# Comparison Test Problem

## Homework Statement

Determine whether the series converges or diverges.

What I would like is some type of information on how to continue the problem.

## Homework Equations

Ʃ √(n+1)/2n^2+n+1
n=1

## The Attempt at a Solution

I was thinking of doing a comparison test by doing

√(n+1)/2n^2+n+1 < √(n+1)/2n^2 or √(n+1)/2n^2+n+1 < 2n(n+1)

After that, I end up being stumped by the problem.

Remember what the Comparison Theorem states:

Suppose that we have two series Ʃa and Ʃb with an,bn≥0 for all n and bn≥ an for all n. Then,

If Ʃb is convergent then so is Ʃa .
If Ʃa is divergent then so is Ʃb .

So let 1/2n2 be Ʃb. Since Ʃb is greater than the given series and we know it converges because of p-series, what can we assume about the original series?

Mark44
Mentor

## Homework Statement

Determine whether the series converges or diverges.

What I would like is some type of information on how to continue the problem.

## Homework Equations

Ʃ √(n+1)/2n^2+n+1
n=1
You need more parentheses. Without them, the expression in your sum does not mean what you think. A literal reading would give this:
$$\frac{\sqrt{n+1}}{2}\cdot n^2 + n + 1$$

Since your intent was that 2n2 + n + 1 was in the denominator, write the expression being summed like this:
√(n+1)/(2n^2+n+1)

## The Attempt at a Solution

I was thinking of doing a comparison test by doing

√(n+1)/2n^2+n+1 < √(n+1)/2n^2 or √(n+1)/2n^2+n+1 < 2n(n+1)

After that, I end up being stumped by the problem.
How did you get this one? √(n+1)/2n^2+n+1 < 2n(n+1)
You're on the right track with the first inequality.