How to proof that a sequence is divergent?

Question

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 $n>N$.

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

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

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

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

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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

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

That is a contradiction.