How the fuck do I test these shits for convergence, Veeky Forums

how the fuck do I test these shits for convergence, Veeky Forums

I do pretty well on simpler series, but when I'm finding the interval of convergence for power series that have fucked forms like this, I just get lost.

Would appreciate the help.

>interval of convergence

???? this isn't a power series

both are trivially divergent since the terms are divergent

Yeah these are the two series yielded from finding the apparent interval of convergence from a power series. What does that mean "trivially divergent since the terms are divergent." How would one prove the divergence of these.

if a series with terms a_n converges, then lim a_n -> 0. use the contrapositive

That's what I'm stuck on though. I'm not sure how to evaluate these limits.

[math]2^n = e^{n ln(2)} > e^n [/math]

which grows faster than any polynomial, which themselves grow faster than [math]\sqrt{n}[/math]

Know what kind of functions grow faster than others. Pic related is helpful.

Okay that all makes a lot of sense.

I'm in a Calc II class though, so I need to be able to mathematically work these out. Can L'Hospital's rule apply to these to show the limit is not zero, and if so, how would it be done?

Ratio test that shit.

That was my thought. I was getting limits not equal to zero but was skeptical of my own work.

Just out of curiosity, are you testing at the ends of the interval of convergence?

Yes. The x term in the power series was [math](x-3)^n[/math] with the endpoints of the interval being 1 and 2.

which category does x^x belong to?

Ratio test, after simplifying will give you a 2 in the numerator, the root will go to 1 (you can move the limit inside since shits continuous and L'Hospital's) 2>1 so both diverge.

Larger than x!

I think you screwed up somewhere, if that's the case. You are supposed to get series where you can't use the ratio test.

therefore exponential > factorial
no?

Is that so...

I redid my work and found that the interval is fucked. Should be [math]5/2 < x < 7/2[/math]. And now, the ratio test can't be used. They're much simpler.

Now looking at pic related, it's a simple comparison test, no?

>[math]2^n > e^n[/math]
>[math]2 > e[/math]
Not even engineering has a theorem for this...

That's more like it. And yes, comparison test does the trick there.

Actually, it's using the limit comp test. I compared both to [math]n^(-1/2)[/math] and the limits exist, therefore both divergent. Easy money.

Nigger how the fuck did you get the second from the first.

n-th root.

[math]n! \approx \sqrt{2n\pi}(\frac{n}{e})^n < n^n [/math]

[math]x^x[/math] doesn't really count as an "exponential" in the context of that chart, so using it to say "exponential > factorial" isn't quite right.

tfw i still can't understand the 'choose' notation after reading the wikipedia page 100 times
sucks to be a brainlet

That's an "n over e" not "n choose e"

...

[math] e^nln2 = e^n [/math] brainlet. Otherwise .

That's clearly false. [math]ln2[/math] is not 1. Also, [math]2^n \equiv e^{nln2} [/math]

It really helps to get used to counting with factorials (!) and then explaining yourself how combinations work.
Step 1) Explain yourself how to count amount of different ways to choose the numbers 1,2,3,...,10. Now generalize this concept to n.
[math]n! [/math]
Step 2) Explain yourself how to count amount of different ways to choose 7 numbers the numbers 1,2,3,...,10. Now generalize this concept to k & n.
[math]\frac{n!}{(n-k)!} [/math]
Step 3) Explain yourself how to count amount of different ways to choose 7 numbers from the numbers 1,2,3,...,10 without considering the order (hint: apply step 1 on the 7 chosen numbers). Now generalize this concept to k & n.
[math]\frac{n!}{k!(n-k)!} [/math]
Boom, you have your own home-cooked binomial bullshit.
Expand your knowledge by generalizing it with choosing two,three,or m colored balls or letters or whatever.

it was fucked formatting m8 the ln didn't stay in the exponent

ln(2) is less than 1, it's still clearly wrong.

true, sorry I'm a faggot I haven't done any math in years

>I need to be able to mathematically work these out
you have a fundamentally terrible approach to these problems. you're trying to just throw recipes at them for them to work. that's not how it goes. you need to use and comparison arguments.

I wish everyone was as cool as you user