Question 1208380
.
find : 1/(2!) + 1/(3!) + 1/(4!) + ... + 1/(100!)
~~~~~~~~~~~~~~~~~~~~


<pre>
Taylor series for {{{e^x}}} is this convergent series

    {{{e^x}}} = 1 + {{{x/1!}}} + {{{x^2/2!}}} + {{{x^3/3!}}} + . . . {{{x^n/n!}}} + . . .     (1)


till infinity,  where  e = 2.71828...  is the base of natural logarithm.


So, e = {{{e^1}}} = 1 + {{{1/1!}}} + {{{1/2!}}} + {{{1/3!}}} + . . . {{{1/n!}}} + . . .     (2)


The given series is the series (2) for number "e" with the first two terms cut
and all the terms after  {{{1/100!}}}  cut, but these later terms are so small 
that they can be neglected for reasonable precision.


Therefore, the given series is  e - 2 = 0.71828  (rounded).    <U>ANSWER</U>
</pre>

Solved, with explanations.