Formal proof of limit/convergence of a sequence

**Pinkk** · April 28th 2009, 06:48 PM

Okay, so I have a problem where $a_{n}=\frac{1}{n}$ and I need to show using the formal definition that $a_{n}$ converges to 0:
$\lim_{n\rightarrow \infty}a_{n}=L=0$ .

So I let $|a_{n}-L| < \varepsilon$ . Simplifying it, I get
$n > \frac{1}{\varepsilon + L}$ , essentially getting within distance $\varepsilon$ of $L$ for any integer $N$ where $n > N$ .

So then I get $n > N \ge \frac{1}{\varepsilon + L}$ .

Doing some manipulation, I eventually get to $\varepsilon > \varepsilon\frac{N}{n}+L\frac{N-n}{n} \ge \frac{1}{n}$ This is where I get lost on showing that $L=0$ .

Now my professor didn't really sufficiently explain how to do a formal proof even when we already assume we know the value of $L$ . For instance, we know $L=0$ for this $a_{n}$ . Knowing this, the inequality would become $\varepsilon > \varepsilon\frac{N}{n} \ge \frac{1}{n}$ . Do we know that this is true because $\varepsilon\frac{N}{n}$ is always less than $\varepsilon$ since $\frac{N}{n} < 1$ , or is there more that needs to be shown? If that is sufficient, than how can that line of thought be applied when we assume we don't know the value of $L$ (like in this problem)?

Sorry if there is already a thread that asks and explains this. Our class pretty much went through sequences today in about 30 minutes so I'm not too sure on this stuff. Thanks!

**NonCommAlg** · April 28th 2009, 06:59 PM

Originally Posted by Pinkk

Okay, so I have a problem where $a_{n}=\frac{1}{n}$ and I need to show using the formal definition that $a_{n}$ converges to 0:
$\lim_{n\rightarrow \infty}a_{n}=L=0$ .

So I let $|a_{n}-L| < \varepsilon$ . Simplifying it, I get
$n > \frac{1}{\varepsilon + L}$ , essentially getting within distance $\varepsilon$ of $L$ for any integer $N$ where $n > N$ .

So then I get $n > N \ge \frac{1}{\varepsilon + L}$ .

Doing some manipulation, I eventually get to $\varepsilon > \varepsilon\frac{N}{n}+L\frac{N-n}{n} \ge \frac{1}{n}$ This is where I get lost on showing that $L=0$ .

Now my professor didn't really sufficiently explain how to do a formal proof even when we already assume we know the value of $L$ . For instance, we know $L=0$ for this $a_{n}$ . Knowing this, the inequality would become $\varepsilon > \varepsilon\frac{N}{n} \ge \frac{1}{n}$ . Do we know that this is true because $\varepsilon\frac{N}{n}$ is always less than $\varepsilon$ since $\frac{N}{n} < 1$ , or is there more that needs to be shown? If that is sufficient, than how can that line of thought be applied when we assume we don't know the value of $L$ (like in this problem)?

Sorry if there is already a thread that asks and explains this. Our class pretty much went through sequences today in about 30 minutes so I'm not too sure on this stuff. Thanks!

for a given $\epsilon > 0,$ we need to find an integer $N > 0$ such that for any $n \geq N: \ \frac{1}{n}=|a_n| < \epsilon,$ which is equivalent to $n > \frac{1}{\epsilon}.$ so any integer $N > \frac{1}{\epsilon}$ will work.

**Pinkk** · April 28th 2009, 07:16 PM

Hmm, so correct me if I'm wrong, so would that imply then that:

$\varepsilon > \frac{1}{n} + L(\frac{1}{\varepsilon n}-1) \ge \frac{1}{n}$

Since $\frac{1}{\varepsilon n} - 1 \le 0$ , $L$ cannot be positive or the inequality does not hold true. But how exactly can we show $L$ is never negative, and therefore $L=0$ .

Also, doesn't the information you provided already assume $L=0$ ? Sorry if these are not the best of questions, but like many others, my teachers and professors have always neglected the formal proofs and definitions (which sucks because I plan on being a pure math major). Thanks again!

**NonCommAlg** · April 28th 2009, 07:28 PM

normally we don't use the "definition of limit" to "find" the limit. this method is used to "prove" that the limit is what we think it is. for that sequence, our guess was that the limit is 0.

then i proved it using the definition.

**Pinkk** · April 28th 2009, 07:29 PM

Ah okay, I thought there was a method to solve for $L$ within the formal definition inequality. Thanks for the help.

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