1 + frac{2}{3} - frac{2}{4} + frac{2}{5} - do | vedclass

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(B) The given series is $S = 1 + \frac{2}{3} - \frac{2}{4} + \frac{2}{5} - \dots \infty$.We can rewrite this as $S = 1 + 2 \left( \frac{1}{3} - \frac{

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$\log_e 4$

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(B) The given series is $S = 1 + \frac{2}{3} - \frac{2}{4} + \frac{2}{5} - \dots \infty$.We can rewrite this as $S = 1 + 2 \left( \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots \right)$.We know the logarithmic expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ for $|x| For $x=1$,$\log_e 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$.Thus,$\frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots = \log_e 2 - (1 - \frac{1}{2}) = \log_e 2 - \frac{1}{2}$.Substituting this back into the expression for $S$:$S = 1 + 2 \left( \log_e 2 - \frac{1}{2} \right) = 1 + 2 \log_e 2 - 1 = 2 \log_e 2 = \log_e 2^2 = \log_e 4$.

$1 + \frac{2}{3} - \frac{2}{4} + \frac{2}{5} - \dots \infty = $

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