango]

Homework Statement

heres the equations: http://img407.imageshack.us/img407/738/untitledzk8.jpg

there are two equations, a and b.

Homework Equations

3/n^2 for a,

and 1/n for b.

The Attempt at a Solution

Using the limit comparison test, i don't understand why to use 3/n^2 and 1/n for these equations.

It was explained that we use the largest n in the numerator and denominator, but what we picked, clearly isn't the largest?

please help. thanks.

also, how does a converge, and b diverge??

 

[Gib Z]

We use 3/n^2 because its a good comparison, a is similar to it. Same for b, if you don't include the 19n, it simplifies to 1/n.

Now, 3/n^2 fulfils all the requirements to converge. The terms converge to zero etc. It can be shown that it is equal to $\frac{\pi^2}{2}$, thought I won't post the proof here.

1/n is the harmonic series, represented by zeta(1) or a p series of p=1.

A p series is of the form $$\sum_{n=1}^{\inf} \frac{1}{n^p}$$. The rule of convergence of these is they converge if p>1, or diverge < or =1.

 

[HallsofIvy]

Homework Statement

heres the equations: http://img407.imageshack.us/img407/738/untitledzk8.jpg

there are two equations, a and b.

Homework Equations

3/n^2 for a,

and 1/n for b.

The Attempt at a Solution

Using the limit comparison test, i don't understand why to use 3/n^2 and 1/n for these equations.

It was explained that we use the largest n in the numerator and denominator, but what we picked, clearly isn't the largest?

please help. thanks.

also, how does a converge, and b diverge??

For large values of n, the highest power of n will "dominate" (10000000³ if far larger than 100000000² [exactly 100000000 times as large!]) so we can ignore lower powers: In a, the numerator is "dominated" by 3n while the denominator is "dominated" by n³. For very large n, the fraction will be close to $$\frac{3n}{n^3}= \frac{3}{n^2}$$.

In b, n²+ 19n is "dominated by n² and the square root of that is n. For very large n, [tex]\frac{1}{\sqrt{n^2+ 19n}}[/itex] will be close to $$\frac{1}{n}$$.<br /> <br /> As for why one converges and the other diverges, use the "integral test":<br /> $$\int \frac{3}{x^2}dx= -\frac{3}{x}$$ <br /> which goes to 0 as x goes to infinity.<br /> On the other hand,<br /> [tex]\int \frac{1}{x}dx= ln(x)[/itex]<br /> which does not converge as x goes to infinity.[/tex][/tex]

 

[rcmango]

awesome job, halls thanks for showing the simplification too. i see it now. thanks.