Testing Convergence/Divergence of Series

[DryRun]

Homework Statement
Test the series for convergence or divergence\sum_{n=1}^{\infty} \frac{n!}{4.7.10...(3n+1)}x^n,\; x>0

The attempt at a solution

I've decided to use the ratio test because of the factorial.
L=\lim_{n\to \infty}\frac{u_{n+1}}{u_n}
I worked it out and i got:
L=\lim_{n\to \infty}\frac{4x(3n+1)(n+1)}{3n+4}=\lim_{n\to \infty}\frac{4x(3n^2+4n+1)}{3n+4} Then, i did long division for the n terms.
\lim_{n\to \infty} 4x \left( n+ \frac{1}{3n+4}\right)=4x\lim_{n\to \infty} \left( n+ \frac{1}{3n+4}\right)=4x(∞+1/∞)=∞
According to the ratio test, since L>1, the series diverges.

Is this correct?

 

[LCKurtz]

You ratio test should have an |x| in there. But something else is wrong with your ratio. Where did the 4 and the (3n+1) come from?

 

[HallsofIvy]

Perhaps if you wrote the multiplication of the fractions out in detail, it would be clearer.

 

[Mark44]

Homework Statement
Test the series for convergence or divergence\sum_{n=1}^{\infty} \frac{n!}{4.7.10...(3n+1)}x^n,\; x>0

The attempt at a solution

I've decided to use the ratio test because of the factorial.
L=\lim_{n\to \infty}\frac{u_{n+1}}{u_n}
I worked it out and i got:
L=\lim_{n\to \infty}\frac{4x(3n+1)(n+1)}{3n+4}

Show us how you arrived at this result. As LCKurtz pointed out, there should be a factor of |x| in it.

 

[DryRun]

u_n=\frac{n!x^n}{4.7.10...(3n+1)}
u_{n+1}=\frac{(n+1)!x^{n+1}}{4.7.10...(3(n+1)+1)}
\frac{u_{n+1}}{u_n}=\frac{(n+1)!x^{n+1}}{4.7.10...(3(n+1)+1)}\times \frac{4.7.10...(3n+1)}{n!x^n}
\frac{u_{n+1}}{u_n}=\frac{(n+1)n!x^nx}{4.7.10...(3(n+1)+1)}\times \frac{4.7.10...(3n+1)}{n!x^n}
OK, i realize that i had made a mistake previously, but I'm still not sure:\frac{u_{n+1}}{u_n}=\frac{(n+1)x}{(3(n+1)+1)}= \frac {(n+1)x} {(3n+4)}Doing long division:\frac {(n+1)x} {(3n+4)}=\frac{1}{3}x-\frac{x}{3(3n+4))}Taking the limit, gives:\lim_{n\to\infty} \left[ \frac{1}{3}x-\frac{x}{3(3n+4)}\right]= \frac{1}{3}x

 

[HallsofIvy]

Yes, that is correct- although again, you have dropped the absolute value sign on x.

Now, do you remember why you divided those fractions? You wanted to use the ratio test. What does the ratio test say?

 

[Mark44]

\frac{u_{n+1}}{u_n}=\frac{(n+1)x}{(3(n+1)+1)}= \frac {(n+1)x} {(3n+4)}
Doing long division:

<snip>
Long division is unnecessary.

$$|x|\frac{n+1}{3n + 4} = |x|\frac{n(1 + 1/n)}{n(3 + 4/n)}$$

 

[DryRun]

Yes, that is correct- although again, you have dropped the absolute value sign on x.

Now, do you remember why you divided those fractions? You wanted to use the ratio test. What does the ratio test say?

I have a few doubts/questions:

1) I conclude that the requirement of having the absolute value sign on x is due to x &gt; 0 in the original problem. In that case, the absolute value sign should appear as from the first line in post #5, as |x^n|, and in the second line, as |x^{n+1}|. Correct?

2) Instead of \frac{1}{3}|x|, could i have alternatively given the answer as \frac{1}{3}x,\; x&gt;0? This should mean the same thing.

3) Now that i know the evaluation of the limit is \frac{1}{3}|x|, i need to find the value of L to know if the series converges or diverges.
However, here is the problem: I do not know the value of x. If x>3, then L>1 and the series diverges, but here is another possibility: if x=3, then L=1 and this means the series can either converge or diverge. A third possibility is if x<3, then L<1 and the series converges.
So, my question is: how do i know which value will x have? Although, x>0 does seem to indicate that i can take it to be as large a value as i like, in which case, L>1.

After some more thinking, i figure that as x increases from 0 to a value greater than 3, the series first converges, then diverges. But I'm not sure what the final answer should be.

