[SOLVED] Is my textbook wrong? Series convergence vs divergence.

**hatsoff** · September 5th 2009, 09:16 AM

The problem:

Define $a_n=\frac{n+(-1)^n}{n(n+1)}$ .

Determine whether $\sum a_n$ converges or diverges.

My solution:

Define

$b_n=\frac{1}{2n}$ .

We will use the comparison test to show that since $\sum b_n$ diverges, so too does $\sum a_n$ diverge.

Observe that if $3<n$ , we may divide both sides by $2$ to yield

$\frac{3}{2}<\frac{n}{2}$ .

Adding $\frac{n}{2}-1$ to both sides gives us

$\frac{n+1}{2}=\frac{3}{2}+\frac{n}{2}-1<\frac{n}{2}+\frac{n}{2}-1=n-1$ .

Dividing by $n(n+1)$ , we have

$\frac{1}{2n}<\frac{n-1}{n(n+1)}$ .

Since $b_n=\frac{1}{2n}$ and $a_n=\frac{n-1}{n(n+1)}$ when $n$ is odd, then we have

$b_n<a_n$

when $n$ is odd.

Now observe that since $\frac{1}{2}<1$ , then dividing by $n$ gives us

$\frac{1}{2n}<\frac{1}{n}=\frac{n+1}{n(n+1)}$ .

Since $b_n=\frac{1}{2n}$ and $a_n=\frac{n+1}{n(n+1)}$ when $n$ is even, then we have

$b_n<a_n$

when $n$ is even.

Thus, whether $n$ is odd or even, we have $0<b_n<a_n$ , with $\sum b_n$ divergent. By the comparison test, therefore $\sum a_n$ is likewise divergent.

The textbook tells me I am wrong, and that this series is actually convergent.

Any ideas what went wrong, here?

Thanks!

**ThePerfectHacker** · September 5th 2009, 12:35 PM

Originally Posted by hatsoff

The problem:
Define $a_n=\frac{n+(-1)^n}{n(n+1)}$ .

Notice that $a_n = \frac{n+(-1)^n}{n(n+1)} \geq \frac{n-1}{n(n+1)} \geq \frac{n - \tfrac{1}{2}n}{n(n+1)} = \frac{1}{2(n+1)} \geq \frac{1}{2(n+n)} = \frac{1}{4n}$ and $\sum_{n=1}^{\infty}\frac{1}{4n} = \text{me}$ .

**hatsoff** · September 5th 2009, 12:41 PM

Originally Posted by ThePerfectHacker

Notice that $a_n = \frac{n+(-1)^n}{n(n+1)} \geq \frac{n-1}{n(n+1)} \geq \frac{n - \tfrac{1}{2}n}{n(n+1)} = \frac{1}{2(n+1)} \geq \frac{1}{2(n+n)} = \frac{1}{4n}$ and $\sum_{n=1}^{\infty}\frac{1}{4n} = \text{me}$ .

So then you agree that this sequence diverges, yes?

Have I somehow misunderstood the comparison test? If $\sum b_n=\infty$ and $0<b_n<a_n$ for large $n$ , then $\sum a_n$ is divergent, right?

**JG89** · September 5th 2009, 12:57 PM

Also note that $\sum \frac{n + (-1)^n}{n(n+1)} = \sum \frac{n}{n(n+1)} + \sum \frac{(-1)^n}{n(n+1)}$ . The first sum diverges while the second one converges, so the original sum must diverge as well.

**putnam120** · September 5th 2009, 08:49 PM

Yes, I think that everyone would agree that $\sum a_n$ diverges. Your book is mistaken, or maybe you are looking at the wrong answer. Do they provide a "proof" for their claim?

**hatsoff** · September 6th 2009, 06:11 AM

Originally Posted by putnam120

Yes, I think that everyone would agree that $\sum a_n$ diverges. Your book is mistaken, or maybe you are looking at the wrong answer. Do they provide a "proof" for their claim?

Checked and double-checked. I'm definitely looking at the right answer.

They do not provide any reasons, or a proof.

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