Ratio Test and Radius of Convergence for ∑ ((n-2)2)/n2, n=1: Homework Solution

[ScreamingIntoTheVoid]

Homework Statement

∞
∑ = ((n-2)²)/n²
n=1

Homework Equations

The ratio test/interval of convergence

The Attempt at a Solution

**NOTE this is a bonus homework and I've only had internet tutorials regarding the ratio test/interval of convergence so bear with me)
lim ((n-1)ⁿ⁺¹)/(n+1)ⁿ⁺¹ * (nⁿ)/(n-2)ⁿ
n→∞
That's all I've got. I know with the ratio test you're supposed to simplify this but I'm not sure how. If anyone could walk me through this, that'd be super helpful

Edit: I was also made aware that the root test exists... Can I use that instead?
n√((n-2)_n/n_n →n-2/n →n/n -2/n →lim as n→∞ 1-2/n = 1
If that is correct, what does that 1 mean? Can I use that to find the interval of convergence?

 

[Ray Vickson]

Homework Statement

∞
∑ = ((n-2)²)/n²
n=1

Homework Equations

The ratio test/interval of convergence

The Attempt at a Solution

**NOTE this is a bonus homework and I've only had internet tutorials regarding the ratio test/interval of convergence so bear with me)
lim ((n-1)ⁿ⁺¹)/(n+1)ⁿ⁺¹ * (nⁿ)/(n-2)ⁿ
n→∞
That's all I've got. I know with the ratio test you're supposed to simplify this but I'm not sure how. If anyone could walk me through this, that'd be super helpful

Edit: I was also made aware that the root test exists... Can I use that instead?
n√((n-2)_n/n_n →n-2/n →n/n -2/n →lim as n→∞ 1-2/n = 1
If that is correct, what does that 1 mean? Can I use that to find the interval of convergence?

If (as your notation suggests) you series has the form $\sum_n t_n$ with $t_n = (n-2)^2/n^2$, can you see immediately why the series cannot possibly converge? No fancy tests are needed at all!

 

[ScreamingIntoTheVoid]

If (as your notation suggests) you series has the form $\sum_n t_n$ with $t_n = (n-2)^2/n^2$, can you see immediately why the series cannot possibly converge? No fancy tests are needed at all!

Ah my computer messed up! Instead of to the 2nd power I meant to the nth power, but I'm assuming that wouldn't make a difference. I'm not really sure why it doesn't converge (sorry). Would it make a difference if 1+n^2 was on the bottom rather than n^2 (that's the second question on my homework)?

 

[Ray Vickson]

Ah my computer messed up! Instead of to the 2nd power I meant to the nth power, but I'm assuming that wouldn't make a difference. I'm not really sure why it doesn't converge (sorry). Would it make a difference if 1+n^2 was on the bottom rather than n^2 (that's the second question on my homework)?

Do you mean "nth" power on both the numerator and denominator---in other words, $[(n-2)/n]^n$? And: if $1+n^2$ is in the denominator (instead of $n^2$), what do YOU think the difference would be, if any? (As you can tell, I am really, really reluctant to give you an answer; you should not need one.)

 

[ScreamingIntoTheVoid]

Do you mean "nth" power on both the numerator and denominator---in other words, $[(n-2)/n]^n$? And: if $1+n^2$ is in the denominator (instead of $n^2$), what do YOU think the difference would be, if any? (As you can tell, I am really, really reluctant to give you an answer; you should not need one.)

I'm thinking that perhaps that (n-2)^n heads towards infinity faster than n^n, which would explain why the equation is divergent? And it's a bit more obvious that (n-2)^n increases to infinity faster than n^2, so that one's obviously divergent... I think I'm wrong with the first one can I have a bit of a hint?

 

[ScreamingIntoTheVoid]

Do you mean "nth" power on both the numerator and denominator---in other words, $[(n-2)/n]^n$? And: if $1+n^2$ is in the denominator (instead of $n^2$), what do YOU think the difference would be, if any? (As you can tell, I am really, really reluctant to give you an answer; you should not need one.)

So I figured my initial statement was wrong, but I still can't figure out why the first problem is divergent. At infinity the limit of the equation would equal to 1. We just learned about geometric series today if that's what your referring to so I'm kind of drawing a blank

Edit: perhaps this has to do with common limits (my teacher hasn't discussed those either, but I found a couple of videos that referenced them). I know that the limit of this equation is 1/e^2 [used a calculator], but I'm not sure why that is or if that implies divergence

 

[FactChecker]

You should try to be careful and specific about what you are asking. You never specifically answered @Ray Vickson 's question of whether you meant [(n-2)/n]ⁿ. And what equation are you referring to in "the limit of this equation is 1/e^2"? Do you mean the summation or the individual terms?

