a_n =n^p * r^n for n>= 1, when is a_n convergent?

[dts5044]

Let a[sub:e6xilvbr]n[/sub:e6xilvbr] = n[sup:e6xilvbr]p[/sup:e6xilvbr]r[sup:e6xilvbr]n[/sup:e6xilvbr] for n >= 1, where p, r are two fixed real numbers. When is the sequence (a[sub:e6xilvbr]n[/sub:e6xilvbr])[sub:e6xilvbr]n>=1[/sub:e6xilvbr] convergent, and what is its limit?

I broke it into [lim n --> infinity] n[sup:e6xilvbr]p[/sup:e6xilvbr] * [lim n --> infinity] r[sup:e6xilvbr]n[/sup:e6xilvbr]. Is this the right way to approach the problem? From there I'm not too sure what to do, can someone help?

 

[royhaas]

Use the ratio test.

 

[dts5044]

I forgot to mention that what I'm dealing with is a sequence, although I'm not sure your answer isn't still valid. In general, if a series is convergent is the sequence of the same function also convergent?

 

[pka]

Do you know the following theorem?
If \(\displaystyle s_n \ne 0\) and \(\displaystyle \lim\left({\left|{\frac{{s_{n+1}}}{{s_n}}}\right|}\right)=L\) then
\(\displaystyle \left\{ \begin{array}{l} L < 1\quad \Rightarrow \quad \left( {s_n } \right) \to 0 \\ L > 1\quad \Rightarrow \quad \left( {s_n } \right) \to \infty \\ \end{array} \right.\)

That is how you use the ratio test.

 

[dts5044]

I know how to do the ratio test. My question is: this problem deals with a sequence. I know the ratio test applies to series, but does it apply to sequences as well? and if so, why?

 

[pka]

I know how to do the ratio test. My question is: this problem deals with a sequence. I know the ratio test applies to series, but does it apply to sequences as well? and if so, why?

Did you read my previous post carefully?
That theorem applies to sequences!
The word series does not even appear in the entire post.

 

[dts5044]

but doesn't s[sub:12ig8tly]n[/sub:12ig8tly] refer to the partial sum of the first n terms, which is a series?

 

[stapel]

but doesn't s[sub:3p7usp9a]n[/sub:3p7usp9a] refer to the partial sum of the first n terms, which is a series?

Why would the source of the terms of the sequence matter? :shock:

If s[sub:3p7usp9a]n[/sub:3p7usp9a] = a[sub:3p7usp9a]1[/sub:3p7usp9a] + a[sub:3p7usp9a]2[/sub:3p7usp9a] + a[sub:3p7usp9a]3[/sub:3p7usp9a] + ... + a[sub:3p7usp9a]n[/sub:3p7usp9a], so what? You're still dealing with the sequence terms s[sub:3p7usp9a]n[/sub:3p7usp9a], and the test still applies, doesn't it? :wink:

Eliz.

 

[dts5044]

ok, so applying the ratio test:
[lim as n goes to infinity]:
(a)[sup:32rgp93r]n+1[/sup:32rgp93r]/ (n + 1)! * n!/(a)[sup:32rgp93r]n[/sup:32rgp93r]

= (a)/ (n+1)

and since the denominator is shooting to infinity, a can be any real number? and the limit is always equal to 0?

Honestly, I am still very confused. Could someone PLEASE take 5 minutes to provide an in-depth explanation of the relationship between sequences and series and why I can use a series test on a sequence?

 

[pka]

Honestly, I am still very confused. Could someone PLEASE take 5 minutes to provide an in-depth explanation of the relationship between sequences and series and why I can use a series test on a sequence?

I must tell you that I had resolved not to try to help you further.
Not at sure this will make sense to you because you still insist upon conflating a sequence problem with series.
Using the theorem about sequences that I posted we get:
\(\displaystyle a_n=n^pr^n\Rightarrow\quad\left|{\frac{{a_{n+1}}}{{a_n}}}\right|=\left|r\right|\left({1+\frac{1}{n}}\right)^p\).
So \(\displaystyle - 1 < r < 1 \Rightarrow \quad \left( {a_n } \right) \to 0\)
\(\displaystyle \left|r\right|>1\Rightarrow\quad\left|{a_n}\right|\to\infty\)

 

[dts5044]

well I understand up to the statement abs(a[sub:1nfa8kig]n+1[/sub:1nfa8kig] / a[sub:1nfa8kig]n[/sub:1nfa8kig]) = abs(r) * (1 + 1/n)[sup:1nfa8kig]p[/sup:1nfa8kig], but not the analysis where you end up with if -1 < r < 1, then a[sub:1nfa8kig]n[/sub:1nfa8kig] goes to infinity.

I consider myself a very capable math student, have been working on this issue all day, and haven't come up with the answer. Is it really so hard to take the time and explain why your analysis holds? I'm not asking you to do a problem for me, just explain a concept so I can do it for myself! That, essentially, is the job of a tutor; even one on a volunteer basis.

 

[stapel]

Is it really so hard to take the time and explain why your analysis holds? I'm not asking you to do a problem for me, just explain a concept so I can do it for myself!

I'm sorry, but it's not a "concept", or an "analysis" thereof. It's a definition: The list of partial sums of a series form a sequence of values, and this sequence of values is a "sequence".

