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InterviewSolution
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The sum of the series `(1)/(1.23)+(1)/(3.45)+(1)/(5.67)+…infty` isA. `log_(e )2-1/2`B. `log_(e)2`C. `log_(e )2+1/2`D. `log_(e )2+1` |
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Answer» Answer: We have `(1)/(1.2.3)+(1)/(3.4.5)+(1)/(5.6.7)+…infty` `=underset(n=1)overset(infty)Sigma(1)/((2n-1)2n(2n+1))` `underset(n=1)overset(infty)Sigma{(1)/(2(2n-1))-(1)/(2n) (1)/(2(2n+1))}` `=1/2underset(n=1)overset(infty)Sigma{(1)/(2n-1)-(2)/(2n)+(1)/(2n+1)}` `=1/2 underset(n=1)overset(infty)Sigma{((1)/(2n-1)-(1)/(2n))-((1)/(2n)-(1)/(2n+1))}` `=1/2log_(e)2+1/2log_(e)2-1/2=log_(e)2-1/2` |
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