I'm sorry to be picky, but these are things that we have to guess at before we can answer your questions.

 

[Mark44]

I'm thinking that perhaps that (n-2)^n heads towards infinity faster than n^n, which would explain why the equation is divergent?

There are a couple of things wrong here. 1) n -2 < n, so $(n - 2)^n < n^n$, for reasonably large n. 2) Both of these expressions approach infinity at the same rate.

You mentioned the n-th root test. Since the general term in your series now seems to be $\frac{(n - 2)^n}{n^n} = (\frac {n - 2} n)^n$, you could use that test.

 

[Mark44]

BTW, your thread title is "Radius of Convergence." That concept has nothing to do with this problem, at least as you have written it. Radius of convergence has to do with power series, which are series consisting of powers of some variable, such as, for instance $\sum_{n = 1}^\infty \frac{(x - 2)^n}{n}$.

 

[Ray Vickson]

I'm thinking that perhaps that (n-2)^n heads towards infinity faster than n^n, which would explain why the equation is divergent? And it's a bit more obvious that (n-2)^n increases to infinity faster than n^2, so that one's obviously divergent... I think I'm wrong with the first one can I have a bit of a hint?

There are some very basic things that must hold if an infinite series is to converge; look at
http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries.aspx
and go the the very first theorem you see there (about 2 or 3 screens down from the beginning).

 

[ScreamingIntoTheVoid]

BTW, your thread title is "Radius of Convergence." That concept has nothing to do with this problem, at least as you have written it. Radius of convergence has to do with power series, which are series consisting of powers of some variable, such as, for instance $\sum_{n = 1}^\infty \frac{(x - 2)^n}{n}$.

The reason it doesn't is because I messed up writing the question twice. Later last night I figured out that I had been mis-writing the problem as "(n-2)^n/n^n" rather than "(x-2)^n/n^n" which also explains why I was struggling to solve the problem itself. That one had a radius of convergence of 2 I believe.

 

[Ray Vickson]

The reason it doesn't is because I messed up writing the question twice. Later last night I figured out that I had been mis-writing the problem as "(n-2)^n/n^n" rather than "(x-2)^n/n^n" which also explains why I was struggling to solve the problem itself. That one had a radius of convergence of 2 I believe.

NO, absolutely not.

Let me ask:
(1) Do you know what it means for an infinite series to "converge" of "diverge"?
(2) Do you know what a "radius of convergence" actually means?

Judging from the nature of your questions I cannot figure out if you really do know the proper answers to those two questions. Getting answers is crucial before we can even start to discuss the issues.

 

[ScreamingIntoTheVoid]

NO, absolutely not.

Let me ask:
(1) Do you know what it means for an infinite series to "converge" of "diverge"?
(2) Do you know what a "radius of convergence" actually means?

Judging from the nature of your questions I cannot figure out if you really do know the proper answers to those two questions. Getting answers is crucial before we can even start to discuss the issues.

1) Yes, convergence implies that the limit of the series is less than infinity/exists/the series is bounded. Divergence implies that the limit of the series goes to infinity/the series is unbounded.
2) Not really. As I mentioned, my teacher never actually mentioned it (and is not planning to before the end of this semester) and my only understanding of it comes from a singular Kahn academy video I watched and a couple of things I have read about it.

Despite the questions I posed last night, I do understand limits and have a general understanding of series (though that's our most recent topic and I'm certainly rough around the edges here). I just am unfamiliar with intervals of convergence and radius of convergence, as this was a topic I only heard of since yesterday.

 

[Ray Vickson]

1) Yes, convergence implies that the limit of the series is less than infinity/exists/the series is bounded. Divergence implies that the limit of the series goes to infinity/the series is unbounded.
2) Not really. As I mentioned, my teacher never actually mentioned it (and is not planning to before the end of this semester) and my only understanding of it comes from a singular Kahn academy video I watched and a couple of things I have read about it.

Despite the questions I posed last night, I do understand limits and have a general understanding of series (though that's our most recent topic and I'm certainly rough around the edges here). I just am unfamiliar with intervals of convergence and radius of convergence, as this was a topic I only heard of since yesterday.