The string of values, being the sums from a[sub:ii4jxi3a]1[/sub:ii4jxi3a] to a[sub:ii4jxi3a]1[/sub:ii4jxi3a] (this partial sum being the value s[sub:ii4jxi3a]1[/sub:ii4jxi3a]), from a[sub:ii4jxi3a]1[/sub:ii4jxi3a] to a[sub:ii4jxi3a]2[/sub:ii4jxi3a] (being s[sub:ii4jxi3a]2[/sub:ii4jxi3a]), from a[sub:ii4jxi3a]1[/sub:ii4jxi3a] to a[sub:ii4jxi3a]3[/sub:ii4jxi3a] (being s[sub:ii4jxi3a]3[/sub:ii4jxi3a]),, from a[sub:ii4jxi3a]1[/sub:ii4jxi3a] to a[sub:ii4jxi3a]4[/sub:ii4jxi3a] (being s[sub:ii4jxi3a]4[/sub:ii4jxi3a]),, ..., and from a[sub:ii4jxi3a]1[/sub:ii4jxi3a] to a[sub:ii4jxi3a]n[/sub:ii4jxi3a] (being s[sub:ii4jxi3a]n[/sub:ii4jxi3a]),, are a sequence. It doesn't matter that the sequence values s[sub:ii4jxi3a]n[/sub:ii4jxi3a] happen to have been derived from a series (from the sum of a[sub:ii4jxi3a]1[/sub:ii4jxi3a] through a[sub:ii4jxi3a]n[/sub:ii4jxi3a]). The string of partial-sum values s[sub:ii4jxi3a]n[/sub:ii4jxi3a] is a sequence of values.

If the values s[sub:ii4jxi3a]n[/sub:ii4jxi3a] in that sequence have a formula, it doesn't matter if that formula happens to have been derived from a summation. The formula gives the values of each term in the sequence, and the rules for testing those sequence values still apply.

I'm sorry, but I don't know how to say it any more clearly: A string of values is a sequence of values, regardless of how those values might have been generated. Any rule, formula, or theorem which applies to other sequences will apply to your partial-sum sequence. The source of your sequence has no bearing on anything.

I apologize for our inability to explain this to you, and for the offence which this has caused you. We really have tried; it is to be regretted that we have caused you such upset by our failure.

My best wishes to you in finding a tutor who can explain the definitions better.

Eliz.

 

[Deleted member 4993]

well I understand up to the statement abs(a[sub:32vunc4a]n+1[/sub:32vunc4a] / a[sub:32vunc4a]n[/sub:32vunc4a]) = abs(r) * (1 + 1/n)[sup:32vunc4a]p[/sup:32vunc4a], but not the analysis where you end up with if -1 < r < 1, then a[sub:32vunc4a]n[/sub:32vunc4a] goes to infinity.

I consider myself a very capable math student, .....

That could be true - you can consider yourself anything - however, here you are forgetting that you are supposed to take the limit of the ratio.

When you do that - what is the limit of the ratio (as n -> inf)?

Is it |r|?

What happens to the sequence when the ratio is < 1? What happens to the nth term of that sequence?

What happens to the sequence when the ratio is > 1? What happens to the nth term of that sequence?

A moderately capable student should have figured that out - let alone 'very' capable........

 

[dts5044]

okay, that was obvious and I missed it. But please, the unnecessary animosity doesn't help anyone. I don't need your approval to know I am a good math student, and it speaks very poorly of you that you insulted me. Everyone, even the smartest people in the world, are capable of making mistakes or missing the obvious.

and to stapel: thank you very much for taking the time to explain it. I didn't mean to offend and I'm sorry if I did. I had spent a long time thinking about the issue off and on and felt everyone who responded hadn't answered the question I was struggling with. I respect very much that you take your time voluntarily to help people who need help and please don't think I was complaining in general about your, or anyone's, ability as a tutor. And again, thank you for taking the time to explain it.

One general question I have left is: A sequence is said to be convergent if the limit of its nth term approaches a real number, and a series can only be convergent if the limit of its nth term equals 0. What effect, if any, does this have on the tests you can apply? (i.e. for the ratio test the series is absolutely convergent if lim n --> infinity |a[sub:1051c6vk]n+1[/sub:1051c6vk]/a[sub:1051c6vk]n|[/sub:1051c6vk] < 1. does the same evaluation apply for a sequence? (if lim n --> infinity |a[sub:1051c6vk]n+1[/sub:1051c6vk]/a[sub:1051c6vk]n|[/sub:1051c6vk] < 1 then the sequence (a[sub:1051c6vk]n[/sub:1051c6vk]) is convergent too?)) I think the answer is yes, but am not 100% sure and don't want to make a mistake.

 

[daon]

One general question I have left is: A sequence is said to be convergent if the limit of its nth term approaches a real number, and a series can only be convergent if the limit of its nth term equals 0. What effect, if any, does this have on the tests you can apply? (i.e. for the ratio test the series is absolutely convergent if lim n --> infinity |a[sub:29a2qeue]n+1[/sub:29a2qeue]/a[sub:29a2qeue]n|[/sub:29a2qeue] < 1. does the same evaluation apply for a sequence? (if lim n --> infinity |a[sub:29a2qeue]n+1[/sub:29a2qeue]/a[sub:29a2qeue]n|[/sub:29a2qeue] < 1 then the sequence (a[sub:29a2qeue]n[/sub:29a2qeue]) is convergent too?)) I think the answer is yes, but am not 100% sure and don't want to make a mistake.

For a series, \(\displaystyle \sum a_n\) is, in general, NOT convergent if \(\displaystyle |\frac{a_{n+1}}{a_n}| < 1\). i.e. Take \(\displaystyle a_n = \frac{1}{n}\). The sum diverges. That works for sequences only.