Calculus

[Anonymous No.6412308]

Determine if the series is convergent or divergent. If convergent(it is) find the sum. (5/2)
I'm so confused to how they solved this and i have a final in a couple days, can anyone please help?? :/

[Anonymous No.6412328]

Did you even try op? I mean really, did you even look at your class notes or the textbook?

[Anonymous No.6412333]

I love grading the tests that your kind turn in. Always worth a chuckle.

[Anonymous No.6412341]

Does anyone actually help on 4chan, or do you guys just put people down?

[Anonymous No.6412345]

>>6412328
I get paid minimum wage to tutor students like this. It is really difficult for them to open their book or listen to a 50 minute lecture that includes examples of how to solve problems like these. They just don't want to try.

[Anonymous No.6412351]

>>6412341

They help, but you have to have a basic knowledge set of mathematics. If someone's doing convergence and divergence and has a "final in a couple of days", and doesn't know this shit, they're not worth helping. This is something that you can look at for half a second and know the answer if you payed attention during the topic. It's essentially the base example of a exponential sum.

[Anonymous No.6412354]

>>6412341
I can't help because I dunno what divergence and convergence mean

[Anonymous No.6412356]

>>6412308

The 1 is irrelevant, thus making it 2^n/3^n.
That makes it, (2/3)^n.
Check your shit about Geometric series, the answer should smack you in the face.

[Anonymous No.6412377]

>>6412356
You're wrong though. The term simplifies to (1/3)^n + (2/3)^n, then if it is convergent you can separate into two series then re-index and you're good to go.

[Anonymous No.6412380]

>>6412377

You're a fucking idiot.

[Anonymous No.6412383]

>>6412380
I'm not the one who thinks (1+2^n)/3^n = (2/3)^n

[Anonymous No.6412382]

>>6412377

(1/3)^n =/= 1/3^n which makes your first term incorrect. Replacing it with the correct term shows that it converges to 0 leaving you with the second term, (2/3)^n, as was previously stated.

[Anonymous No.6412386]

>>6412382
>(1/3)^n =/= 1/3^n
I think I'm being rused.

[Anonymous No.6412385]

>>6412383

In terms of n going to infinity it does. I seriously hope you're trolling and aren't this stupid.

[Anonymous No.6412392]

>>6412385
>not noticing the giant summation sign
That's the only excuse for being so retarded.

[Anonymous No.6412393]

>>6412377

Oh god the stupidity.

[Anonymous No.6412396]

>mfw this thread

[Anonymous No.6412408]

>>6412308
I never understood why so much of this was covered in AP Calc BC, instead of linear algebra. They should make an AP Linear Algebra btw.

[Anonymous No.6412416]

>>6412408
The reason why I question it is because I never encountered convergence/non convergence in any of my other courses, nor did I see any sort engineers discuss it

Even the lower division calc for math majors only spent 1 lecture on it

[Anonymous No.6412450]

shiiiit... I'm taking BC next quarter and I'm pretty sure this crap is on there so I'm just going to guess.
So basically it's increasing "n" to infinity (starting at n=1), and adding the sums with diminishing returns?
What's the "secret"? this thread is the productive equivalent of monkeys flinging shit at each other so far

What do?

[Anonymous No.6412456]

>>6412450
As far as it's taught

memorize this
http://en.wikipedia.org/wiki/Convergence_tests

[Anonymous No.6412470]

>>6412450

There's 10 (mostly interchangeable) tests that help you determine convergence or divergence of an infinite series.

They are easy as piss because there's almost no math to be done. For the ones that do require math, it's basically simplifying fractions.

I say they're incredibly easy because a lot of the convergence tests are interchangeable and/or it's REALLY easy to recognize which test to use. Got a fraction? Try a geometric series! Is it slightly different because you've got a plus one? Do the limit comparison test with a geometric series. Got a power or an nth root? Probably should use the ratio/root test!

It's really not hard.

[Anonymous No.6412482]

>/sci/ actually arguing over this shit

I don't think I'm coming back

[Anonymous No.6412483]

>>6412356
Lmao
>>6412377
Correct
>>6412380
Top kek
>>6412382
Sum(1/3^n) from 1 to infinity equals 0 !! Call the fields medal comittee!
>>6412385
I think you should stop maths and try something easier, like gender studies.
>>6412393
If you are refering to your own stupidity then yes
>>6412308
Ok OP, it's easy the sum of a geometric series converges if and only if its term is smaller than 1. For example sum(1/2^n) converges but sum(3/2^n) doesn't.

Now, there is a formula that you should know by heart : sum(q^n) for n from 0 to N equals (1-q^(N+1))/(1-q) which, when N tends to infinity, equals 1/(1-q).

Be careful when solving your sum though! The sum starts at 1 and not 0, so you need to "shift" your sum by one index before using the formula!

[Anonymous No.6412488]

>>6412308

in particular...

<span class="math">\displaystyle \sum_{n=1}^{\infty} \left(\frac{1+2^n}{3^n}\right) = \sum_{n=1}^{\infty} \frac{1}{3^n} + \sum_{n=1}^{\infty} \frac{2^n}{3^n}[/spoiler]

The right summation converges, the left one is the divergent harmonic series. (i.e., a p-series with <span class="math">\displaystyle n \leq 1[/spoiler]. Therefore, the whole sum diverges.

