# Thread: Test the convergence of the series

1. It should be $\displaystyle (-1)^{2n}$, which equals 1 for all n.

2. ## Is this correct

Originally Posted by redsoxfan325
It should be $\displaystyle (-1)^{2n}$, which equals 1 for all n.
I have done something could u please check if it is correct or no

= $\displaystyle \lim_{n \to \infty} \frac{(-1)^{2n}}{(n + 2)\sqrt{n + 1}} . n^{\frac{3}{2}}$

= $\displaystyle \lim_{n \to \infty} \frac{(-1)^{2n}}{\sqrt{(n + 2)^2 (n + 1)}} . n^{\frac{3}{2}}$

= $\displaystyle \lim_{n \to \infty} \frac{(-1)^{2n}}{\sqrt{(n + 4n + 4)(n + 1)}} . n^{\frac{3}{2}}$

= $\displaystyle \lim_{n \to \infty} \frac{(-1)^{2n}}{\sqrt{(n^3 + 5n^2 + 8n + 4)}} . n^{\frac{3}{2}}$

= $\displaystyle \lim_{n \to \infty} \frac{(-1)^{2n}}{n^{\frac{3}{2}}\sqrt{( 1 + \frac{5}{3n} + \frac{8}{3n^2} + \frac{4}{3n^3})}} . n^{\frac{3}{2}}$

= $\displaystyle \lim_{n \to \infty} \frac{(-1)^{2n}}{ \sqrt{( 1 + \frac{5}{3n} + \frac{8}{3n^2} + \frac{4}{3n^3})}}$

= $\displaystyle \frac{1}{\sqrt{1}}$

= 1 Is this Correct ????????

3. Originally Posted by zorro
I have done something could u please check if it is correct or no

= $\displaystyle \lim_{n \to \infty} \frac{(-1)^{2n}}{(n + 2)\sqrt{n + 1}} . n^{\frac{3}{2}}$

= $\displaystyle \lim_{n \to \infty} \frac{(-1)^{2n}}{\sqrt{(n + 2)^2 (n + 1)}} . n^{\frac{3}{2}}$

= $\displaystyle \lim_{n \to \infty} \frac{(-1)^{2n}}{\sqrt{(n + 4n + 4)(n + 1)}} . n^{\frac{3}{2}}$

= $\displaystyle \lim_{n \to \infty} \frac{(-1)^{2n}}{\sqrt{(n^3 + 5n^2 + 8n + 4)}} . n^{\frac{3}{2}}$

= $\displaystyle \lim_{n \to \infty} \frac{(-1)^{2n}}{n^{\frac{3}{2}}\sqrt{( 1 + \frac{5}{3n} + \frac{8}{3n^2} + \frac{4}{3n^3})}} . n^{\frac{3}{2}}$

= $\displaystyle \lim_{n \to \infty} \frac{(-1)^{2n}}{ \sqrt{( 1 + \frac{5}{3n} + \frac{8}{3n^2} + \frac{4}{3n^3})}}$

= $\displaystyle \frac{1}{\sqrt{1}}$

= 1 Is this Correct ????????
Originally Posted by redsoxfan325
So far, yes. Now calculate the limit. You should find that it equals 1, which tells you...
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4. ## Is this correct

Originally Posted by Defunkt
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Since the result is 1 the series convergences ....Am i correct

5. Yes, it converges because $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$ does.

6. ## Thanks u very much

Originally Posted by redsoxfan325
Yes, it converges because $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$ does.

Thank u redsoxfan325 , Defunkt and Mr fantastic

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