Question

Summation of 2^n/n! with limit 1 to infinity

Answer 1

Method

Consider&#xA0;&#x2211;n=0&#x221E;xn=11&#x2212;x,&#xA0;where&#xA0;|x|&lt;1." role="presentation">Consider ∑n=0∞xn=11−x, where |x|<1.Consider ∑n=0∞xn=11−x, where |x|<1.

Then differentiating both sides w.r.t. x, we have

&#x2211;n=1&#x221E;nxn&#x2212;1=1(1&#x2212;x)2" role="presentation">∑n=1∞nxn−1=1(1−x)2∑n=1∞nxn−1=1(1−x)2

Putting x=12" role="presentation">x=12x=12 gives

&#x2211;n=1&#x221E;n2n&#x2212;1=1(1&#x2212;12)2=4" role="presentation">∑n=1∞n2n−1=1(1−12)2=4∑n=1∞n2n−1=1(1−12)2=4

12&#x2211;n=1&#x221E;n2n&#x2212;1=2" role="presentation">12∑n=1∞n2n−1=212∑n=1∞n2n−1=2

&#x2211;n=1&#x221E;n2n=2" role="presentation">