Series convergence test $\sum_{n=1}^\infty 2^{-n-(-1)^n}$

Question

I want to determine the convergence of $\displaystyle \sum_{n=1}^\infty 2^{-n-(-1)^n}$.

My approach to this question:
I tried the root test so far.
We know that $2^x>0~\forall x\in\mathbb{R}$, hence $\left|2^{-n-(-1)^n}\right|=2^{-n-(-1)^n}$ (I'm not so sure of this step.)
\begin{align*} \lim_{n\rightarrow\infty} \sqrt[n]{2^{-n-(-1)^n}}&=\lim_{n\rightarrow\infty} 2^{\frac{-n-(-1)^n}{n}} \\&=2^{\lim_{n\rightarrow\infty}\frac{-n-(-1)^n}{n}} \\&=2^{-1}=\frac{1}{2} \end{align*} Because $\frac{1}{2}<1$, therefore $\displaystyle \sum_{n=1}^\infty 2^{-n-(-1)^n}$ converges absolutely.

Someone please check my algebra.

Edit: I wanted to test for absolute / conditional convergence.
Corrected typos.

Answer 1

This is good. Easier is just to say $0 \lt 2^{-n-(-1)^n} \le 2^{-n+1}$ so $$\displaystyle \sum_{n=1}^\infty 2^{-n-(-1)^n}\le \displaystyle \sum_{n=1}^\infty 2^{-n+1}$$ which is a geometric series that we know converges, so the original series converges by the comparison test.

Answer 2

Another approach is to just compute the powers: $$\begin{array}{c|rr}n&1&2&3&4&5&6&7&8&9&10&\ldots\\\hline-n-(-1)^n&0&-3&-2&-5&-4&-7&-6&-9&-8&-11&\ldots\end{array}$$ So $$\sum_{n=1}^{\infty}2^{-n-(-1)^{n}}=-\frac12+\sum_{k=0}^{\infty}2^{-k}=-\frac12+\frac{1}{1-\frac12}=1.5$$

Answer 3

$\sum_{n=1}^\infty 2^{-n-(-1)^n} $

Sum to $2m$ terms:

$\begin{array}\\ s(2m) &=\sum_{n=1}^{2m}2^{-n-(-1)^n}\\ &=\sum_{n=1}^{m}\left(2^{-(2n-1)+1}+2^{-2n-1}\right)\\ &=\sum_{n=1}^{m}\left(2^{-2n+2}+2^{-2n-1}\right)\\ &=\sum_{n=1}^{m}2^{-2n}\left(2^{2}+2^{-1}\right)\\ &=\frac92\sum_{n=1}^{m}2^{-2n}\\ &=\frac92\sum_{n=1}^{m}4^{-n}\\ &=\frac92\dfrac{4^{-1}-4^{-n-1}}{1-1/4}\\ &=\frac92\dfrac{1-4^{-n}}{3}\\ &=\frac32(1-4^{-n})\\ &\to \frac32 \qquad\text{since all terms }\to 0\\ \end{array} $