Convergence of $\sum_{n=1}^{\infty} t_n$ means that the sequence of partial sums $S_N = \sum_{n=1}^N t_n$ converges to a finite limit as $N \to \infty$. Divergence means that the $S_N$ do not have a finite limit; however, absence of a limit does not always mean that the sums go to $+\infty$ or $-\infty$. For example, the simple series $S = \sum_{n=0}^{\infty} (-1)^n = 1 -1 +1 -1 + \cdots$ is divergent, but its partial sums are either equal to 0 or +1, and so do not become unbounded at all; they just oscillate and never settle down near some fixed value.

As for radius of convergence: a power series---that is, a series of the form
$$\sum_{n=0}^{\infty} a_n x^n $$
has radius of convergence $r \; (\text{with} \; r > 0)$ if it is convergent for all values of $x$ with magnitude $< r ;$ that is, it converges if $|x| < r .$ It might, or might not converge when $x=r$ or $x=-r$: some series do, and some don't.

If a series has radius of convergence $r = 0$ that means that it converges only when $x = 0$---in which case it is just the series $0+0+0+ \cdots$. Some power series have infinite radii of convergence, which means that they converge for all values of $x$.

 

[WWGD]

1) Yes, convergence implies that the limit of the series is less than infinity/exists/the series is bounded. Divergence implies that the limit of the series goes to infinity/the series is unbounded.
.

First of all, excellent handle, you must get this one right, to honor the handle ;). At any rate, this test is nice to have in your bag of tricks: a series will converge iff the difference between consecutive terms becomes indefinitely small *. Notice Ray's sum 1-1+1-1+... violates this condition.

* There may be other "environments" where this is not true, but don't worry for now.
EDIT: Please ignore . This is a necessary but , as Mark pointed out, not sufficient; my bad for trying to cut corners. Please see my correction in post #17 .

 

[Mark44]

a series will converge iff the difference between consecutive terms becomes indefinitely small *.

Not true. A counterexample is $\sum_{n = 1}^\infty \frac 1 n$. The difference between $S_{n + 1}$ and $S_n$ is $\frac 1 {n + 1}$, which can be made arbitrarily* small.

* I believe "arbitrarily small" was the expression you were looking for, not indefinitely small.

 

[WWGD]

Not true. A counterexample is $\sum_{n = 1}^\infty \frac 1 n$. The difference between $S_{n + 1}$ and $S_n$ is $\frac 1 {n + 1}$, which can be made arbitrarily* small.

* I believe "arbitrarily small" was the expression you were looking for, not indefinitely small.

Yes, I was thinking of the n-th term, but this is obviously wrong. So much for trying to popularize terms like Cauchy sequence and complete metric spaces without getting tangled up in details :

You need the condition that, given any $\epsilon >0$ ,beyond a certain term $N$ , $| a_j -a_k| < \epsilon ; \forall j,k >N$

 

[Mark44]

Yes, I was thinking of the n-th term, but this is obviously wrong. So much for trying to popularize terms like Cauchy sequence and complete metric spaces without getting tangled up in details :

You need the condition that, given any $\epsilon >0$ ,beyond a certain term $N$ , $| a_j -a_k| < \epsilon ; \forall j,k >N$

Which fairly obviously does not apply to the Harmonic series of my counterexample.

 

[WWGD]

Which fairly obviously does not apply to the Harmonic series of my counterexample.

Yes, it is pretty tricky to popularize and simplify without making these types of mistakes; at least for me, and , from what I have seen, for others. Ian Stewart is one of the few ( 2-3) that I have seen pull this off. EDIT: Maybe most of the fat has already been cut off from Mathe explanations, and little else can be removed without damage.

 

[StoneTemplePython]

Yes, it is pretty tricky to popularize and simplify without making these types of mistakes; at least for me, and , from what I have seen, for others. Ian Stewart is one of the few ( 2-3) that I have seen pull this off. EDIT: Maybe most of the fat has already been cut off from Mathe explanations, and little else can be removed without damage.

I think you need to ask: What Would Polya Do?

 

[WWGD]

I think you need to ask: What Would Polya Do?

Nice plug! EDIT: You're beating me at this, you should be worried :).

 

[FactChecker]

Yes, it is pretty tricky to popularize and simplify without making these types of mistakes; at least for me, and , from what I have seen, for others. Ian Stewart is one of the few ( 2-3) that I have seen pull this off. EDIT: Maybe most of the fat has already been cut off from Mathe explanations, and little else can be removed without damage.

I like and agree with your edit comment.

