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

• Sep 5th 2009, 09:16 AM
hatsoff
[SOLVED] Is my textbook wrong? Series convergence vs divergence.
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, 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

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

when $n$ is even.

Thus, whether $n$ is odd or even, we have $0, 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!
• Sep 5th 2009, 12:35 PM
ThePerfectHacker
Quote:

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 $\sum_{n=1}^{\infty}\frac{1}{4n} = \text{me}$.
• Sep 5th 2009, 12:41 PM
hatsoff
Quote:

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 for large $n$, then $\sum a_n$ is divergent, right?
• Sep 5th 2009, 12:57 PM
JG89
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.
• Sep 5th 2009, 08:49 PM
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?
• Sep 6th 2009, 06:11 AM
hatsoff
Quote:

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.