# Sequence problem

1. Aug 17, 2013

### tylersmith7690

1. the nsider, for n → 1, the sequence an given by

an = n log (n/n+1)

Determine the limit of the sequence as n→∞, If it exists , or explain why the sequence diverges. In your answers include the names of any rules, theorems or limits you have used.

2. Relevant equations

3. The attempt at a solution

lim n→∞ an = lim n→∞ n log(n/n+1)

embed sequence in function f(x)= x log (x/(x+1))

now limx→∞ x log(x/x+1) which is (∞ . ∞) form, can rearange to give

= limx→∞ log(x/(x+1)) / (1/x) which gives ( 0 / 0 ) indeterminate form, can then use
log laws on the top function to rewrite log(x/(x+1)) as log(x)-log(x+1)

= limx→∞ log(x) - log (x+1) / (1/x)

now differntiate numerator and denominator by l'hopital rule.

= limx→∞ ((1/x) -(1/(1+x)) / (-1 /x^2) The x(1+x) terms cancel out when you flip and multiply. Leaving

= limx→∞ 1 x (-1) = -1

Therefore the limit approaches -1 and converges to -1.
Using theorem limx→∞ f(x) = L => lim n→∞ an = L.

So limn→∞ an converges to -1.

Any help or tips would be appreciated as my Maths is really bad.

Last edited by a moderator: Aug 17, 2013
2. Aug 17, 2013

### Goa'uld

You have arrived at the correct answer.

3. Aug 17, 2013

### LCKurtz

Your work looks OK. Another simpler approach would be to write$$a_n = n\log \frac n {n+1} = \log\left( \frac n {n+1}\right)^n = \log\left(\frac 1 {1 +\frac 1 n}\right)^n$$Do you see how to finish easily?

4. Aug 17, 2013

### Staff: Mentor

Please remember to ALWAYS quote the OP's posts.