 

[Mark44]

I have a few doubts/questions:

1) I conclude that the requirement of having the absolute value sign on x is due to x &gt; 0 in the original problem.

There's nothing in the original problem that says x > 0. In the Ratio Test, you look at this limit:
$$\lim_{n \to \infty}\left| \frac{a_{n+1}}{a_n}\right|$$

In that case, the absolute value sign should appear as from the first line in post #5, as |x^n|, and in the second line, as |x^{n+1}|. Correct?

Yes, but you can write these as |x|ⁿ and |x|ⁿ⁺¹.

2) Instead of \frac{1}{3}|x|, could i have alternatively given the answer as \frac{1}{3}x,\; x&gt;0? This should mean the same thing.

Bad idea since you are potentially losing half the interval of convergence.

3) Now that i know the evaluation of the limit is \frac{1}{3}|x|, i need to find the value of L to know if the series converges or diverges.

However, here is the problem: I do not know the value of x. If x>3, then L>1 and the series diverges, but here is another possibility: if x=3, then L=1 and this means the series can either converge or diverge. A third possibility is if x<3, then L<1 and the series converges.
So, my question is: how do i know which value will x have? Although, x>0 does seem to indicate that i can take it to be as large a value as i like, in which case, L>1.

If |x|/3 < 1 the series converges absolutely. What interval does that represent?
If |x|/3 = 1, the Ratio Test is inconclusive. Substitute the two values of x into the original series, and use some other test to determine the convergence/divergence of each of these two endpoints.

 

[DryRun]

If |x|/3 < 1 the series converges absolutely. What interval does that represent?

Since x cannot be a negative value and the original problem states that x>0, 0&lt;|x|&lt;3

If |x|/3 = 1, the Ratio Test is inconclusive. Substitute the two values of x into the original series, and use some other test to determine the convergence/divergence of each of these two endpoints.

The two values of x must be: <3 and >3. I choose x=1 and x=4. I'm not sure what you mean here.

 

[Mark44]

Apologies - I missed that it says x > 0 in your OP.

|x|/3 < 1 ==> -3 < x < 3
Since we also have x > 0, the interval reduces to 0 < x < 3, so I will dispense with absolute values.

If x/3 > 1, the Ratio Test says the series diverges (i.e., if x > 3).

If x/3 = 1, or equivalently, if x = 3, the Ratio Test is inconclusive. Use some other test to determine whether the series converges or diverges at this value.

 

[DryRun]

If x/3 = 1, or equivalently, if x = 3, the Ratio Test is inconclusive. Use some other test to determine whether the series converges or diverges at this value.

So, my final answer will have to comprise of 3 parts: when 0<x<3, x>3 and x=3. Correct?

Now, at x=3, the series becomes: \sum_{n=1}^{\infty} \frac{n!}{4.7.10...(3n+1)}3^nI would usually immediately go for the Ratio Test, but since i have to use another test... I would say the comparison test or the limit comparison test. But i don't know with what expression to compare it to, due to the n! term. I think the nth-term test might be easier, but then i would need to differentiate the numerator which contains n! and what's the derivative of n! w.r.t.n?

 

[Mark44]

If you're thinking "derivative" you're probably thinking L'Hopital's Rule, which is the wrong direction to go, IMO.

Break up the general term of the series like this...
$$\frac{1}{4} \frac{2}{7} \frac{3}{10}... \frac{n}{3n + 1}\cdot~3^n $$

Each of the fractions is bounded above and below, and there are n of them. Write an inequality with 3 members, with the above in the middle, and then take the limit of all three members.

 

[DryRun]

In your expression above, the numerator should be n! not just n, right?

The sequence of partial sums:S_n=\frac{3}{4}+\frac{18}{28}+\frac{162}{280}+...
The sequence:a_n=\frac{n!}{3n + 1}\cdot~3^n=\frac{3}{4},\frac{18}{7},\frac{54}{10},...
I have no idea what I'm doing.

 

[Mark44]

In your expression above, the numerator should be n! not just n, right?

No. I have broken n! up into 1*2*3* ... *n.

 

[DryRun]

OK, just to clarify your suggestion, for n=3:
\frac{1}{4} \frac{2}{7} \frac{3}{10}... \frac{n}{3n + 1}\cdot~3^n=\frac{1}{4}\times \frac{2}{7}\times \frac{3}{10}... \frac{n}{3n + 1}\cdot~3^n But in your expression, 3^n is not multiplied into the product. It should have been:\left( \frac{1}{4}\times \frac{2}{7}\times \frac{3}{10}\right)\cdot~3^3... \frac{n}{3n + 1}\cdot~3^n

 

[Mark44]

OK, just to clarify your suggestion, for n=3:
\frac{1}{4} \frac{2}{7} \frac{3}{10}... \frac{n}{3n + 1}\cdot~3^n=\frac{1}{4}\times \frac{2}{7}\times \frac{3}{10}... \frac{n}{3n + 1}\cdot~3^n

Yes, those are the same. I debated with myself about putting parentheses around each fraction, but decided it wasn't necessary.

