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Whether the given series is absolutely convergent, conditionally convergent or divergent:

$a)\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{\cos nx}{n^2}:|(-1)^n\frac{\cos nx}{n^2}|\le\dfrac{1}{n^2}\implies$ Absolutely convergent.

$b)\displaystyle\sum_{n=1}^{\infty}(-1)^n\dfrac{n}{n+2}:(-1)^n\dfrac{n}{n+2}\to0\implies\dfrac{n}{n+2}\to0!\implies$ Divergent.

$c)\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{1}{2n+3}:u_n=\dfrac{1}{2n+3}\\\implies\dfrac{u_n}{u_{n+1}}=\dfrac{2n+3}{2n+5}=\dfrac{1+\dfrac{3}{2n}}{1+\dfrac{5}{2n}}=({1+\dfrac{3}{2n}})(1+\dfrac{5}{2n})^{-1}=({1+\dfrac{3}{2n}})(1-\dfrac{5}{2n}+\cdots)=1-\dfrac{1}{n}+O(\dfrac{1}{n^2})\\\implies \text{Not absolutely convergent. So conditionally convergent by Leibnitz test.}$

Am i right?

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closed as unclear what you're asking by Najib Idrissi, user91500, Daniel, user147263, tired Oct 25 '15 at 17:10

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Too many questions => unclear. $\endgroup$ – Najib Idrissi Oct 25 '15 at 8:34
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Looks good to me! Did you have any serious doubts, or are you just looking for confirmation? (If in serious doubt, why?) Assuming you mean that in $(b)$ that the limit does not approach $0$, therefore, divergent (fails absolute and conditional convergence) everything is good!

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Your solution to (c) is incomplete. The Leibniz test is frequently used incorrectly by students. It has three parts:

  1. Terms alternate (usually easy)

  2. Terms approach $0$ (fairly easy)

  3. Absolute value of terms is monotone decreasing (!)

The third part is often hard to check, but is essential if you wish to apply the test. Fortunately for you, in your problem it's not so bad. $2n+5>2n+3$ so $\frac{1}{2n+3}>\frac{1}{2n+5}$ and you're done.

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Yes, though I don't know why you would choose to write $0!$ instead of $1$.

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For (b), $\frac{n}{n+2} \to 1$, so the series is absolutely fairly violently divergent.

The sum of consecutive even-odd terms is $\frac{2n}{2n+2} - \frac{2n+1}{2n+3} = \frac{2n(2n+3)-(2n+2)(2n+1)}{(2n+2)(2n+3)} = \frac{4n^2+6n-(4n^2+3n+2)}{(2n+2)(2n+3)} = \frac{3n-2}{(2n+2)(2n+3)} $, and the sum of consecutive odd-even terms is $-\frac{2n+1}{2n+3}+\frac{2n+2}{2n+4} - = \frac{-(2n+1)(2n+4)+(2n+2)(2n+3)-}{(2n+3)(2n+4)} = \frac{-(4n^2+7n+4)+(4n^2+6n+6)}{(2n+3)(2n+4)} = \frac{-n+2}{(2n+3)(2n+4)} $, so the series is not even conditionally convergent.

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