How do you determine convergence of a series?

[arl146]

I just need to know how you determine if a series of convergent or divergent. I have this example in which I know is divergent I just don't know why: summation (n=1 to infinity) 1/(2n)

The first couple of terms are 1/2 + 1/4 + 1/6 + 1/8 + ...

Up until that point, it's already beyond equaling 1. Dont know if that means anything.

Another example is summation (k=2 to infinity) (k^2)/((k^2)-1) also divergent

Hoping if someone explains it to me with these examples that I'll understand better. Please help so I can learn this!

 

[clamtrox]

I just need to know how you determine if a series of convergent or divergent. I have this example in which I know is divergent I just don't know why: summation (n=1 to infinity) 1/(2n)

The first couple of terms are 1/2 + 1/4 + 1/6 + 1/8 + ...

Up until that point, it's already beyond equaling 1. Dont know if that means anything.

You should use the integral test for convergence for this kind of series.

Another example is summation (k=2 to infinity) (k^2)/((k^2)-1) also divergent

Hoping if someone explains it to me with these examples that I'll understand better. Please help so I can learn this!

Well, I'm not sure if you made a typo here... But you can immediately see that the terms a_k>1 for all k, so you know that
$$\sum_{k=2}^N a_k \ge \sum_{k=2}^N 1$$

 

[HallsofIvy]

Your series is
$$\sum \frac{1}{2}\frac{1}{n}= \frac{1}{2}\sum\frac{1}{n}$$
That's the "harmonic series" which is well known to be divergent (by the integral test as clamtrox suggests).

For all k, $k^2/(k^2- 1)> k^2/k^2= 1$ and, of course, $\sum 1$ diverges so by the comparison test the original series diverges.

 

[arl146]

well up to this point, this is the first section on series. therefore, we haven't met the integral test for convergence just yet. and not sure what you mean on a typo ...

For all k, $k^2/(k^2- 1)> k^2/k^2= 1$ and, of course, $\sum 1[/itex[ diverges so by the comparison test the original series diverges.$

$<br /> i don't understand what youre saying$

 

[Mark44]

I don't see that you had a typo - it looked fine to me.

For your second series, HallsOfIvy is saying that each term of your series is larger than 1, and the series $$\sum_{n = 1}^{\infty}1 = 1 + 1 + 1 + ... + 1 + ...$$ diverges, because the sequence of partial sums keeps increasing. Since that series diverges, and since each term of the series you're interested in is larger, then your series diverges, too. This is the comparison test in action.

 

[arl146]

so is it always that if the sequence of partial sums increases, the series in the problem diverges. but does it work the other way around, like if the sequence of partial sums decreases, the series in the problem converges?