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The sum of the series `(1)/(1!)+(2)/(2!)+(3)/(3!)+(4)/(4!)`+…..to `infty` isA. eB. 2eC. `(1)/(2)e`D. none of these |
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Answer» Answer: We have `1+(1)/(1!)+(2)/(2!)+(3)/(3!)+…= underset(n=1)overset(infty)Sigma(nb)/(n!)=underset(n=1)overset(infty)Sigma(1)/(n-1)!=e` |
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