# ConvergenceWatch

This discussion is closed.
#1
Hi
I need help with some sequences questions.

Determine with justification, which of the following generate sequences that converge and where a limit exists, determine it.

(a) an := (1 + n -2n²)/(3 - 7n + n²)
(b) an := cos(n pi)
(c) an := (7+(-1)^n)/(log(n+1))

I've got a few more as well but I want to see how these are done then I'll attempt the rest myself.
Many many thanks
0
#2
anyone?
0
13 years ago
#3
(a)
a(n) = (n^(-2) + n^(-1) - 2) / (3n^(-2) - 7n^(-1) + 1)

numerator -> -2
denominator -> 1

So a(n) -> -2.

(b)
a(n) = (-1)^n

So a(n) doesn't converge.

(c)
The numerator is bounded (between 6 and 8), and the denominator tends to infinity.

So a(n) -> 0.
0
#4
(Original post by Jonny W)
(a)
a(n) = (n^(-2) + n^(-1) - 2) / (3n^(-2) - 7n^(-1) + 1)

numerator -> -2
denominator -> 1

So a(n) -> -2.

(b)
a(n) = (-1)^n

So a(n) doesn't converge.

(c)
The numerator is bounded (between 6 and 8), and the denominator tends to infinity.

So a(n) -> 0.
I am immensely grateful for these solutions but although I really dont understand how you obtained these results. Could you please elaborate a little more?
0
13 years ago
#5
(a)
I got a(n) = (n^(-2) + n^(-1) - 2) / (3n^(-2) - 7n^(-1) + 1) by dividing the top and bottom of the given expression by n^2.

I then worked out the limits of the three terms in the numerator and the three terms in the denominator.

n^(-2) -> 0
n^(-1) -> 0
-2 -> -2
Therefore (n^(-2) + n^(-1) - 2) -> 0 + 0 - 2 = -2.

3n^(-2) -> 0
-7n^(-1) -> 0
1 -> 1
Therefore (3n^(-2) - 7n^(-1) + 1) -> 0 + 0 + 1 = 1.

Finally I used the theorem that says that if (p(n)) and (q(n)) are sequences such that p(n) -> P and q(n) -> Q, where Q is not zero, then p(n)/q(n) -> P/Q. In our case, p(n) = n^(-2) + n^(-1) - 2, q(n) = 3n^(-2) - 7n^(-1) + 1, P = -2 and Q = 1.

(b)
The sequence goes 1, -1, 1, -1, 1, -1, ... It is almost obvious that it doesn't converge, but if your lecturer is pedantic you might have to write out a proof.

(c)
Let epsilon > 0 be given. Choose a positive integer N such that 8/log(N + 1) < epsilon. Then for all n >= N we have

| (7 + (-1)^n) / log(n + 1) |
= |7 + (-1)^n| / log(n + 1)
<= 8/log(n + 1)
<= 8/log(N + 1)
< epsilon

So (7 + (-1)^n) / log(n + 1) -> 0.
0
#6
Thanks,
I have another question from the following years paper.

Determine with justification, which of the following generate sequences that converge and where a limit exists, determine it.
7.
(a) an:= (3n³+2n²+n+1)/(1000n²+n+1)

Is this right?

Multiply the sequence by (1/n³ / 1/n³)
to get

an:= [ 3 + 2n^(-1) + n^(-2) + n^(-3) ] / [ 1000n^(-1) + n^(-2) + n^(-3) ]

The limit of p(x) = 3
The limit of q(x) = 0 so the sequence does not converge ?
0
13 years ago
#7
You're right, although I would prefer to write the answer in a different way. Dividing top and bottom by n^2 (not n^3) gives

a(n) = =

You can prove as I did above that tends to 3/1000.

So, for large n, a(n) equals n times something very close to 3/1000. So a(n) does not converge.

--

If you don't have to be formal, you can say:

"The dominant term on the top is 3n^3. The dominant term on the bottom is 1000n^2. For large n, 3n^3 is much bigger than 1000n^2. So a(n) does not converge."
0
X
new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of the Arts London
MA Design: Ceramics, Furniture or Jewellery Open Day Postgraduate
Fri, 25 Jan '19
• Coventry University
Sat, 26 Jan '19
• Brunel University London
Sat, 26 Jan '19

### Poll

Join the discussion

Yes (34)
97.14%
No (1)
2.86%