Does $$ (a_{n}) = \sqrt{(4- (b_{n}) ^2} $$, where $$ (b_{n}) =\frac{1}{2}( 1+ \frac{1}{n}) ^{n} $$ converge towards $ \sqrt{4- (e/2) ^2} $? Because $(b_{n})$ converges towards $\frac{e}{2}$?
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Where do you get $e/2$? – kccu Feb 06 '19 at 15:25
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Sorry I corrected it – user15269 Feb 06 '19 at 15:27
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1Short answer: yes – Wojowu Feb 06 '19 at 15:29
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Using $$\sqrt{4-b_n^2} = \sqrt{2 - b_n}\sqrt{2 + b_n}$$ then the limit is $$\lim_{n\to\infty} a_n = \lim_{n\to\infty} c^{-}_nc^{+}_n = \lim_{n\to\infty} c^{-}_n \cdot \lim_{n\to\infty} c^{+}_n $$ we then have $$ \lim_{n\to\infty} c^{\pm}_n = (\lim_{n\to\infty} 2\pm b_n)^{1/2} $$ Then you can apply the limits.
Chinny84
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