Use the Ratio Test to determine whether sum{n=1, infinity} [ (-1/n)^(2n) ] converges or diverges
- Thread starter armn91
- Start date
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:
\(\displaystyle \lim_{n\rightarrow \infty}(1+\frac{1}{n})^{2n} \neq 0\)
\(\displaystyle \lim_{n\rightarrow \infty}\left|\frac{1}{(1+\frac{1}{n})^{2n}(n + 1)^{2}}\right| = \left|\frac{\lim_{n\rightarrow \infty}\frac{1}{(n + 1)^{2}}}{\lim_{n\rightarrow \infty}(1+\frac{1}{n})^{2n}}\right| \neq \infty\)
Here was your mistake:
\(\displaystyle \lim_{n\rightarrow \infty}(1+\frac{1}{n})^{2n} \neq 0\)
BigBeachBanana
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[math]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}}\\[/math]
Proceed with L'Hopital and undo the log.
Proceed with L'Hopital and undo the log.
skeeter
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note [imath]\left(\dfrac{n}{n+1}\right)^{2n} = \left(\dfrac{n+1}{n}\right)^{-2n}[/imath]
[imath]\displaystyle \lim_{n \to \infty} \left(\dfrac{n+1}{n}\right)^{-2n}[/imath]
[imath]\displaystyle {\color{red}\lim_{n \to \infty}} \bigg[{\color{red}\left(1 + \dfrac{1}{n}\right)^{n}}\bigg]^{-2}[/imath]
are you familiar with that basic limit ?
[imath]\displaystyle \lim_{n \to \infty} \left(\dfrac{n+1}{n}\right)^{-2n}[/imath]
[imath]\displaystyle {\color{red}\lim_{n \to \infty}} \bigg[{\color{red}\left(1 + \dfrac{1}{n}\right)^{n}}\bigg]^{-2}[/imath]
are you familiar with that basic limit ?