frac{1}{3} + frac{1}{2 cdot 3^2} + frac{1}{3 | vedclass

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(B) The given series is $S = \frac{1}{3} + \frac{1}{2 \cdot 3^2} + \frac{1}{3 \cdot 3^3} + \frac{1}{4 \cdot 3^4} + \dots \infty$. We know the logarith

Vedclass · Accepted Answer

$\log_e 3 - \log_e 2$

Vedclass · Answer

(B) The given series is $S = \frac{1}{3} + \frac{1}{2 \cdot 3^2} + \frac{1}{3 \cdot 3^3} + \frac{1}{4 \cdot 3^4} + \dots \infty$. We know the logarithmic expansion: $-\log_e(1 - x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots \infty$ for $|x| Comparing the given series with the expansion,we set $x = \frac{1}{3}$. Thus,$S = -\log_e(1 - \frac{1}{3}) = -\log_e(\frac{2}{3})$. Using the property $-\log_e(\frac{a}{b}) = \log_e(\frac{b}{a})$,we get $S = \log_e(\frac{3}{2})$. Using the property $\log_e(\frac{a}{b}) = \log_e a - \log_e b$,we get $S = \log_e 3 - \log_e 2$.

$\frac{1}{3} + \frac{1}{2 \cdot 3^2} + \frac{1}{3 \cdot 3^3} + \frac{1}{4 \cdot 3^4} + \dots \infty = $

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