Infinite Factorial Summation with a Twist

Calculus Level 4

$\large 1 + \dfrac{1 + \frac{1}{1!}}{2} + \dfrac{1 + \frac{1}{1!} + \frac{1}{2!}}{2^2} + \dfrac{1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!}}{2^3} + \ldots$

If the above series can be expressed as $S$ , find $\big \lfloor 100S \rfloor$ .

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The answer is 329.

1 solution

Rajen Kapur
Jun 17, 2015

Collecting terms the summation in this way: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . .$ $+ \frac{1}{1!}\large [ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . ]$ $+ \frac{1}{2!} \large [\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + . . . ]$ ans so on. $= 2 + 1.\frac{1}{1!} + \frac{1}{2} . \frac{1}{2!} + \frac{1}{4} . \frac{1}{3!} + . . .$ , $= 2. \large [ 1 + \frac{1}{2} + \frac{1}{2!} . \frac{1}{2^2} + . . . . ]$ which equals 2 e^0.5.

Infinite Factorial Summation with a Twist

Try more problems of this type here .

Try my set .

The answer is 329.

1 solution

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