# Thread: Trying to pin down a divergence proof

1. ## Trying to pin down a divergence proof

Hi,

I would like to show that the sequence $\{1+(-2)^n\}$ diverges. Here is my attempt.

The sequence converges to some element $L \in \mathbb{R}$ iff for all $\epsilon > 0,\,\, \exists N \in \mathbb{N}$ such that $n > N \implies |a_n - L| < \epsilon$. My aim is to show that there is an epsilon for which no positive integer N satisfies this requirement.

Choose $\epsilon = 1$. Then there exists an N such that if n > N then $|1+(-2)^n - L| < 1$.

Now let k be a positive odd integer such that k > N, then

$|1+(-2)^{k+1} - L| < 1$
so $|1+2^{k+1} - L| < 1$

that is $-1 < 1+2^{k+1} - L < 1$

so $-2 < 2^{k+1} - L < 0,\,\,\implies L > 2^{k+1}$

and by a similar argument we can show that if we take k to be some even integer such that k > N then

$L > 2^{k}$

But we can always find an even AND odd k such that the following inequalities are false for any real number L.

However there's something in the back of my mind which I'm not happy with. I don't like how I finished the proof, it doesn't seem quite sound to me. I was wondering if someone could point me in the right direction to finish it off or let me know where I've gone wrong and perhaps suggest a better approach.

Observe though that $(s_n)_{n=1}^\infty$ diverges iff $(s_n - c)_{n=1}^\infty$ diverges, so you can subtract off the 1 and get the sequence $((-2)^n)_{n=1}^\infty$. Then observe that a sequence diverges iff it has a divergent subsequence, so you can restrict to even n, in which case you have just $(4^n)_{n=1}^\infty$, whose divergent character is obvious.