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

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

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

Determine whether $\displaystyle \sum a_n$ converges or diverges.

My solution:

Define

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

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

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

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

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

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

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

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

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

$\displaystyle b_n<a_n$

when $\displaystyle n$ is odd.

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

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

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

$\displaystyle b_n<a_n$

when $\displaystyle n$ is even.

Thus, whether $\displaystyle n$ is odd or even, we have $\displaystyle 0<b_n<a_n$, with $\displaystyle \sum b_n$ divergent. By the comparison test, therefore $\displaystyle \sum a_n$ 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 $\displaystyle a_n=\frac{n+(-1)^n}{n(n+1)}$.

Notice that $\displaystyle a_n = \frac{n+(-1)^n}{n(n+1)} \geq \frac{n-1}{n(n+1)} \geq \frac{n - \tfrac{1}{2}n}{n(n+1)} = \frac{1}{2(n+1)} \geq \frac{1}{2(n+n)} = \frac{1}{4n}$ and $\displaystyle \sum_{n=1}^{\infty}\frac{1}{4n} = \text{me}$.
• Sep 5th 2009, 12:41 PM
hatsoff
Quote:

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

So then you agree that this sequence diverges, yes?

Have I somehow misunderstood the comparison test? If $\displaystyle \sum b_n=\infty$ and $\displaystyle 0<b_n<a_n$ for large $\displaystyle n$, then $\displaystyle \sum a_n$ is divergent, right?
• Sep 5th 2009, 12:57 PM
JG89
Also note that $\displaystyle \sum \frac{n + (-1)^n}{n(n+1)} = \sum \frac{n}{n(n+1)} + \sum \frac{(-1)^n}{n(n+1)}$. 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 $\displaystyle \sum a_n$ 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 $\displaystyle \sum a_n$ 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.