Answer
$\{a_{n}\}$ diverges to negative infinity.
Work Step by Step
We interpret the sequence
$1, -1,-5,-13,-29,$
as
$(3-2), (3-4), (3-8), (3-16),...,(3-2^{n}),....$
$a_{1}=1\qquad (=-2^{n}+3$ when n=$1)$
$a_{2}=2(1)-3=-1\qquad (=-2^{n}+3$ when n=$2)$
$a_{3}=2(-1)-3=-5\qquad (=-2^{n}+3$ when n=$3)$
$a_{4}=2(-5)-3=-13\qquad (=-2^{n}+3$ when n=$4)$
$a_{5}=2(-13)-3=-29\qquad (=-2^{n}+3$ when n=$5)$
$...$
$a_{n}=-2^{n}+3$
We see:
$\qquad\{a_{n}\}$ is unbounded from below because $2^{n}$ grows without bound,
$\qquad\{a_{n}\}$ is monotonic, non-increasing
This sequence is divergent.
To check, use the definition for "diverges to negative infinity".
Let m be a negative number. We want to find N such that
$n\gt N\Rightarrow a_{n}\lt m$
$-2^{N}+3=m$
$3-m=2^{N} \quad $
We see that $3-m$ is positive, so we can take the logarithm
$\log_{2}(3-m)=N$
If $n\gt N$, then
$a_{n}=-2^{n}+3\lt-2^{\log_{2}(3-m)}+3=-(3-m)+3=m,$
$a_{n}\lt m$
