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Sequences - Salguod - 2011-04-06
I'm doing the basics of sequences like converging, diverging, monotonic or not, upper/lower bounds and I need help.
Determine if the sequence converges or diverges.
1. a[SIZE="1"]n[/SIZE] = 1/n^2 + 2/n^2 + ... + n/n^2
My first thought was to multiply top and bottom by 1/n^2 to put all the n's on the bottom but that gave me 0 instead of 1/2, so my approach is probably wrong.
Getting that wrong shook my confidence, so I'd appreciate it if you guys can check my other answers, especially my reasoning/explanation since my teacher makes a big deal about that.
2. a[SIZE="1"]n[/SIZE] = (-1)^n sin (1/n)
This one I was pretty unsure of. What I did was write statements for two cases (n is even/odd). Since they both go to 0 as n-> infinity , the sequence converges to 0.
3. a[SIZE="1"]n[/SIZE] = ln (n+1) - ln n
First, I set f(x) = y = ln ((x+1)/x). Then I did e^y = x+1/x. Taking the limit x-> infinity for both sides, I got e^y -> 1, so y -> ln 1 = 0.
Therefore, lim a[SIZE="1"]n[/SIZE] as n->infinity is 0 and a[SIZE="1"]n[/SIZE] converges to 0.
3. a[SIZE="1"]n[/SIZE] = (1 + 3n)^ (1/n)
First, I set f(x) = y = (1 + 3x)^ (1/x). Then I did ln y = (1/x) ln (1+3x). Taking the limits of both sides again, I get ln y -> 0 and so y -> e^0 = 1.
Therefore, a[SIZE="1"]n[/SIZE] converges to 1
Additionally, there is a problem on the bounds of a sequence.
4. Find the least upper bound and the greatest lower bound of a[SIZE="1"]n[/SIZE] = (e^n)/ (n+1)!
The sequence is decreasing, so I think the least upper bound would simply be a[SIZE="1"]1[/SIZE]. I don't know what to do for the lower bound.
Sequences - 2147483647 - 2011-04-06
Salguod Wrote:I'm doing the basics of sequences like converging, diverging, monotonic or not, upper/lower bounds and I need help.
Determine if the sequence converges or diverges.
1. a[SIZE="1"]n[/SIZE] = 1/n^2 + 2/n^2 + ... + n/n^2
My first thought was to multiply top and bottom by 1/n^2 to put all the n's on the bottom but that gave me 0 instead of 1/2, so my approach is probably wrong.
Getting that wrong shook my confidence, so I'd appreciate it if you guys can check my other answers, especially my reasoning/explanation since my teacher makes a big deal about that.
You probably shouldn't use the same variable to represent the nth term of your sequence for the constant. n implies that you can plug in the nth term of the sequence. For example, if your sequence is:
∑ 1/n^2 = 1/1 + 1/4 + 1/9 + 1/16 + ...
Then you're definitely not going to get 1/2. Let's say you're counting this:
a[SIZE="1"]n[/SIZE] = 1/n^2 + 2/n^2 + ... + k/n^2
For this, to determine convergence, you should use the ratio test.
lim ((k+1) /(n+1)^2) / (k /(n^2)) as n approaches ∞
lim ((k+1)/k * n^2 /(n+1)^2 ) as n approaches ∞
You're left with (k+1)/k, so the series diverges since (k+1)/k >1
Salguod Wrote:2. a[SIZE="1"]n[/SIZE] = (-1)^n sin (1/n)
This one I was pretty unsure of. What I did was write statements for two cases (n is even/odd). Since they both go to 0 as n-> infinity , the sequence converges to 0.
You want to be really careful of this. Just because the limit at infinity is 0 doesn't mean the series converges. The most obvious example of this is 1/n. The limit of 1/n as n approaches infinity is 0, but 1/n fails the the p-series test.
In this case, you really want to apply the alternating series test, which states that convergence is only proven when the limit at infinity is 0, AND the series is alternating. You already found that the limit as n approaches ∞ is 0, so all you need to further mention is that the (-1)^n dictates that the series alternates. Thus, it converges.
Salguod Wrote:3. an = ln (n+1) - ln n
First, I set f(x) = y = ln ((x+1)/x). Then I did e^y = x+1/x. Taking the limit x-> infinity for both sides, I got e^y -> 1, so y -> ln 1 = 0.
Therefore, lim an as n->infinity is 0 and an converges to 0.
For this series:
an = ln (n+1) - ln(n)
= ln ((n+1)/n)
= ln (1+1/n)
You would want to use the ratio test.
lim ln(1+1/(n+1)) / ln(1+1/n)
Applying l'Hopital's rule:
= lim [-1/(n+1)^2/(1+1/(n+1))] / [-1/n^2 /(1+1/n)]
= lim [(n^2)*(1+1/n)] / [(n+1)^2 * (1+1/(n+1))]
= 1 as n approaches ∞
Thus, you need to apply Raabe's test.
lim n*[((n^2)*(1+1/n)) / ((n+1)^2 * (1+1/(n+1))) -1] = 0 > -1 as n approaches ∞
Therefore, the series diverges.
Salguod Wrote:3. a[SIZE="1"]n[/SIZE] = (1 + 3n)^ (1/n)
First, I set f(x) = y = (1 + 3x)^ (1/x). Then I did ln y = (1/x) ln (1+3x). Taking the limits of both sides again, I get ln y -> 0 and so y -> e^0 = 1.
