frac{1}{1 cdot 3} + frac{1}{2 cdot 5} + frac{ | vedclass

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(B) ધારો કે $S = \sum_{n=1}^{\infty} \frac{1}{n(2n+1)}$.આંશિક અપૂર્ણાંકનો ઉપયોગ કરતા,$\frac{1}{n(2n+1)} = \frac{1}{n} - \frac{2}{2n+1}$.તેથી,$S = \sum

Vedclass · Accepted Answer

$2 - \log_e 2$

Vedclass · Answer

(B) ધારો કે $S = \sum_{n=1}^{\infty} \frac{1}{n(2n+1)}$.આંશિક અપૂર્ણાંકનો ઉપયોગ કરતા,$\frac{1}{n(2n+1)} = \frac{1}{n} - \frac{2}{2n+1}$.તેથી,$S = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{2}{2n+1} \right) = \left( 1 - \frac{2}{3} \right) + \left( \frac{1}{2} - \frac{2}{5} \right) + \left( \frac{1}{3} - \frac{2}{7} \right) + \dots$$S = 1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots$$\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ શ્રેણીનો ઉપયોગ કરતા,$x=1$ માટે $\log_e 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$.તેથી,$S = 1 - (\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots) = 1 - (1 - \log_e 2) = 2 - \log_e 2$.

$\frac{1}{1 \cdot 3} + \frac{1}{2 \cdot 5} + \frac{1}{3 \cdot 7} + \frac{1}{4 \cdot 9} + \dots$ ની કિંમત શોધો.

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