frac{1}{2} + frac{3}{2} cdot frac{1}{4} + fra | vedclass

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Question

(A) The given series is $S = \sum_{n=1}^{\infty} \frac{2n-1}{n} \cdot \frac{1}{2^n}$.We can rewrite the general term as $\left( 2 - \frac{1}{n} \right

Vedclass · Accepted Answer

$2 - \log_e 2$

Vedclass · Answer

(A) The given series is $S = \sum_{n=1}^{\infty} \frac{2n-1}{n} \cdot \frac{1}{2^n}$.We can rewrite the general term as $\left( 2 - \frac{1}{n} \right) \frac{1}{2^n} = 2 \cdot \frac{1}{2^n} - \frac{1}{n} \cdot \frac{1}{2^n}$.Thus,$S = 2 \sum_{n=1}^{\infty} \frac{1}{2^n} - \sum_{n=1}^{\infty} \frac{(1/2)^n}{n}$.Using the sum of an infinite geometric series $\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}$,we have $2 \sum_{n=1}^{\infty} (1/2)^n = 2 \cdot \frac{1/2}{1 - 1/2} = 2(1) = 2$.Using the logarithmic series expansion $-\ln(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$,for $x = 1/2$,we have $\sum_{n=1}^{\infty} \frac{(1/2)^n}{n} = -\ln(1 - 1/2) = -\ln(1/2) = \ln 2$.Therefore,$S = 2 - \ln 2$.

$\frac{1}{2} + \frac{3}{2} \cdot \frac{1}{4} + \frac{5}{3} \cdot \frac{1}{8} + \frac{7}{4} \cdot \frac{1}{16} + \dots \infty = $

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