Use the Ratio Test to determine whether sum{n=1, infinity} [ (-1/n)^(2n) ] converges or diverges

[armn91]

Hello all,
Could anybody let me know if what I have been doing in this activity is correct?
Thank you!

 

[nasi112]

The caluclation in last step was wrong.

$\displaystyle \lim_{n\rightarrow \infty}\left|\frac{1}{(1+\frac{1}{n})^{2n}(n + 1)^{2}}\right| = \left|\frac{\lim_{n\rightarrow \infty}\frac{1}{(n + 1)^{2}}}{\lim_{n\rightarrow \infty}(1+\frac{1}{n})^{2n}}\right| \neq \infty$

Here was your mistake:

$\displaystyle \lim_{n\rightarrow \infty}(1+\frac{1}{n})^{2n} \neq 0$

 

[armn91]

Thank you for the indications. Could you please let me know if this is correct now?

 

[nasi112]

The final answer is correct but the calculations are wrong.

$\displaystyle \lim_{n\rightarrow \infty}\left(\frac{n}{n+1}\right)^{2n} \neq 1$

 

[BigBeachBanana]

$$y = \left(\frac{n}{n+1}\right)^{2n}\\ \ln y =2n \ln \left(\frac{n}{n+1}\right)\\ \lim_{n \to \infty} \ln y = \lim_{n \to \infty} 2n \ln \left(\frac{n}{n+1}\right)\\ \lim_{n \to \infty} \ln y = 2 \lim_{n \to \infty} \dfrac{ \ln \left(\frac{n}{n+1}\right)}{\dfrac{1}{n}}\\$$
Proceed with L'Hopital and undo the log.

 

[skeeter]

note $\left(\dfrac{n}{n+1}\right)^{2n} = \left(\dfrac{n+1}{n}\right)^{-2n}$

$\displaystyle \lim_{n \to \infty} \left(\dfrac{n+1}{n}\right)^{-2n}$

$\displaystyle {\color{red}\lim_{n \to \infty}} \bigg[{\color{red}\left(1 + \dfrac{1}{n}\right)^{n}}\bigg]^{-2}$

are you familiar with that basic limit ?