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Which of the following series are convergent?

$$\sum_{n=1}^\infty \frac{1/2+(-1)^n}{n} \tag{a}$$ $$\sum_{n=1}^\infty (-1)^n (\sqrt{n+1}-\sqrt{n}) \tag{b}$$ $$\sum_{n=1}^\infty \frac{\sin (n^{3/2})}{n^{3/2}} \tag{c}$$

My thoughts:

(a) true by alternating series test.
(b) true by alternating series test.
(c) true by comparison test.
Are my thoughts correct?

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  • $\begingroup$ I was just wondering can u break (a) into $\sum \frac{1}{2n}+$? $\endgroup$ – Marso Jan 19 '13 at 18:56
  • $\begingroup$ I am a real fool.thanks for your comment.each year I am doing this types of silly mistakes in exam hall. $\endgroup$ – poton Jan 19 '13 at 19:05
  • $\begingroup$ Dont worry, It is happening to other students too. $\endgroup$ – Marso Jan 19 '13 at 19:07
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  1. $$\sum_{n=1}^{\infty} \dfrac{1/2 + (-1)^n}n = \sum_{n=1}^{\infty} \dfrac1{2n} + \sum_{n=1}^{\infty} \dfrac{(-1)^n}n$$ diverges since the first sum diverges and the second one converges to $- \log 2$.
  2. $$\sum_{n=1}^{\infty} (-1)^n (\sqrt{n+1} - \sqrt{n}) = \sum_{n=1}^{\infty} \dfrac{(-1)^n}{(\sqrt{n+1} + \sqrt{n})}$$ converges by alternating test.
  3. $$\sum_{n=1}^{\infty} \dfrac{\sin(n^{3/2})}{n^{3/2}}$$ converges since $\vert \sin(n^{3/2}) \vert \leq 1$.
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For (a), $\displaystyle \sum\limits_{n \geq 1} \left( \frac{1}{2n}+ \frac{(-1)^n }{n} \right)$ is convergent iff $\displaystyle \sum\limits_{n \geq 1} \frac{1}{2n}$ is convergent, because $\displaystyle \sum\limits_{n \geq 1} \frac{(-1)^n}{n}$ converges. So (a) is in fact false.

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