Determine whether the following series converges absolutely , conditionally or diverges:
(i) $ \ \sum_{n=0}^{\infty} (-1)^n \frac{k^2-2}{k^2+6} \ $
(ii) $ \ \sum_{n=0}^{\infty} (-1)^n \frac{k^2}{(3k)!} \ $
Answer:
(i)
The series is $ \ \sum_{n=0}^{\infty} (-1)^n \frac{k^2-2}{k^2+6} \ $
$ a_k=(-1)^n \frac{k^2-2}{k^2+6} \ $
The series is alternating series .
But the term of series does not decrease by its absolute value. So how can we conclude about the convergence of the series.
further taking absolute value, we get
$ \ |a_k|=\frac{k^2-2}{k^2+6} \ $
For absolute convergence,
$ \lim_{n \to \infty} |\frac{a_{n+1}}{a_n} |=1 \ $
so we can say whether converges or not
(ii) The given series is $ \ \sum_{n=0}^{\infty} (-1)^n \frac{k^2}{(3k)!} \ $
This is an alternatic series test.
The absolute value of each term decreases from the previous term.
Thus by Alternating series test , the series converges.
But the series does not converges absolutely.
Help with the part $ \ (i) \ $ question