Math Quiz 7

[Ether]

Determine whether this series converges or diverges using any one or a combination of the following tests/rules/theorems:

Comparison Test

Limit Comparison Test

Squeeze Theorem

Divergence Test

Ratio Test

Root Test

Harmonic Series

Geometric Series

Integral Test

L'Hopital's Rule

Show your reasoning.

[Rachel Lim]

Using the Divergence Test and l'Hopital's Rule (and the rule that lets me take the exponent outside the limit), I got that the series diverged: http://goo.gl/KVYEKZ


Edited to add:

Graph of the series (to confirm my working):


After thinking about this problem a bit more, I was wondering if this logic would work: as x→∞, the sum will resemble Σ1^(2x), which is just Σ1^(x), which diverges by the Geometric Series test (since it's just 1+1+1+1 infinite times). I get the same result, but I feel like I'm messing up something by applying l'Hopital's rule like that...

[Caleb Sykes]

Originally Posted by Rachel Lim

After thinking about this problem a bit more, I was wondering if this logic would work: as x→∞, the sum will resemble Σ1^(2x), which is just Σ1^(x), which diverges by the Geometric Series test (since it's just 1+1+1+1 infinite times). I get the same result, but I feel like I'm messing up something by applying l'Hopital's rule like that...

This is not a valid approach. As a counterexample, consider the following:
sum((x/(x+3))^(x^2))

By your reasoning, as x goes to infinity, this sum will also resemble sum(1^(x^2))=sum(1^x), so it should also diverge. However, this series does not diverge.

[Jared]

It doesn't converge because the limit as k goes to infinity of your function is equal to e^-6. This is the divergence test.


Here's my work. The -6 I got as my limit is the exponent for e^-6, which means that the function is always changing, and the sum cannot converge.

[Ether]

Nice work everybody !

Originally Posted by Caleb Sykes

...consider the following:

sum((x/(x+3))^(x^2))

...this series does not diverge.

Somebody check my work. Is there a shorter way?

[iRobot_]

I'm not familiar with your root test method Ether. When I used the root test, it was inconclusive. ie the limit for the kth root of (k/(k+3))^2k = 1

[Ether]

Originally Posted by iRobot_

I'm not familiar with your root test method Ether. When I used the root test, it was inconclusive. ie the limit for the kth root of (k/(k+3))^2k = 1

You're confusing two different series being discussed in this thread.

In the post you are referring to, I quoted and was referring to the series that Caleb mentioned in post#3:

Originally Posted by Caleb Sykes

consider the following:
sum((x/(x+3))^(x^2))

The Root Test can be used to show that Caleb's series converges.

The Root Test, however, is inconclusive for the series I posted in post#1.

[GeeTwo]

Originally Posted by Ether

Is there a shorter way?

For all k>2, 0 < k/(k+3) < 1

For all k>2, k[sup]2 > 2k

Therefore, for all k>2, (k/(k+3))^k2 < (k/(k+3))^2k.

Already proven: S_k=1^OO (k/(k+3))^2k converges

Therefore, S_k=1^OO (k/(k+3))^k2 converges by the Direct Comparison Test.

[Caleb Sykes]

Originally Posted by GeeTwo

Already proven: S_k=1^OO (k/(k+3))^2k converges

This statement is false. It has been proven that S_k=1^OO (k/(k+3))^2k diverges.

Originally Posted by Rachel Lim

Using the Divergence Test and l'Hopital's Rule (and the rule that lets me take the exponent outside the limit), I got that the series diverged: http://goo.gl/KVYEKZ

[GeeTwo]

Originally Posted by Caleb Sykes

This statement is false. It has been proven that S_k=1^OO (k/(k+3))^2k diverges.

Oops, that's what I get for not going back and refreshing my memory.

How about this one:

• JaredRachel has shown that (k/(k+3))^2k (the terms, not the series sum) converges to e^-6 ~ .00248 ~ 1/403.
• This means that for every k sufficiently large, (k/(k+3))^2k < 1/4 (actually I believe this is true for all of them)
• As k/(k+3) is positive, (k/(k+3))^k < 1/2 for sufficiently large k.
• Therefore, for every k sufficiently large, (k/(k+3))^k2 < 1/2^k
• We know S_k=1^OO 1/2^k, converges at 1.0.
• By the comparison test, S_k=1^OO (k/(k+3))^k2 converges.

[Ether]

Originally Posted by GeeTwo

Jared has shown that (k/(k+3))^2k (the terms, not the series sum) converges to e^-6

Give credit where credit is due: Rachel was the first to post that proof (4 hours earlier).

[Ether]

Originally Posted by GeeTwo

How about this one:

• [as shown earlier] (k/(k+3))^2k (the terms, not the series sum) converges to e^-6 ~ .00248 ~ 1/403.
• This means that for every k sufficiently large, (k/(k+3))^2k < 1/4 (actually I believe this is true for all of them)
• As k/(k+3) is positive, (k/(k+3))^k < 1/2 for sufficiently large k.
• Therefore, for every k sufficiently large, (k/(k+3))^k2 < 1/2^k
• We know S_k=1^OO 1/2^k, converges at 1.0.
• By the comparison test, S_k=1^OO (k/(k+3))[sup]k^2[/sup converges.

Unless I'm missing something, I see no flaw there. Nice.

[Strants]

I'm perhaps a little late, but it's not actually necessary to use l'Hospital's: we can just use the limit definition of the exponential function.

We have that (1 - a/n)ⁿ converges to e^-a. Thus, the sequence

a_k = (1-3/(k+3))^2k+6 = ((1-3/(k+3))^k+3)²

will converge to (e^-3)² = e^-6. The sequence of terms we have is

b_k= a_k / (1-3/(k+3))⁶

and (1-3/(k+3))⁶ goes to 1 as k becomes large, so in the limit b_k converges to e^-6/1 = e^-6.

[Ether]

Originally Posted by Strants

we can just use the limit definition of the exponential function.

Most sources call that a property rather than a definition.

See for example
http://mathworld.wolfram.com/ExponentialFunction.html

But let's call it a definition. Then since it's a definition you can use it even though it's not in the list of allowable list of test/rules/theorems specified in the original problem statement

I like your answer, nice work.

[M1KRONAUT]

I went into a really intricate solution, and got a quadratic: the limit as k goes to infinity of (k/(k+3))^k = 0, 1. Very helpful. So now I know that the series either converges or diverges. My previous attempts at a solution have had logical errors, unfortunately.