But in your expression, 3^n is not multiplied into the product. It should have been:\left( \frac{1}{4}\times \frac{2}{7}\times \frac{3}{10}\right)\cdot~3^3... \frac{n}{3n + 1}\cdot~3^n

No it shouldn't. That's not the way multiplication works. You are apparently thinking that there is some distributive product for multiplication. There isn't.

IOW, this is NOT TRUE: (ab)c = (ac)(bc)

Multiplication distributes over addition ( a(b + c) = ab + ac) , but it doesn't distribute over multiplication!

 

[DryRun]

But i don't understand how to write an inequality out of these. Here's an attempt:
\left( \frac{1}{4}&lt;\frac{2}{7}&lt;\frac{3}{10}&lt;\frac{n}{3n + 1}\right)3^n

 

[Mark44]

For a start, let's focus on the fractions:
(1/4)(2/7)(3/10)...(n/(3n + 1))

1. How many fractions are in the above?
2. Every fraction above is >= what number?
3. Hence, the product above is >= what number?
4. Every fraction above is < what number?
5. Hence, the product above is < what number?

Now can you write an inequality involving (1/4)(2/7)(3/10)...(n/(3n + 1))3ⁿ?

 

[DryRun]

For a start, let's focus on the fractions:
(1/4)(2/7)(3/10)...(n/(3n + 1))

1. How many fractions are in the above?
2. Every fraction above is >= what number?
3. Hence, the product above is >= what number?
4. Every fraction above is < what number?
5. Hence, the product above is < what number?

Now can you write an inequality involving (1/4)(2/7)(3/10)...(n/(3n + 1))3ⁿ?

1. There are 4 fractions.

2. \frac{2}{7}&gt;\frac{1}{4},\; \frac{3}{10}&gt;\frac{2}{7},\; \frac{n}{3n + 1}&gt;\frac{3}{10}

3. Hence, \frac{n}{3n + 1}&gt;\frac{1}{4}

4. Every fraction above is less than 3^n

5. Hence, \frac{n}{3n + 1}&lt; 3^n

The inequality is:
\frac{1}{4}&lt; \frac{n}{3n + 1}&lt;3^n
If the above inequality is correct, then taking the limits for all 3 members:
\lim_{n \to \infty}\frac{1}{4}=\frac{1}{4}
\lim_{n \to \infty}\frac{n}{3n + 1}=\lim_{n \to \infty}\frac{1}{3 + 1/n}=\frac{1}{3}
\lim_{n \to \infty}3^n=∞The first two limits converges and the third limit diverges. What does this mean? The squeeze theorem fails in this case.

 

[Mark44]

For a start, let's focus on the fractions:
(1/4)(2/7)(3/10)...(n/(3n + 1))

1. How many fractions are in the above?
2. Every fraction above is >= what number?
3. Hence, the product above is >= what number?
4. Every fraction above is < what number?
5. Hence, the product above is < what number?

Now can you write an inequality involving (1/4)(2/7)(3/10)...(n/(3n + 1))3ⁿ?

1. There are 4 fractions.

That's literally true, but you are ignoring the ellipsis (...). If I write 1, 2, 3, ..., n, how many numbers am I implying?

2. \frac{2}{7}&gt;\frac{1}{4},\; \frac{3}{10}&gt;\frac{2}{7},\; \frac{n}{3n + 1}&gt;\frac{3}{10}

Yes, and since 1/4 < 2/7 < 3/10 < ... < n/(3n + 1), every fraction is greater than or equal to 1/4.

3. Hence, \frac{n}{3n + 1}&gt;\frac{1}{4}

Yes.

4. Every fraction above is less than 3^n

That's true, but not relevant, considering that we're focusing on just the fractions.

Considering just the fractions, each fraction is >= 1/4, so the product of these fractions is >= ? (Consider how many of them there are, which is why I asked question #1.

Now, can you get an upper bound on each fraction?
Can you get an upper bound on their product?

Everything below here is nonproductive...

5. Hence, \frac{n}{3n + 1}&lt; 3^n

The inequality is:
\frac{1}{4}&lt; \frac{n}{3n + 1}&lt;3^n
If the above inequality is correct, then taking the limits for all 3 members:
\lim_{n \to \infty}\frac{1}{4}=\frac{1}{4}
\lim_{n \to \infty}\frac{n}{3n + 1}=\lim_{n \to \infty}\frac{1}{3 + 1/n}=\frac{1}{3}
\lim_{n \to \infty}3^n=∞The first two limits converges and the third limit diverges. What does this mean? The squeeze theorem fails in this case.

