Well, Veeky Forums? Does it converge or diverge?

And?

Just because the limit is zero does not necessarily mean it converges. It means the test is inconclusive.

what if the first term is the convergent one

What's the sum of 1/n?

It's convergent. See the Limit comparison test or direct comparison test for proof. Apparently most people on this thread don't remember the
p-series rule in that if the degree of the denominator of the expression in the summand is greater than one then the series converges

Accidentally typed converge. I meant it diverges

0.5 > 1 ?

Yeah thats not how inverses, converses, and contrapositives work... Just google the calculus for yourself. Your method is called the DIVERGENCE test and can only have two results: divergence or inconclusive. The correct answer is achieved through comparison to a p-series and since p=0.5, the p-series (a.k.a integral test) we are comparing to diverges. Therefore they both diverge.

ITT: people who don't know the P Series test.

OP, your series is divergent. I can't formally prove it but it's clearly divergent by the p series test.

Yeah whenever you have a known convergent or divergent series and if you can't just do the direct comparison test, I don't think the Limit comparison test with the "parent function" ever screwed me over.

I did the lim k ---> inf (ak/bk) test and got the limit equal to 1 which means it will diverge like 1/(k^1/2)

desu whenever you just tack on something small like a +2, it's not going to really change anything but the Limit Comparison Test is probably what a professor would want when you show your work.

uh yeah and it didn't pass the divergence test so it doesn't diverge

if something isn't false then it's true. don't see what your problem is

[math] {\frac{\sqrt{k}}{\sqrt{k}+2}}\to 1[/math]

So

[math].5\leq {\frac{\sqrt{k}}{\sqrt{k}+2}}[/math]

for all k sufficiently large

and

[math]{\frac{.5}{\sqrt{k}}}\leq {\frac{1}{\sqrt{k}+2}}[/math]

for all k sufficiently large

makes u think

This is some seriously effective bait

Just compare to 1/√n and diverge by p series, what is the point of this problem

To put it simply. For your series to converge, then this must converge:

[eqn] \frac{1}{2} + \frac{1}{3} + \frac{1}{\sqrt{2} + 2} + \frac{1}{\sqrt{3} + 2} +\sum_{n=4}^{\infty} \frac{1}{\sqrt{k} + 2} [/eqn]

But given that most of that is just a finite sum, you care only about the convergence of

[eqn]\sum_{n=4}^{\infty} \frac{1}{\sqrt{k} + 2} \geq\sum_{n=4}^{\infty} \frac{1}{k} \text{ In terms of partial sums}[/eqn]

And then by the comparison test, it diverges.

I remember that in the analysis textbook that taught me how to not be retarded, it went a little deep into this concept in a paragraph. The author said that when you talk about limits, you really do not care about the first elements. In fact, you don't even care about the first 30 quadrillion elements. Those "first" elements do not determine the value of the limit. What really determines it is how it behaves after the arbitrarily big. That is why in cases like this you can disregard where a series start. You don't care where the series start. If I had wanted I could have chopped off the first 3000 parts of the sum, and when working with tighter limits you may want to literally do just that. All depends on how far you have to go inside the terms of a series to find the terms that are really representative of the limit.

I don't know whether you can use a p-series comparison from k=0 as opposed to from k=1

[math]\sqrt{j}+2 \le 2\sqrt{j}[/math] whenever [math]j \ge 4[/math]
Hence [math]\sqrt{k+1} - 2 = int_2^{k+1}\frac{dt}{2\sqrt{t}} \le \sum_{j=4}^k \frac{1}{2\sqrt{j}} \le \sum_{j=4}^k \frac{1}{\sqrt{j}+2}[/math]
Hence the series diverges.

First terms don't matter for convergence

If the series minus the first term isn't convergent then the series isn't going to be convergent
Just think about it

Apply Raabe-Duhamel

>What is the difference between an implication and an equivalence: The post
The limit test can never be used to determine if a series converges, it can only determine whether it diverges

Intuitively I'd say it's divergent because for large k you can neglect the constant term which just makes it the harmonic series with extra terms.

Each term is strictly greater than the corresponding term in the sum of 1/(x+2) which is just the harmonic series with its first term deleted. The latter is known to diverge, hence this series diverges

literally the only correct answer with a valid proof in the thread

Or you could just realize that as n grows to infinity the function behaves like 1/sqrt(n), which is a p series with p

So, we're all agreed it diverges, but what does it equal?

How can this diverage if the denominator gets bigger and bigger?

Brainlet here, wanna know the same. Pls halp Veeky Forums

The denominator doesn't get bigger fast enough for convergence.

It doesn't matter what happens with finitely many terms.
Off yourself.
Also, what is the harmonic series?
Nothing. That's what DIVERGENT means.
Look at the harmonic series.

