Answer
Converges absolutely
Work Step by Step
Let us consider an alternating series $\Sigma_{k=1}^\infty(-1)^k a_k$ that can be convergent , when it satisfy the following two conditions such as:
1) The magnitude of terms must form a non-increasing sequence.
2) $\lim\limits_{k \to \infty} a_k=0$
The series will converge conditionally when it is a convergent series , but the condition is when the series of its absolute value diverges and the series will converge absolutely but the condition is when the series of its absolute value converges.
Here, in the problem we have $a_k=e^{-k}$
a) In the given sequence, $\lim\limits_{k \to \infty }e^{-k}=0$ . This implies that $\Sigma (-1)^k a_k $ converges , so the terms are a non-increasing sequence.
Also, $\sum e^{-k} =\Sigma (\dfrac{1}{e})^k$ shows a geometric series with ratio $r =\dfrac{1}{e} \lt 1$ . This implies that the given series is a geometric series and converges as well.
Thus, the given series converges absolutely.
