# A(n+1) = An + 1/(5^n) sequence

1. Apr 7, 2009

### Sonic7145

So, couple sequence questions!

Converge or Diverge?
1.
A(n+1) = An + 1/(5^n) (recursive sequence)

That should read a sub n plus one equals a sub n plus one over 5 to the n

I believe it converges, but would like to be sure.

2. sigma from 1 to infinity of:
(1+5/n)^n
That should read the summation from n equals 1 to infinity of one plus 5 over n quantity to the n-th power. I believe this diverges, but would like to be sure.

2. Apr 8, 2009

### Focus

Re: Sequences!

3. Apr 8, 2009

### JG89

Re: Sequences!

For a) If $$A_{n+1} = A_n + \frac{1}{5^n}$$ then $$A_{n+1} - A_n = \frac{1}{5^n}$$. Can you prove that this is a Cauchy sequence?

For b) First prove that $$\lim_{n \rightarrow \infty} (1 + c/n)^n = e^c$$. Now, if the summand of the series doesn't have a limit of zero, what can you say about the convergence/divergence of the series?

Last edited: Apr 8, 2009
4. Apr 8, 2009

### Sonic7145

Re: Sequences!

So doesn't b diverge if you use the n term test?

And a converges to what, 6? Not sure what a Cauchy sequence is

5. Apr 8, 2009

### mvantuyl

Re: Sequences!

For b, are you familiar with the root test?

6. Apr 8, 2009

### Sonic7145

Re: Sequences!

Nope, but doesn't e^inf not equal 0 meaning it diverges?

7. Apr 8, 2009

### mvantuyl

Re: Sequences!

The root test says that if the limit of the sequence raised to the power 1/n gives a result which is < 1, the series converges.
If the result is > 1, the series diverges.
If the result = 1, it can't be determined by this test whether the series converges or diverges.

8. Apr 8, 2009

### Sonic7145

Re: Sequences!

Doesn't n-term work?

9. Apr 9, 2009

### JG89

Re: Sequences!

I don't know if the test I'm talking about is usually called the n-term test but if the summand doesn't have a limit of zero, then the series diverges.

And a Cauchy sequence is a sequence a_n such that for all epsilon > 0, there is an integer N such that for m,n > N, the following relation holds: |a_m - a_n| < epsilon.

In your case, you can take m = n+1 and so you must show that $$|a_{n+1} - a_n| = \frac{1}{5^n} < \epsilon$$