Sequences

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

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.

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.

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.