Critique my proof?

**Ares_D1** · August 23rd 2010, 04:58 AM

Claim: The series $\sum_{n=1}^{ \infty} \frac{2}{n^2 + n}$ converges.

By definition:
The series $\sum_{n=1}^{ \infty} \frac{2}{n^2 + n} = \sum_{n=1}^{ \infty} a_n$ converges if and only if for every $\epsilon > 0$ , there exists some $N \in \mathbb{N}$ such that for all $n, m$ with $n > m \geq N$ , we have $|a_{m+1} + a_{m+2}+ ...+ a_n| < \epsilon$ .

(Attempted) proof:
Let $m \in \mathbb{N}$ be arbitrary but fixed, and let $S_m = \sum_{n=1}^{m} \frac{2}{n^2 + n}$ . Then the sequence $(S_m)$ is convergent, and it necessarily follows that $(S_m)$ is Cauchy. That is, for all $\epsilon > 0$ , there exists some $N \in \mathbb{N}$ such that for all $n, m \geq N$ , we have $|S_{n} - S_{m}| < \epsilon$ .

But if $n > m \geq N$ , we have $|S_n - S_m| = |a_{m+1} + a_{m+2}+ ...+ a_n| < \epsilon$

which shows, directly from the definition, that the series $\sum_{n=1}^{ \infty} \frac{2}{n^2 + n}$ is convergent. $\square$

Is this satisfactory? Or is there a more simplistic approach that I should be taking?

**Defunkt** · August 23rd 2010, 05:06 AM

1) Do you have to do this by definition, or have you already seen the comparison test?

2) If you want to do this by definition, then you didn't prove anything. You stated that $\displaystyle S_m = \sum_{n=1}^{m} \frac{2}{n^2 +n}$ is convergent, which means $\displaystyle \lim_{m \to \infty} S_m = \sum_{n=1}^{m} \frac{2}{n^2+n}$ converges, but that is what you have to prove! What you said is "this series converges because it converges" - this is not a proof.

**Ares_D1** · August 23rd 2010, 05:18 AM

Originally Posted by Defunkt

1) Do you have to do this by definition, or have you already seen the comparison test?

2) If you want to do this by definition, then you didn't prove anything. You stated that $\displaystyle S_m = \sum_{n=1}^{m} \frac{2}{n^2 +n}$ is convergent, which means $\displaystyle \lim_{m \to \infty} S_m = \sum_{n=1}^{m} \frac{2}{n^2+n}$ converges, but that is what you have to prove! What you said is "this series converges because it converges" - this is not a proof.

Well, I've stated that $S_m$ is convergent since m is finite and fixed. The rest of the "proof" was intended to show that this extends to the case in which n goes to infinity.

I did try my hand at the comparison test, but wasn't very successful there either. I'll give it another shot and see what I come up with.

**Defunkt** · August 23rd 2010, 05:23 AM

If m is finite and fixed then $S_m$ is just a number. It does not depend on n.

The easiest way to prove this is by using the comparison test. I assume you have proved that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}$ converges. Now,

$\frac{2}{n^2+n} \le \frac{2}{n^2} \ \forall n \in \mathbb{N}$

and so

$\displaystyle \sum_{n=1}^{\infty} \frac{2}{n^2 + n}$ converges iff $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, since they both have positive terms.

**Ares_D1** · August 23rd 2010, 05:35 AM

Wow. I literally just made the connection with that particular series before coming back to see your post.

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