Solve. e ≈ 1 + 1 + 1 2 2 ! + 1 3 3 ! + 1 4 4 ! + 1 5 5 !

Question

Solve.   e  ≈  1  +  1  +            1              2                    2      !            ﻿    +            1              3                    3      !        +            1              4                    4      !        +            1              5                    5      !

Jaelyn Rosario · Accepted Answer

This is the lower bound given. Since in this   (  1  +  1      /        e          n      +      1        ) is a good converging replace for 1 for n&amp;gt;=1. The difference between the lower and the upper bound is   1.9      10          −      20       for n=51.Taylor&#039; s theorem now garantees that the truncated expansions converges to the infinite and so approximates e=2.78128.

Solve. e approx 1+1 + 1^2/(2!)+1^3/(3!)+1^4/(4!)+1^5/(5!)

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