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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

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  • 3
    $\begingroup$ In (1) we have that $a_n$ does NOT converge to zero. $\endgroup$ – Robert Z May 8 '18 at 6:48
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    $\begingroup$ Remember that if $\lim\limits_{k\to\infty}a_k\neq 0$ then $\sum\limits_{k=1}^\infty a_k$ does not converge, and vice versa if $\sum\limits_{k=1}^\infty a_k$ does converge then $\lim\limits_{k\to\infty}a_k=0$ $\endgroup$ – JMoravitz May 8 '18 at 6:53
  • $\begingroup$ So the series diverges in $ \ (i) \ $ $\endgroup$ – M. A. SARKAR May 8 '18 at 6:53
  • $\begingroup$ Am I correct about the series in part $ \ (ii) \ $ ? $\endgroup$ – M. A. SARKAR May 8 '18 at 6:53
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    $\begingroup$ Why, and how, did you determine the series (ii) does not converge absolutely? $\endgroup$ – DonAntonio May 8 '18 at 6:58

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