Use the formal definition to prove that the following sequence diverges

[alexmahone]

$\displaystyle s_n=\left(\frac1n-1\right)^n$

My attempt:

For large $n$, the sequence oscillates between $e^{-1}$ and $-e^{-1}$ and therefore diverges. Now for the proof.

Assume, for the sake of argument, that the sequence converges to $L$.

$\exists N\in\mathbb{N}$ such that $|s_n-L|<0.1$ whenever $n\ge N$

$\displaystyle\left|\left(\frac1n-1\right)^n-L\right|<0.1$ whenever $n\ge N$

$\displaystyle\implies\left|\left(\frac1{n+1}-1\right)^{n+1}-L\right|<0.1$ whenever $n\ge N$

We can rewrite these 2 equations as

$\displaystyle\left|(-1)^n\left(1-\frac1n\right)^n-L\right|<0.1$ whenever $n\ge N$ --------------- (1)

$\displaystyle\left|(-1)^{n+1}\left(1-\frac1{n+1}\right)^{n+1}-L\right|<0.1$ whenever $n\ge N$

$\displaystyle\implies\left|(-1)^n\left(1-\frac1{n+1}\right)^{n+1}+L\right|<0.1$ whenever $n\ge N$ --------------- (2)

How do I get a contradiction from equations (1) and (2)?

 

[Evgeny.Makarov]

$\displaystyle\left|(-1)^n\left(1-\frac1n\right)^n-L\right|<0.1$ whenever $n\ge N$ --------------- (1)

$\displaystyle\left|(-1)^{n+1}\left(1-\frac1{n+1}\right)^{n+1}-L\right|<0.1$ whenever $n\ge N$

$\displaystyle\implies\left|(-1)^n\left(1-\frac1{n+1}\right)^{n+1}+L\right|<0.1$ whenever $n\ge N$ --------------- (2)

How do I get a contradiction from equations (1) and (2)?

If it is known that $(1-1/n)^n\to1/e$, then you can reason as follows. Suppose $|(1-1/n)^n-1/e|\le0.1$ for all $n\ge N$ (we can achieve this increasing $N$ if necessary). Then (1) with an even $n$ implies that $|L-1/e|<0.2$ and therefore $L>1/e-0.2>0$ and (2) implies that $|(-L)-1/e|<0.2$ and therefore $L<-1/e+0.2<0$, which is a contradiction.

 

[math771]

To get a contradiction, we can use the property of limits that if a sequence converges to a certain value, then any subsequence of that sequence must also converge to the same value.

Let's consider the subsequence $\displaystyle s_{2n}=\left(\frac{1}{2n}-1\right)^{2n}$, which is just the even terms of the original sequence. From equation (1), we know that $\displaystyle \left|(-1)^{2n}\left(1-\frac{1}{2n}\right)^{2n}-L\right|<0.1$ whenever $2n\ge N$. This can be simplified to $\displaystyle \left(\frac{1}{2n}\right)^{2n}-L<0.1$.

Now, let's consider the subsequence $\displaystyle s_{2n+1}=\left(\frac{1}{2n+1}-1\right)^{2n+1}$, which is the odd terms of the original sequence. From equation (2), we know that $\displaystyle \left|(-1)^{2n+1}\left(1-\frac{1}{2n+1}\right)^{2n+1}+L\right|<0.1$ whenever $2n+1\ge N$. This can be simplified to $\displaystyle \left(\frac{1}{2n+1}\right)^{2n+1}+L<0.1$.

Now, if we take the limit as $n$ approaches infinity for both of these subsequence equations, we get

$\displaystyle \lim_{n\to\infty}\left(\frac{1}{2n}\right)^{2n}-L=0$

and

$\displaystyle \lim_{n\to\infty}\left(\frac{1}{2n+1}\right)^{2n+1}+L=0$.

But this is a contradiction, as the first limit converges to $0$, while the second limit converges to $2L$. Therefore, our assumption that the original sequence converges to a finite value $L$ must be false, and the sequence must diverge.