# How to proof that a sequence is divergent?

For this proof, I know that the sequence is divergent and I am trying to use proof by contradiction to prove it is divergent. Here is what I have so far. Can anyone please help me out?

Determine whether the sequence is convergent or divergent. $\{(-2)^n + \pi\}$

Let $\epsilon > 0$ be arbitrary. Suppose that $n>N$.

$|(-2)^n + \pi -L|<\epsilon$

$(-2)^n + \pi -L >N$

$n\log(-2) + \pi-L>\log(N)$

$n> \frac{\log(N) - \pi +L}{\log(-2)}$

• If a sequence converges, all its subsequences converges to the same limit. What can you tell about the subsequence $((-2)^{2n}+\pi )_{n \in \mathbb{N}}$? Otherwise, you could also show that this is not Cauchy, meaning that the distance between $2$ elements far is not decreasing for $n$ large. – Maxime Scott Mar 23 '18 at 17:42
• Explain the transition from line 1 to line 2. You removed the absolute values, swapped the direction of the inequality for some reason, and changed your epsilon to an N, and I know your teacher didnt teach you to do that. – CogitoErgoCogitoSum Mar 23 '18 at 18:16
• Dont assume the existence of N. This is what youre trying to prove; ie that it doesnt exist. – CogitoErgoCogitoSum Mar 23 '18 at 18:17
• We are supposing $n>N$ by contradiction aren't we? – dg123 Mar 23 '18 at 18:17
• $N$ and $\varepsilon$ are not the same thing. – CogitoErgoCogitoSum Mar 23 '18 at 18:34

You should always avoid taking the logarithm of negative numbers. You probably haven't defined that yet.

As per Maxime's comment, do you have theorems at your disposal that say things about subsequences of convergent sequences?

If you do not, then you probably need to proceed with a proof by contradiction. In this case, you should be very careful with your variables; as written, I don't know what $N$ is. Also, when you took the logarithm of both sides of your inequality, you violated several rules.

Try something like the following:

Suppose, for contradiction, that the sequence $\{(-2)^{n}+\pi\}$ is convergent, and converges to some real number $L$. Then, by definition, we know that for any $\epsilon > 0$ there exists a natural number $N$ such that for all $n>N$ we have $|(-2)^{n}+\pi-L|<\epsilon$. But, if $n=2m$ and $m=\max(N,\lceil\log_{4}(L)\rceil)$, then...

You fill in the rest of the proof.

• Where did the m come from and why are we assuming n = 2m? – dg123 Mar 23 '18 at 18:06
• m is a dummy variable that I created, and we can assume that n=2m because the stated inequality is supposed to work for all n. In retrospect, I need to be slightly more careful, and also demand that m be greater than N, so the real definition should be m=max(N,ceil(log(L)/log(4)) – Valborg Mar 23 '18 at 18:18
• I am so confused now – dg123 Mar 23 '18 at 18:24
• Our definition of convergence says that for any positive epsilon, there is a cutoff point N where all n>N work in that inequality, yes? Well, lets pick one. If they all work, then the one I choose must also work. Now, I don't like (-2), i like positive numbers, so lets pick n to be even, hence n=2m. But I have to pick my n larger than N, so that I am inside the cutoff range. m will then need to be >N/2, but I don't like division, so lets just demand that m>N. But I also want to be able to compare 4^m and L, so I can also demand that m>log(L)/log(4), and I took the ceiling to keep it an integer – Valborg Mar 23 '18 at 18:28

Note that

• $(-2)^{2k} + \pi\to +\infty$
• $(-2)^{2k+1} + \pi\to -\infty$

Then the limit doesn't exist.

For the case $2k$ indeed fix wlog $M>\pi$

• $(-2)^{2k}+\pi=4^k+\pi>M\iff4^k>M-\pi \\\iff k\log4>\log(M-\pi)\iff k>\frac{\log(M-\pi)}{\log4}$

For the case $2k+1$ the proof is almost the same.

• Is this proof by induction? – dg123 Mar 23 '18 at 18:04
• @dg123 it is just by definition of limit, at first we claim that limit doesn't exist since the limits of 2 subsequences are different, the we show that the 2 limits are divergent – gimusi Mar 23 '18 at 18:06
• So like an even case and an odd case? – dg123 Mar 23 '18 at 18:06
• The even diverges at $+\infty$ the odd at $-\infty$ then the sequence has not limit. – gimusi Mar 23 '18 at 18:11
• Ok so basically since it converges at two different numbers, the limit doesn't exist, and therefore it diverges? – dg123 Mar 23 '18 at 18:14

Let ϵ>0 be arbitrary

okay.

Suppose that n>N.

What is $N$? And what significance does it have? If it's just some number like $9,317$ it's got nothing to do with anything.

$|(−2)^n+π−L|<ϵ$

Why? Why should that be true? And what is $L$?

Okay, I think you you are trying to do is to say. "Assume that the sequence converges to $L$. Then for any $\epsilon > 0$ there will exist an $N$ so that for all $n$ it will be true that $|(-2)^n + \pi -L| < \epsilon$. Let $\epsilon$ be an arbitrary positive value. Let's let $N$ be the valus so that $|(-2)^n + \pi -L| < \epsilon$ for all $n > N$. ANd let $n > N$, then $|(-2)^n + \pi -L| < \epsilon$"

$(−2)^n+π−L>N$

Now you've gone off the deep end. That's nonsense and doesn't follow and if $\epsilon < N$ is obviously not true.

