An Approach

to

C A L C U L U S

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15

EVALUATING e

IN THE previous lesson, we saw the following definition of e:

  On changing the variable from x to  1
n
, we have:
e =
  By letting n take on larger and larger values in  we can come

closer and closer to a decimal value for e.

 =  2.25
 
 =  2.489
 
 =  2.594
 
 =  2.6534
 
 =  2.705
 
 =  2.7169

2.7169 is an approximate value for e.

As a more efficient approach, we can derive a sequence that converges

   to e more rapidly.  To   let us apply the binomial theorem

(Topic 24 of Precalculus):

(a + b)n  =

an  +  nan − 1b  + an − 2b2  +  an − 3b3 + . . .

On putting a = 1 and b =  1
n
, we get

Now, e is the limit of that sum as n becomes infinite.  When that happens, each fraction that depends on n approaches 1, because 1 is the quotient of the leading coefficients.  (Lesson 4.)

Therefore, on taking the limit of that sum as n becomes infinite:

Notice:  Each term can be derived from the previous term.  The second term follows from the first by dividing it by 1.  The next term follows by dividing by 2.  The next term, by dividing by 3.  The next, by 4.  And so on. e is the limit of the sequence of partial sums.  Here is the sum of the first 10 terms expressed as decimals:

11st term     1.000000
 
12nd term (dividing by 1)   1.000000
 
13rd term (dividing by 2)   0.500000
 
14th term (dividing by 3)   0.166667
 
15th term (dividing by 4)   0.041667
 
16th term (dividing by 5)   0.008333
 
17th term (dividing by 6)   0.001389
 
18th term (dividing by 7)   0.000198
 
19th term (dividing by 8)   0.000025
 
10th term (dividing by 9)   0.000003
 
  Sum   2.718282

And so after only 10 terms, we obtain a value of e accurate to 6 decimal digits.  That is an example of a rapidly converging series.

e however, like π, is an irrational number.

Problem.   In this term of the binomial theorem,

an − 2b²

  show that, on putting a = 1 and b 1
n
,  the term becomes

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!

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