Sequence Convergence and Divergence
Examples

Today in class, I proved that the sequence $\left\{ \int_{1+\frac{1}{n}}^{2-\frac{1}{n}} \frac{1}{x} \: dx \right\} ^{\infty}_{n=k}$ converges to a limit of $\ln{2}$.

There are two steps in working such a problem:
1. Find the limit, L (i.e. using calculus techniques).
2. Prove that the limit exists, and is equal to L.

Step 1: Find the limit.

$$\begin{eqnarray*} \lim_{n \to \infty} \int_{1+\frac{1}{n}}^{2-\frac{1}{n}} \frac... ... \infty}\frac{2n - 1}{n + 1} \Biggr) } \\ [.1 in] & = & \ln{2}. \end{eqnarray*}$$

Now that we have computed what the limit is (or rather, at this point we ought to say ``what the limit would be, if it exists''), we have to prove that the limit exists, and that it is in fact, $\ln{2}$.

Step 2: Prove the limit is $\ln{2}.$

Let us recall the definition of convergence: a sequence $\{a_n\}_{n=k}^{\infty}$ converges to a limit, L, if, for every $\epsilon > 0$, there exists a real number, M (a function of $\epsilon$), such that if n > M, then $\vert a_n - L\vert < \epsilon$.

What this means is that in order to prove our sequence converges, we need to demonstrate that when someone hands us an $\epsilon$ (think: really small number), we can give them back an M (think: really large number, usually expressed in terms of $\epsilon$, and NOT involving n), so that as long as we are looking at terms of the sequence past the Mth (i.e. as long as n > M), then we can guarantee that $\vert a_n - L\vert < \epsilon$.

So how do we go about finding such an M? Well, we assume that what we want to be true is true, for the time being. What we want is for

$$\Biggl\vert \int_{1+\frac{1}{n}}^{2-\frac{1}{n}} \frac{1}{x} \: dx - \ln{2} \Biggr\vert < \epsilon,$$

when n is greater than some M (which we get to pick). How do we find such an M?

Well, the inequality above tells us exactly what restriction we need to place on n to ensure we are no less than $\epsilon$ away from our limit. We just need to solve the inequality for n so that we can see it explicitly.

Having already determined that

$$\lim_{n \to \infty} \int_{1+\frac{1}{n}}^{2-\frac{1}{n}} \frac{1}{x} \: dx = \ln{\Biggl( \frac{2n - 1}{n + 1} \Biggr) },$$

we begin working backwards to find n:

$$\Biggl\vert \ln{\Biggl( \frac{2n - 1}{n + 1} \Biggr) } - \ln{2} \Biggr\vert < \epsilon$$

Now, as I said in class, the stuff inside the absolute value bars is actually a negative number (since $\frac{2n-1}{n+1} < 2$), so we switch the order of subtraction to make it positive...

$$\Biggl\vert \ln{2} - \ln{\Biggl( \frac{2n - 1}{n + 1} \Biggr) } \Biggr\vert < \epsilon$$

Now, use the logarithm subtraction formula ( $\ln{a} - \ln{b} = \ln{\frac{a}{b}}$), noting that absolute value bars are no longer needed:

$$\ln{\Biggl( \frac{2n+2}{2n-1} \Biggr)} < \epsilon$$

Next, exponentiate both sides

$$\frac{2n+2}{2n-1} < e^{\epsilon}$$

and perform long division on the left-hand side:

$$1 + \frac{3}{2n-1} < e^{\epsilon}$$

and finally solve for n:

$$\frac{3}{2n-1} < e^{\epsilon} - 1$$

$$\frac{3}{e^{\epsilon} - 1} < 2n - 1$$

$$n > \frac{3}{2(e^{\epsilon} - 1)} + \frac{1}{2}$$

What does this last statement say? It says that if we assume for the moment that we have some value of n such that

$$\Biggl\vert \int_{1+\frac{1}{n}}^{2-\frac{1}{n}} \frac{1}{x} \: dx - \ln{2} \Biggr\vert < \epsilon,$$

then it logically follows that

$$n > \frac{3}{2(e^{\epsilon} - 1)} + \frac{1}{2}.$$

Hence, if given some $\epsilon > 0$, we set

$$M = \frac{3}{2(e^{\epsilon} - 1)} + \frac{1}{2},$$

then whenever n > M, we will have

$$\Biggl\vert \int_{1+\frac{1}{n}}^{2-\frac{1}{n}} \frac{1}{x} \: dx - \ln{2} \Biggr\vert < \epsilon.$$

This is exactly what we were trying to prove.

