First slide

Special Series in Sequences and Series

Question

If $L = lim_{n \to \infty} (1 + 3^{- 1}) (1 + 3^{- 2}) (1 + 3^{- 4}) (1 + 3^{- 8}) \dots$ $(1 + 3^{- 2^{n}}),$ then

Moderate

A

$L = \frac{2}{3}$
B

$L = \frac{3}{2}$
C

$L = 1$
D

$L = 2$

Solution

$\begin{array}{l} (1 - 3^{- 1}) L = lim_{n \to \infty} (1 - 3^{- 1}) (1 + 3^{- 1}) (1 + 3^{- 2}) (1 + 3^{- 4}) \dots (1 + 3^{- 2^{n}}) \\ \frac{2}{3} L = lim_{n \to \infty} (1 - 3^{- 2}) (1 + 3^{- 2}) (1 + 3^{- 4}) \dots (1 + 3^{- 2^{n}}) \\ = lim_{n \to \infty} (1 - 3^{- 2^{2}}) (1 + 3^{- 4}) \dots (1 + 3^{- 2^{n}}) \\ = lim_{n \to \infty} (1 - 3^{- 2^{n + 1}}) = 1 \\ \Rightarrow L = 3 / 2 \end{array}$

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