 

[DryRun]

That's literally true, but you are ignoring the ellipsis (...). If I write 1, 2, 3, ..., n, how many numbers am I implying?

Infinite numbers, from 1, up to number, n. So, in this problem, there is an infinite number of fractions, from \frac{1}{4} \; to \frac{n}{3n + 1}.

4. Every fraction above is < what number?

\frac{n}{3n + 1}

5. Hence, the product above is < what number?

\left( \frac{1}{4}\times \frac{2}{7}\times \frac{3}{10}\times ... \times \frac{n}{3n + 1}\right)&lt;\frac{n}{3n + 1}
So,
\frac{n}{3n + 1}&gt;\left( \frac{1}{4}\times \frac{2}{7}\times \frac{3}{10}\times ... \times \frac{n}{3n + 1}\right)
And, \left( \frac{1}{4}\times \frac{2}{7}\times \frac{3}{10}\times ... \times \frac{n}{3n + 1}\right)≥\frac{1}{4}

 

[Mark44]

No, 1, 2, 3, ..., n is NOT an infinitely long list of numbers.

If I had written 1, 2, 3, ..., n, ... then that would have been a correct answer, but my list stops (i.e., is finite) at n. The ellipsis indicates that the pattern repeats indefinitely.

Go back to post #21 and answer the questions I asked.

 

[DryRun]

No, 1, 2, 3, ..., n is NOT an infinitely long list of numbers.

If I had written 1, 2, 3, ..., n, ... then that would have been a correct answer, but my list stops (i.e., is finite) at n. The ellipsis indicates that the pattern repeats indefinitely.

Go back to post #21 and answer the questions I asked.

Yes, it's a list of n numbers.

1. How many fractions are in the above?

There are n fractions.

4. Every fraction above is < what number?

\frac{n+1}{3(n+1) + 1}

5. Hence, the product above is < what number?

\left( \frac{1}{4}\times \frac{2}{7}\times \frac{3}{10}\times ... \times \frac{n}{3n + 1}\right)&lt;\frac{n+1}{3(n+1) + 1}

 

[Mark44]

OK, now we're getting somewhere.

1/4 <= 1/4
1/4 < 2/7
1/4 < 3/10
...
1/4 < n/(3n + 1)
So ? < (1/4)(2/7)(3/10)...(n/(3n+1))
Please don't tell me that ? = 1/4

 

[DryRun]

OK, now we're getting somewhere.

1/4 <= 1/4
1/4 < 2/7
1/4 < 3/10
...
1/4 < n/(3n + 1)
So ? < (1/4)(2/7)(3/10)...(n/(3n+1))
Please don't tell me that ? = 1/4

The product of proper fractions gives increasingly smaller fractions, then... the answer can only be 0, unless it's a negative number, which is not possible, as n increases from 1 in the series. So, 0 < (1/4)(2/7)(3/10)...(n/(3n+1))

 

[Mark44]

You're still thinking in terms of infinitely many factors. The product is not 0, and is in fact larger than 0. All of the fractions are positive, so I don't see how a negative product even enters into the discussion.

Try again...

 

[DryRun]

How about a geometric series, with r=\frac{n}{3n+1} and a=\frac{1}{4}
Then, S_n= \frac {a}{1-r}= \frac{\frac{1}{4}}{1- \frac{n}{3n+1}}
So, S_n&lt;(1/4)(2/7)(3/10)...(n/(3n+1))

 

[Mark44]

In a geometric series, the terms are added, not multiplied.

Since we're already 29 posts into this, here's a clue:

1/4 <= 1/4
1/4 < 2/7
so (1/4)² < (1/4)(2/7)

 

[DryRun]

In a geometric series, the terms are added, not multiplied.

Since we're already 29 posts into this, here's a clue:

1/4 <= 1/4
1/4 < 2/7
so (1/4)² < (1/4)(2/7)

I didn't see the problem from that angle. Here it is:
(1/4)^n&lt;(1/4)(2/7)(3/10)...(n/(3n+1))
So, the inequality becomes:
(1/4)^n&lt;(1/4)(2/7)(3/10)...(n/(3n+1))&lt;n/(3n+1)
Now, finding the limits:
\lim_{n \to \infty}\left( \frac{1}{4}\right)^n=0
\lim_{n \to \infty} \frac{n}{3n+1}=\frac{1}{3}

 

[Mark44]

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

YES![/color]

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