Several people have queried me about what would happen if we tried to prove that the limit of this sequence was $\ln{3}$. How would you know you went wrong? Well, let's try to repeat the calculations, this time using $\ln{3}$.

$$\Biggl\vert \ln{\Biggl( \frac{2n - 1}{n + 1} \Biggr) } - \ln{3} \Biggr\vert < \epsilon$$

Again, since $\frac{2n-1}{n+1} < 3$, the stuff inside the absolute value bars is negative, so we reverse the order...

$$\Biggl\vert \ln{3} - \ln{\Biggl( \frac{2n - 1}{n + 1} \Biggr) } \Biggr\vert < \epsilon$$

Now, use the logarithm subtraction formula ( $\ln{a} - \ln{b} = \ln{\frac{a}{b}}$), noting that absolute value bars are no longer needed:

$$\ln{\Biggl( \frac{3n+3}{2n-1} \Biggr)} < \epsilon$$

Next, exponentiate both sides

$$\frac{3n+3}{2n-1} < e^{\epsilon}$$

and perform long division on the left-hand side:

$$\frac{3}{2} + \frac{9}{2(2n-1)} < e^{\epsilon}$$

and finally solve for n:

$$\frac{9}{4n-2} < e^{\epsilon} - \frac{3}{2}$$

$$\frac{9}{e^{\epsilon} - \frac{3}{2}} < 4n - 2$$

$$n > \frac{9}{4e^{\epsilon} - 6} + \frac{1}{2}$$

Let's call the right-hand side of this inequality M'.

At first glance, this may look just as legitimate as the last example, but it isn't. Here's a guideline that should be pretty easy to use. As we said before, someone is handing you an $\epsilon$, and you have to find an M that's going to guarantee the sequence terms are closer than $\epsilon$ to the limit value. It should make sense, then, that the smaller the value of $\epsilon$, the higher you need to make M in order to get close enough to the limit value. In fact, as $\epsilon \rightarrow 0$, you should expect that $M \rightarrow \infty$. Let's try this for our original M:

$$\lim_{\epsilon \to 0} \frac{3}{2(e^{\epsilon} - 1)} = \infty$$

The denominator goes to zero (from the positive side), hence we get $\infty$ for the limit.

Now, let's try it for the value M' we computed using $\ln{3}$ for the limit:

$$\lim_{\epsilon \to 0} \frac{9}{4e^{\epsilon} - 6} + \frac{1}{2} = \frac{9}{4(1) - 6} + \frac{1}{2} = -4.$$

Hmmm...that doesn't seem to make much sense. If this were true, then if $\epsilon$ is really small, we should be able to pick any n value greater than -4 and find we are within $\epsilon$ of our limit value, $\ln{3}$. Suppose $\epsilon = .001$. Then

$$M' = \frac{9}{4e^{.001} - 6} \approx -4.509$$

So we should be able to pick any n > -4.509, and have our nth term (and beyond) be within .001 of $\ln{3}$. So let's try it for n = 2.

$$\Biggl\vert \ln{\Biggl( \frac{2(2) - 1}{(2) + 1} \Biggr) } - ... ... \Biggr\vert = \vert\ln{1} - \ln{3}\vert = \ln{3} \nless .001$$

If you've lasted this long, you may be wondering ``but so what? You just chose a bad value for $\epsilon$ and n.'' But if we got back and look at the definition of covergence, the M we choose must work for any choice of $\epsilon$, and for any n > M. Since it doesn't, we must conclude that our original assumption, that the limit of the sequence was $\ln{3}$, must be wrong.