# Convergence/divergence test

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1. Nov 30, 2014

### ironman

1. The problem statement, all variables and given/known data
Show if this sequence (with n=1 to infinity) diverge or converge

2. Relevant equations

3. The attempt at a solution

If I use the Limit Comparison Test:

compare with so you get that equals

lim n -> inf => inf.

Can I use the Test like this? What does this tell me? (If anything at all...)

2. Nov 30, 2014

### Orodruin

Staff Emeritus
I suggest doing limit comparison with $(2/3)^n$ instead.

3. Nov 30, 2014

### ironman

Ah the limit is now 1 .
So:
Limit < infinity.
And $(2/3)^n$ converges.
So the original function must converge as well.

4. Nov 30, 2014

### Ray Vickson

Right, and a direct comparison is easier. First, for $a > 1$ the function $a^n$ always swamps $n^k$ for any fixed power $k$, meaning that for large enough $n$ we always have $a^n > n^k$. So, if
$$t_n = \frac{2^n + n^2}{3^n + n^3}$$
we have that the numerator is $2^n + n^2 < 2^n + 2^n = 2 \cdot 2^n$ for all $n > n_0$ (where you can even find $n_0$ if you want). The denominator is $3^n + n^3 > 3^n$, so $0 < t_n < \frac{2 \cdot 2^n}{3^n} = 2(2/3)^n$, and now you are almost done.

Last edited: Nov 30, 2014