Nah many. If the Limit to infinity is 0 then the nth term test for divergence is inconclusive which doesn't necessarily mean that the series isn't divergent

>I remember that in the analysis textbook that taught me how to not be retarded, it went a little deep into this concept in a paragraph. The author said that when you talk about limits, you really do not care about the first elements. In fact, you don't even care about the first 30 quadrillion elements.
I should think this is obvious though. Most books put the topological definition of sequence convergence and show its equivalent to the metric space notion in metric spaces. That whole "every neighborhood of L contains all but finitely many points of x_n" thing. Since epsilon-pushing is kind of a roundabout way to understand what's going on, and it's such an elementary consequence of the definition.

No goddamnit, this rule applies to Sequencies, not to series. sum of 1/n also goes to zero, but it's not convergent.

By comparison:

1/(k^(1/2) + 2) > 1/(k + 2)

---

1/(k+2) diverges, therefore, the function also diverges.

You can also do the integral method and the reason one.

Well, everyone is born a brainlet. The book was Abbot's Understanding Analysis. Which is literally analysis 101: How to not be retarded edition.

n-th term can't be used to state that something converges

>simple homework bait thread
>43 posts of brainlets who clearly never took calc 2

quintessentially Veeky Forums.

>Integral test
No.

Related question, what's the slowest converging series? Like what series has the minimum growth rate for convergence?

P series with large p converge extremely slowly

Also in addition there is no series with minimum growth rate

There is no "minimum growth rate for convergence," just a max growth rate for divergence.

Thanks for repeating what I said brainlet

for any convergent sequence s_i the Cauchy subsequence s_{2^i} will converge more slowly, so there's no minimum growth rate

there is a finite n for which all k>n, 1/(sqrt(k)+2) > 1/k
since the sum 1/k diverges, the other sum diverges, too

The harmonic series (infinite sum of 1/n, n natural) diverges.
1/sqrt(n) is greater than 1/n for all natural n.
For large enough n, the constant 1 in sqrt(n)+1 is negligible.
weierstrass criterion => the sum of 1/(sqrt(n)+1) diverges.

how do you prove that

[eqn] \frac{1}{sqrt{k}+2} > \frac{1}{k} [/eqn]

?

Given a numerator of 1, what is the slowest growing function you can have in the denominator for convergence?

its obvious that after some k>0 it will be true.

n^(1+epsilon), for epsilon>0

meant n- epsilon

This. I can't believe the thread is still going on.

>it's obvious

doesn't sound very mathematical to me

it's obvious means TRY SOME NUMBERS FUCKING BRAINLET

>numerical solutions

prove it algebraically

Use the root test or ratio test when direct comparison or limit comparison is inconclusive

This looks like a calc II HW problem desu

>doesn't sound very mathematical to me
okay

we want [math]\frac{1}{\sqrt{k}+2}> \frac{1}{k}[/math]

We start with [math]k^2>k \ \ \forall k>1[/math]

Because the root function is strictly monotonous:

[math]k>\sqrt k \ \ \forall k>1[/math]

Then we know that some k>0 exists such that (since it is true for 5 and we have the previous inequality)

[math]k>\sqrt k +2 \ \ \forall k>1[/math]

But that means

[math]\frac{1}{\sqrt{k}+2}> \frac{1}{k}[/math]

Because you can just divide each side by the other.

Better?

you're fucking retarded
I'm not telling you to check some results
I'm telling you to CHECK SOME FUCKING LOWER BOUNDS
YOU WERE ALREADY TOLD HOW TO PROVE IT: IT'S TRUE FROM SOME LOWER BOUND L SO CHECK SOME VALUES OF L FOR THE PROOF

better
eternal brainlet, lmao, kys.

>someone has to hold his hand and literally try the value 5
>he says "better"
>still can't be assed to do anything himself
you'll get fucking nowhere asshole

>Nothing. That's what DIVERGENT means.
It doesn't mean the infinite sum can't necessarily be assigned a meaningful value.

1+2+3+...=-1/12 after all.

What would be the value of [math]\zeta(1)[/math]?

[math]\frac{1}{\sqrt k+2} > \frac{1}{k} \equiv k > \sqrt k + 2 \equiv k-2 > \sqrt k\equiv k^2 -2k +4 > k\equiv k^2 +4 > 3k[/math]

Let k >= 3, then it reduces to k+4/k > 3 which is trivially true

That's why I said "necessarily".

Some sums can't be given a value, but being divergent isn't a sufficient condition.

I thought they were clear enough here:

Why is this autistic discussion still going?

I already explained the proof you stupid PIGGOT

>It doesn't mean the infinite sum can't necessarily be assigned a meaningful value.
Yes.

>1+2+3+...=-1/12
No.
That is COMPLETELY WRONG. if this were true Analysis would break down.
You can assign a meaningful value to a divergent series BUT THAT DOES NOT MEAN THAT THIS SERIES EQUALS THAT VALUE.

That's completely different. He was talking about the comparison test... Yes, op, you can restrict the domain like that to get a series that will converge, as long as you can prove that every point after it will converge by the test.