How do you calculate Euler's Number?

Question

Samantha Adkison · Accepted Answer

Note
This solution provides an estimate for Euler's Number (or #e#, the base of Natural (or Naperian) logarithms) rather than Euler's Constant. There may be differing opinions, but I would of thought that simplest derivation is to use the Maclaurin Series for #e^x# The Maclaurin Series for #e^x# is as follows: # e^x = 1 + x + (x^2)/(2!) + (x^3)/(3!) + (x^4)/(4!) + (x^5)/(5!) + ... # So the power series sought for #e# can be gained by putting #x=1#, viz: # e = 1 + 1 + (1)/(2!) + (1)/(3!) + (1)/(4!) + (1)/(5!) + ... # We can compute an approximating fraction by truncating the series as we see fit.  For example, If we restrict ourself to the first five terms, we get: # e ~~ 1 + 1 + (1)/(2!) + (1)/(3!) + (1)/(4!)#
# \ \ \  =  1 + 1 + 1/2+1/6+1/24#
# \ \ \  =  65/24 #
# \ \ \  =  2.708333 # We can compare this to a calculator answer for the Euler Constant, #e#: # e = 2.7182818  ... # And for further comparison, if we take further terms we get: # 6 # terms: # e~~65/24+1/120 #
# "           " = 163/60 #
# "           " = 2.716666 ... # # 7 # terms: # e~~ 163/60 + 1/720 #
# "           " = 1957/720 #
# "           " = 2.718055 ... # # 8 # terms: # e~~ 1957/720 + 1/5040 #
# "           " =  685/252#
# "           " = 2.718253 ... #

Isabella Ackerman · Answer

The Euler-Mascheroni constant is defined as:

#gamma = lim_(n->oo) (sum_(k=1)^n 1/k -lnn)#

that is as the difference between the #n#-th partial sum of the harmonic series and #lnn#.

Since the limit is in indeterminate form we must prove it is convergent. Write #ln n# as:

#lnn = ln n - ln(n-1) + ln(n-1) -...-ln1 = sum_(k=1)^(n-1) (ln(k+1) -lnk )= sum_(k=1)^(n-1) ln (1+1/k)#

Then:

#gamma = sum_(k=1)^oo 1/k-ln(1+1/k)#

Now note that using Taylor's formula with Lagrange's rest and #0 < x < 1#:

#ln(1+x) = sum_(k=0)^oo (-1)^k x^(k+1)/(k+1) = x - xi^2/2# with #xi in (0,x)#

So:

#1/k-ln(1+1/k) = xi^2/2# with #xi in (0,1/k)#

and:

#0 < 1/k-ln(1+1/k) < 1/(2k^2) #

As the series:

#sum_(k=1)^oo 1/(2k^2)#

is convergent based on the #p#-series test, then by direct comparison also the series:

#gamma = sum_(k=1)^oo 1/k-ln(1+1/k)#

is convergent and #gamma# is a finite real number.

Incidentally this proves that the partial sums of the harmonic series diverge with the same order as #lnn#

Ezekiel Alexander · Answer

undefined Euler's number, denoted as $ e $, is calculated using the following infinite series:

$$ e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \cdots $$

Alternatively, it can be calculated using the limit of the expression $ \left(1 + \frac{1}{n}\right)^n $ as $ n $ approaches infinity.