Testing Convergence/Divergence of Series

[Mark44]

I didn't see the problem from that angle. Here it is:
(1/4)^n<(1/4)(2/7)(3/10)...(n/(3n+1))

YES![/color]

So, the inequality becomes:
(1/4)^n<(1/4)(2/7)(3/10)...(n/(3n+1))<n/(3n+1)

On the right, not what I was looking for. Every fraction in the product, bar none, is smaller than what specific number (not a variable)?

 

[DryRun]

On the right, not what I was looking for. Every fraction in the product, bar none, is smaller than what specific number (not a variable)?

It could be any fraction bigger than 1/4, like 1/3, 1/2 or 1.

 

[Mark44]

1/3 will do handsomely, which you will see shortly.

Here's where we are:

(1/4)ⁿ < (1/4)(2/7)(3/10) ... (n/(3n + 1)) < _____?

Once you get that, then multiply all three members by 3ⁿ. What does that inequality look like?

Now take the limit as n --> ∞. By the n-th term test, what can you say about the series that you're investigating? (This is really all about that series you started the thread with.)

 

[DryRun]

(1/4)^n&lt;(1/4)(2/7)(3/10)...(n/(3n+1))&lt;1/3
Multiplying throughout by 3^n:
(3/4)^n&lt;(1/4)(2/7)(3/10)...(n/(3n+1)).3^n&lt;(1/3).3^n
\lim_{n \to \infty}(3/4)^n=0
\lim_{n \to \infty}(1/3).3^n=∞
\lim_{n \to \infty}(1/4)(2/7)(3/10)...(n/(3n+1)).3^n=∞
I think i get it. The first limit converges, which is the same as when 0 < x < 3 and the second limit diverges which corresponds to x > 3. Since the third limit is ∞, therefore, according to the nth-term test, the series diverges at x = 3.

 

[Mark44]

(1/4)^n&lt;(1/4)(2/7)(3/10)...(n/(3n+1))&lt;1/3

What's missing here?

 

[DryRun]

(1/4)^n≤(1/4)(2/7)(3/10)...(n/(3n+1))&lt;1/3

 

[Mark44]

That's not what's missing. The inequality was fine with <.

 

[DryRun]

Well, in post #34, i already multiplied throughout by 3^n. I'm not sure what else is missing.

Did you mean rewriting the inequality into this form?
(1/4)^n&lt;\frac{n!}{4.7.10...(3n+1)}&lt;1/3

 

[Mark44]

No, you off on the wrong track. Isn't there something glaringly different between your lower bound and upper bound for the product of the fractions?

 

[DryRun]

No, you off on the wrong track. Isn't there something glaringly different between your lower bound and upper bound for the product of the fractions?

Really, i have no idea.

 

[Mark44]

Let me make it simpler for you. Here are your lower and upper bounds. Don't you notice something very different about them?

(1/4)^n&lt; \text{snip}&lt;1/3

 

[DryRun]

Do you mean that the lower bound is raised to the power of n? I guess i could eliminate that power, since the limit converges either way.
1/4&lt;\frac{n!}{4.7.10...(3n+1)}&lt;1/3

 

[Mark44]

Do you mean that the lower bound is raised to the power of n?

Yes, I mean exactly that.

I guess i could eliminate that power

Why would you do that? Don't you remember why you put it in there?

The question is, why isn't there an exponent on the other side of the inequality? Each fraction is strictly less than 1/3, so the product of those n fractions will be < _______?

, since the limit converges either way.
1/4&lt;\frac{n!}{4.7.10...(3n+1)}&lt;1/3

You keep taking the limit before you get the correct answers to the previous steps. IOW, you keep jumping the gun, and coming up with limit values that are bogus and have nothing to do with this problem.

After you get the correct lower and upper bounds, then multiply all three members by 3ⁿ. Finally, take the limit.

 

[DryRun]

This has to be it:(1/4)^n&lt;\frac{n!}{4.7.10...(3n+1)}&lt;(1/3)^n
I'll wait for confirmation before finding the limits.

 

[Mark44]

Bingo.

This has to be it:(1/4)^n&lt;\frac{n!}{4.7.10...(3n+1)}&lt;(1/3)^n
I'll wait for confirmation before finding the limits.

So the following is also true.
$$(1/4)^n 3^n<\frac{3^n n!}{4.7.10...(3n+1)}<(1/3)^n 3^n$$

Since the inequality is true for each n >= 1, it's true in the limit as n --> ∞.
What do you conclude from this?

 

[DryRun]

The inequality reduces to:
(3/4)^n&lt;\frac{3^n n!}{4.7.10...(3n+1)}&lt;(1)^n
\lim_{n\to \infty}(3/4)^n=0
\lim_{n\to \infty}(1)^n=1
I would use the sandwich theorem to find the limit of the middle term in the inequality, but the limits of the lower and upper bounds are different, so i don't think that the sandwich theorem applies in this case.

 

[Mark44]

Since $\lim_{n \to \infty}\frac{3^ n!}{4\cdot 7 \cdot 10 \cdot \cdot \cdot (3n + 1)} > 0$, (in other words, it's not equal to 0), then by the n-th term test, the series diverges when x = 3.

 

[DryRun]

Well, it's been a long journey with this one. I believe we have finally reached an agreeable conclusion.

As for the answer to this problem, i believe we dug deep into it and fleshed it out completely, but the actual answer (this past-exam question is only worth 5 points) relates to x>0, and by that, i assume it suggests x>>0, which would be x>3, therefore the series diverges.

Thank you for your infinite patience, Mark44. As always, your help is much appreciated.

 

[Mark44]

We already know that for x > 3, the series diverges. The concern for the past 20 or so posts has been what happens at x = 3, and this is what we found out. This allows us to write the interval of convergence as (0, 3), and not (0, 3].

 

[DryRun]

And the interval of divergence is [3,∞)

Thanks again.