[Anonymous No.6412491]

>>6412488
Ok now I don't understand. Were all the retarded posts(including this one) troll posts to confuse op? Or /sci/ really this bad at maths?

[Anonymous No.6412492]

>>6412491
Lol ninja delete. He claimed that sum(1/3^n) was the harmonic series and thus the series diverges...

[Anonymous No.6412493]

>>6412308

in particular...

<span class="math">\displaystyle \sum_{n=1}^{\infty} \left(\frac{1+2^n}{3^n}\right) = \sum_{n=1}^{\infty} \frac{1}{3^n} + \sum_{n=1}^{\infty} \frac{2^n}{3^n} = \sum_{n=1}^{\infty} \left( \frac{1}{3} \right)^n + \sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n[/spoiler]

Both are a geometric series which converge. Therefore the whole series converges.

[Anonymous No.6412497]

>>6412493
>mfw you claim sum(1/3^n) is divergent.
nice save haha

[Anonymous No.6412498]

>>6412497

lol shit nigga i don need to no dis no more. i passed cal 2

[Anonymous No.6412510]

>>6412498
That's why it's cool of you to break it down.

you the man, dog!

[Anonymous No.6412513]

>>6412493
You can only separate the series' when the whole thing converges though. While it is technically all true, the logic here is circular.

[Anonymous No.6412517]

>>6412513
Not that anon but no.

You can seperate a converging series if each seperate part converges.

Here the series converges because it is O((2/3)^n).
Each seperate series converges because they are geometric series with a term<1.
So you can seperate.

[Anonymous No.6412521]

>>6412513

It really all depends. Obviously this wouldn't work with something like:

<span class="math">\displaystyle \sum_{n=0}^{\infty} \left( \frac{3^n - 4^n}{2^n} \right) = \sum_{n=0}^{\infty} \left( \frac{3}{2} \right)^n - \sum_{n=0}^{\infty} \left( 2\right)^n = \infty - \infty \Rightarrow[/spoiler] indeterminant.

[Anonymous No.6412522]

I had made an algebraic error, but i had done the problem correctly. Thank you for everyone who actually tried to help me, and everyone who just responded with narcissistic bullshit about how superior you are to the world, i hope you carry that mentality when you're all alone at the age of 70. Good thing you always do practice problems, you'll need a strong hand. Deuces fuckheads.

[Anonymous No.6412523]

>>6412522
Nobody here was actually trying to help you. We were all taking the piss out of one another.

[Anonymous No.6412525]

>>6412521
This wouldn't even converge in the first place...

A better example is simply sum(a^n-a^n), which equals 0, whereas sum(a^n)-sum(a^n) equals infinity - infinity for a bigger than 1

[Anonymous No.6412527]

>>6412521
Doesn't that diverge anyway?
>>6412517
What are you objecting to? Of course you can separate, but you can't just do it without justifying that the whole series is convergent. The post I was responding to ended with concluding that the whole series converges, after they needed to make use of the fact that the series converges. Hence why I said it was circular.

And obviously the separated series' have to converge, otherwise addition is meaningless.

[Anonymous No.6412528]

>>6412527
Oh ok I didn't see that the guy hadn't proved that the whole series converged.

My bad

[Anonymous No.6412531]

>>6412525

Right. It's just like L'Hopital's Rule where if the problem does work in one form, you must rewrite it until you get an indeterminate form.

This was like one of the most basic first tips my calculus book gave me when we first started doing infinite series. Use properties of series in order to come up with an answer.

<span class="math">\displaystyle \sum_{n=1}^{\infty} n^2 - \sum_{n=1}^{\infty} n^2 = (\infty - \infty) \Rightarrow[/spoiler] indeterminate whereas you can rewrite the series as <span class="math">\displaystyle \sum_{n=1}^{\infty} \left(n^2 - n^2\right) [/spoiler] to discover it is zero.

[Anonymous No.6412534]

>>6412531
I meant to put "doesn't work in one form"

[Anonymous No.6412533]

>>6412308
The answer is 2.5
Jesus.

[Anonymous No.6412536]

>>6412534
And I'm guessing you also meant "until you don't get an indeterminate form" !

[Anonymous No.6412538]

>>6412536

Yes.

[Anonymous No.6412540]

>"I'm so confused to how they solved this and i have a final in a couple days"
Don't worry. The world needs ditch diggers, too.

[Anonymous No.6412541]

>>6412534

Well rather I should say a special type of indeterminate form. <span class="math">\displaystyle \frac{\infty}{\infty}[/spoiler] or <span class="math">\displaystyle \frac{0}{0}[/spoiler]

[Anonymous No.6412549]

>>6412525
>>6412531
A lot of problems in these posts suggest you guys haven't taken real analysis yet?

[Anonymous No.6412557]

>>6412549
Uh? What do you mean?

[Anonymous No.6412561]

>>6412557
sum(n^2)-sum(n^2) isn't defined, it certainly isn't infinity-infinity or sum(n^2-n^2).

[Anonymous No.6412563]

>>6412561
Well sure I meant that it wasn't defined, I said infinity-infinity just to show why, I didn't mean to say that it was equal to that.