Therefore, a[SIZE="1"]n[/SIZE] converges to 1
For this one, apply the ratio test:
a[SIZE="1"]n[/SIZE] = (1 + 3n)^ (1/n)
lim (4+3n)^(1/(n+1)) / (1+3n)^(1/n)
To make this easier for yourself, apply the comparison test. The above series is always going to be greater than the below series:
lim (4+3n)^(1/(n+1)) / (4+3n)^(1/n)
= lim (4+3n)^(1/(n+1) - 1/n)
= lim (4+3n)^((n-1) / n*(n+1))
= 0 as n approaches ∞
Therefore, by comparison, a[SIZE="1"]n[/SIZE] = (1 + 3n)^ (1/n) converges.
Sequences - Salguod - 2011-04-06
For #1, I think you mistook it for a series instead of a sequence? I'm still on sequences. Before I started finding the limits, I tried to find a[SIZE="1"]1[/SIZE], a[SIZE="1"]2[/SIZE], a[SIZE="1"]3[/SIZE], etc.
a[SIZE="1"]1[/SIZE] = 1/1 = 1
a[SIZE="1"]2[/SIZE] = 1/4 + 2/4 = 3/4
a[SIZE="1"]3[/SIZE] = 1/9 + 2/9 + 3/9 = 2/3
a[SIZE="1"]4[/SIZE] = 1/16 + 2/16 + 3/16 + 4/16 = 5/8
so the sequence converging to 1/2 would make sense.
For #2, I haven't learned those tests yet but I'll keep that in mind. Thanks.
Sequences - 2147483647 - 2011-04-07
Salguod Wrote:For #1, I think you mistook it for a series instead of a sequence? I'm still on sequences. Before I started finding the limits, I tried to find a[SIZE="1"]1[/SIZE], a[SIZE="1"]2[/SIZE], a[SIZE="1"]3[/SIZE], etc.
A sequence is just a set of numbers. For example, {1, 2, 3, 4}. You're not really doing anything to them. A series implies that you're adding all the terms in a sequence together, and that's exactly what you're doing here.
Salguod Wrote:4. Find the least upper bound and the greatest lower bound of a[SIZE="1"]n[/SIZE] = (e^n)/ (n+1)!
The sequence is decreasing, so I think the least upper bound would simply be a[SIZE="1"]1[/SIZE]. I don't know what to do for the lower bound.
Upper and lower bound is codeword for: check the end behavior.
lim e^n / (n+1)! as n approaches ∞
It's not immediately obvious to see this limit, but you can compare it to:
lim e^n / (n+1)^n = 0 as n approaches ∞
because e^n / (n+1)^n is always going to be greater than e^n / (n+1)! . Therefore, by comparison, lim e^n / (n+1)! is also 0 as n approaches ∞, so the lower bound is 0.
Sequences - Panacea - 2011-04-07
Sequence =! Series.
The sequence calls for him to add the previous term to something else, but he's not actually adding all the values to find the sum. He just wants to determine whether they converge or not.
Gimme a few to sort these out.
Sequences - 2147483647 - 2011-04-07
Panacea Wrote:I can tell you now the first one converges by p-series. (p = 2 > 1).Sec.
It's not a true p-series. The numerator is increasing at a constant rate, so it's almost like:
an = n/n^2 = 1/n
which fails the p-series.
Salguod Wrote:3. an = ln (n+1) - ln n
First, I set f(x) = y = ln ((x+1)/x). Then I did e^y = x+1/x. Taking the limit x-> infinity for both sides, I got e^y -> 1, so y -> ln 1 = 0.
Therefore, lim an as n->infinity is 0 and an converges to 0.
I just realized that there's a completely easier way to prove divergence for this one. If the derivative of an diverges, then an should also diverge, since an has to be "larger" than its own derivative. Therefore,
an = ln (n+1) - ln n
an' = 1/(n+1) - 1/n
an' fails the p-series test, so both an' and an diverge.
Sequences - Panacea - 2011-04-07
I know, I lied. For some reason I'm able to do the AST/Ratio Test no problem, but when it comes to the basics I have to struggle.
I'mma leave this one to you, since you have a better handle on it. Sorry for the intrusion.
Sequences - Salguod - 2011-04-07
Thank you, I really appreciate your efforts to help me. Unfortunately, I don't understand all those tests you're doing and I have to go for tonight, so I'll have to look at these another day.
Sequences - 2147483647 - 2011-04-07
Salguod Wrote:Thank you, I really appreciate your efforts to help me. Unfortunately, I don't understand all those tests you're doing and I have to go for tonight, so I'll have to look at these another day.
Sequences - Stereo - 2011-04-07
1. an = 1/n^2 + 2/n^2 + ... + n/n^2
Ask yourself: what is
sum (i = 1 to n) { i }
That is, the series 1, 2, ..., n.
The denominators are all the same, so you just add up the numerators.
Comes to n(n+1)/2
So a_n = [n(n+1)/2]/n^2
a_n = (n+1) / (2n)
Which pretty clearly diverges, as it's greater than 1/2 for all n.
Sequences - Salguod - 2011-04-07
Yes, someone showed me that today, except they took it further and separated the fraction into n/2n + 1/2n , which becomes 1/2 as n approaches infinity.
We'll be doing series next so I should see more of what you were doing for the other ones soon. I don't see any mention of Raabe's test though.