# Convergence by Comparison Test

1. Feb 2, 2014

### vanceEE

1. The problem statement, all variables and given/known data
Use $$\sum\limits_{n=1}^∞ \frac{1}{n^2}$$ to prove by the comparison test that $$\sum\limits_{n=1}^∞ \frac{n+1}{n^3}$$ converges.

2. Relevant equations
$$\sum\limits_{n=1}^∞ \frac{n+1}{n^3} \equiv \sum\limits_{n=1}^∞ \frac{1}{n^2} + \sum\limits_{n=1}^∞ \frac{1}{n^3}$$

3. The attempt at a solution
$$\frac{n+1}{n^3} < \frac{1}{n^2}$$
$$n + 1 < n$$
$$1 < 0$$ which is false.
I cannot prove the convergence of the series by direct comparison with 1/n2, correct? The series is a sum of 2 p-series (p > 1) so it is clearly a convergent series and can be shown by limit comparison with 1/n2, but I cannot see how I can prove this by direct comparison because there is no way that 1/n2 + 1/n3 is less than 1/n2 if n > 0...

2. Feb 2, 2014

### AlephZero

True, but can you show 1/n2 + 1/n3 is less than some constant times 1/n2?

3. Feb 2, 2014

### jackarms

Looks like direct comparison test doesn't work, but it's possible that "comparison test" in the problem could also refer to the limit comparison test -- see if that works instead.

4. Feb 2, 2014

### vanceEE

The problem was referring to direct comparison; limit comparison does in fact work as I stated in my original post. I'm aware that I'm dealing with a convergent series, but I just wanted to confirm whether or not direct comparison with 1/n2 is inconclusive.

5. Feb 2, 2014

### vanceEE

Yes, that seems feasible.
$\frac{n+1}{n^3} < c(\frac{1}{n^2})$
$\frac{n+1}{n^3} < \frac{c}{n^2}$
$1 + \frac{1}{n} < c$
$\frac{1}{n} < c-1$ if c > 2.

Let c = 3
$\frac{n+1}{n^3} < 3(\frac{1}{n^2})$
$n^3 + n^2 < 3n^3$
$1 + \frac{1}{n} < 2$
$\frac{1}{n} < 1$ for all n > 1.

Therefore, $\frac{n+1}{n^3} < \frac{c}{n^2}$ where c > 2,

Last edited: Feb 2, 2014
6. Feb 2, 2014

### exclamationmarkX10

assume the sum 1/n^2 converges. for each positive integer n, (n+1)/n^3 = 1/n^2 + 1/n^3 which is less than or equal to 2/n^2. and the sum of 2/n^2 converges since each term in the sequence of partial sums equals 2 times the corresponding term in the sequence of partial sums for 1/n^2. (there is a theorem that states that if (xn) is a sequence of real numbers that converges to L, then the sequence (a * xn) where a is a real number converges to a * L).

7. Feb 2, 2014

### AlephZero

Well, we don't know how pedantic your math lecturer is.

But I would have thought that if you are told that $\sum\frac{1}{n^2}$ converges, it seems perfectly reasonable to use the fact that $\sum\frac{2}{n^2}$ also converges.