 

[ScreamingIntoTheVoid]

Convergence of $\sum_{n=1}^{\infty} t_n$ means that the sequence of partial sums $S_N = \sum_{n=1}^N t_n$ converges to a finite limit as $N \to \infty$. Divergence means that the $S_N$ do not have a finite limit; however, absence of a limit does not always mean that the sums go to $+\infty$ or $-\infty$. For example, the simple series $S = \sum_{n=0}^{\infty} (-1)^n = 1 -1 +1 -1 + \cdots$ is divergent, but its partial sums are either equal to 0 or +1, and so do not become unbounded at all; they just oscillate and never settle down near some fixed value.

As for radius of convergence: a power series---that is, a series of the form
$$\sum_{n=0}^{\infty} a_n x^n $$
has radius of convergence $r \; (\text{with} \; r > 0)$ if it is convergent for all values of $x$ with magnitude $< r ;$ that is, it converges if $|x| < r .$ It might, or might not converge when $x=r$ or $x=-r$: some series do, and some don't.

If a series has radius of convergence $r = 0$ that means that it converges only when $x = 0$---in which case it is just the series $0+0+0+ \cdots$. Some power series have infinite radii of convergence, which means that they converge for all values of $x$.

Right, that makes sense, thank you for taking the time to provide me with a more extensive responce. And r would be found using the ratio test correct? So just to make sure that I'm doing this right, if I had my original problem (except written correctly):
∞
Σ (x-2)ⁿ/nⁿ , using the ratio test I would get (x-2)ⁿ⁺¹/nⁿ⁺¹ * (nⁿ)/(x-2)ⁿ which I
n→1
could turn into [(x-2)(x-2)ⁿ] * (nⁿ)/(x-2)ⁿ , which by applying a limit and simplifying that would turn into:
|x-2| lim n→∞ (nⁿ/(n+1)ⁿ⁺¹) which would turn into |x-2|>0, leaving the interval of convergence to be -2>n>2 and the r value to be 2?

(note: That's probably a somewhat messy version of the process if I did it right because I also taught myself intervals of convergence/the ratio of convergence last night. Though I feel like I have somewhat of a grasp on those, I've probably done it in a somewhat messy manner)

**ALSO thanks for responding to this thread even though it's just about a day old**

 

[Ray Vickson]

Right, that makes sense, thank you for taking the time to provide me with a more extensive responce. And r would be found using the ratio test correct? So just to make sure that I'm doing this right, if I had my original problem (except written correctly):
∞
Σ (x-2)ⁿ/nⁿ , using the ratio test I would get (x-2)ⁿ⁺¹/nⁿ⁺¹ * (nⁿ)/(x-2)ⁿ which I
n→1
could turn into [(x-2)(x-2)ⁿ] * (nⁿ)/(x-2)ⁿ , which by applying a limit and simplifying that would turn into:
|x-2| lim n→∞ (nⁿ/(n+1)ⁿ⁺¹) which would turn into |x-2|>0, leaving the interval of convergence to be -2>n>2 and the r value to be 2?

(note: That's probably a somewhat messy version of the process if I did it right because I also taught myself intervals of convergence/the ratio of convergence last night. Though I feel like I have somewhat of a grasp on those, I've probably done it in a somewhat messy manner)

**ALSO thanks for responding to this thread even though it's just about a day old**

Sometimes the ratio test can be used to determine the radius of convergence, but sometimes other tests must be used. Sometimes we really do not need any such tests at all, but can just rely on a bounding property.

The radius of convergence of $\sum (x-2)^n/n^n$ is $\infty$, not 2 or 3 or any other finite number.

First: simplify writing by putting $x-2 = y$, so the series is $\sum y^n/n^n$. If we set $t_n = y^n/n^n$ we have the ratio
$$ \frac{|t_{n+1}|}{|t_n|} = \frac{|y|^{n+1}}{|y|^n} \frac{n^n}{(n+1)^{n+1}} = |y| \left( \frac{n}{n+1}\right)^n \frac{1}{n+1}. $$
Now $n/(n+1) < 1$ so $(n/(n+1))^n < 1$ for all $n > 0$, so we have
$$\frac{|t_{n+1}|}{|t_n|} < |y| \, \,\frac{1}{n+1} \to 0\; \text{as} \; n \to \infty.$$
The limiting ratio is <1 for any $|y|$---and, in fact, is < 0.1 or < 0.000000000001 or ... or less than any positive number at all. Therefore, the sum will converge, no matter what is the value of $y = x-2$, so will converge no matter what is the value of $x$. It converges for $x = -1,000,000$ or for $x = 175,000$ or whatever other value you choose to employ.

The radius of convergence cannot be 2 because you would be saying that the series diverges if $|x-2| > 2$, and that is definitely not the case here.