$nlog(−2)+π−L>log(N)$

If you are going to take logs you must take logs of the whole thing $\log((-2)^n + \pi - L) > \log (N)$. You can just leave the $\pi - L$ out. Also you can't take logs of possibly negative values.

....

What you should do demonstrate a contradiction is to show that there is an $m > n > N$ so that $|(-2)^m + \pi - L| \ge \epsilon$>

Consider $m = n+1$.

If $|(-2)^n + \pi - L | < \epsilon$ then $|(-2)^{n+1} + \pi - L| = |(-2)^n*(-2) + \pi -L|$. Can we work with that? One thing to note is that if $(-2)^n$ is positive/negative then $(-2)^{n+1}$ is negative/postive.

Let $(-2)^n + \pi - L = K$ and Le $(-2)^{n+1} + \pi - L =M$.

Notice $M - K = (-2)^{n+1} - (-2)^n = -2*(-2)^n - (-2)^n = -3*(-2)^n$.

So if $|(-2)^n + \pi - L|=|K| < \epsilon$

And if $|(-2)^{n+1} + \pi - L| =|M| < \epsilon$.

Then $|M - K| \le |M| + |K| < 2\epsilon$.

But $|M-K| = |-3*(-2)^n| = 3*2^n$.

So $3*2^n < 2\epsilon$.

That's not necessarily a contradiction but it can be made into one.

Let $n > \max (\log_2(\frac 23 \epsilon), N)$ then

$2\epsilon < 3*2^n = |((-2)^{n+1} +\pi - L)-((-2)^{n} +\pi - L)|\le |(-2)^{n+1} +\pi - L| + |(-2)^{n} +\pi - L| < 2\epsilon$.

Let us assume for contradiction taht the limit exists and is $L$. Then by definition $\exists N\in\mathbb{N}$ such that $\forall n\ge N$, we have $|(-2)^n+\pi-L|<\varepsilon$

So let us break up into cases: evens and odds. When $n=2k$ we have $|(-2)^{2k}+\pi-L|<\varepsilon$, which is equal to $|4^k+\pi-L|<\varepsilon$. Moreover this implies $-\varepsilon<4^k+\pi-L<\varepsilon$.

And when $n=2k+1$, we observe that $|-2\cdot 4^k+\pi-L|<\varepsilon$, or alternatively $-\varepsilon<-2\cdot 4^k+\pi-L<\varepsilon$. Let us add $3\cdot 4^k$ across the board. We get $3\cdot 4^k-\varepsilon<4^k+\pi-L<3\cdot 4^k +\varepsilon$.

Look at those intervals and realize it must hold $\forall n\ge N$, if $N$ exists. The inner part of each inequality is the same, $4^k+\pi-L$. Notice that no $L$ can satisfy both intervals simultaneously. In one case, $4^k+\pi-L$ exists in an epsilon neighborhood of $0$, and in the other case, $4^k+\pi-L$ exists in an epsilon neighborhood of $3\cdot 4^k$. Any large $N$ we might choose, the inequalities are completely mutually exclusive.

To make this clearer, note that when $n$ is even, $-\varepsilon<4^k+\pi-L<\varepsilon$ can be rewritten $|(4^k+\pi)-(L)|<\varepsilon$, and that can also be rewritten $L\in N(4^k+\pi;\varepsilon)$.

Whereas when $n$ is odd, $-\varepsilon<-2\cdot 4^k+\pi-L<\varepsilon$ is rewritten as $|(-2\cdot 4^k+\pi)-(L)|<\varepsilon$, or altneratively, $L\in N(-2\cdot 4^k+\pi;\varepsilon)$.

It should be clear that for large enough $k$ (i.e. $n$) and for small enough $\varepsilon$, the hypothetical limit $L$ is in mutually exclusive sets simultaneously. For example, if $n=2$ (i.e. $k=1$) and $\varepsilon=1$ then $L\in (3+\pi,5+\pi)$. When $n=3$ (i.e. $k=1$), then $L\in(-9+\pi,-7+\pi)$. The larger $n$ gets and the smaller $\varepsilon$ gets, the more drastic the disparity.

The value $L$ bounces back and forth between two mutually exclusive intervals as $n$ grows larger. Therefore $L$ never settles on a single constant value. The two mutually exclusive sets that $L$ has to "be in" IS the contradiction. The assumption made that led to this contradiction is that $L$ exists. It clearly doesnt.

• So do we break it up into even and odd cases whenever a sequence diverges or is it just for this specific example? – dg123 Mar 23 '18 at 18:38
• There is no hard and fast rule. By breaking it up into cases we show that two different and exclusive intervals exist. $L$ cannot exist in both. Therefore no $N$ can satisfy the definition of the limit. Breaking it up into even and odd was just easy for this problem. – CogitoErgoCogitoSum Mar 23 '18 at 18:41
• Why are we adding $3*4^k$? – dg123 Mar 23 '18 at 19:29
• To make the inner part of both inequalities the same, for comparison. – CogitoErgoCogitoSum Mar 23 '18 at 19:30
• Is this using subsequences? – dg123 Mar 24 '18 at 18:26