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

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(A) The general term of the series is $T_n = \frac{n+1}{n} \cdot \frac{1}{3^n} = \left( 1 + \frac{1}{n} \right) \frac{1}{3^n} = \frac{1}{3^n} + \frac{

Vedclass · Accepted Answer

$\frac{1}{2} - \log_e \frac{2}{3}$

Vedclass · Answer

(A) The general term of the series is $T_n = \frac{n+1}{n} \cdot \frac{1}{3^n} = \left( 1 + \frac{1}{n} \right) \frac{1}{3^n} = \frac{1}{3^n} + \frac{1}{n \cdot 3^n}$.The sum $S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{1}{3^n} + \sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n}$.The first part is a geometric series: $\sum_{n=1}^{\infty} \frac{1}{3^n} = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = \frac{1}{2}$.The second part uses the logarithmic expansion $-\log_e(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$. For $x = \frac{1}{3}$,we get $\sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n} = -\log_e(1 - 1/3) = -\log_e(2/3)$.Thus,$S = \frac{1}{2} - \log_e \left( \frac{2}{3} \right)$.

$\frac{2}{1} \cdot \frac{1}{3} + \frac{3}{2} \cdot \frac{1}{9} + \frac{4}{3} \cdot \frac{1}{27} + \frac{5}{4} \cdot \frac{1}{81} + \dots \infty = $

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