Question #a67d3

1 Answer
sente
May 19, 2016

#lim_(x->oo)1/(n!) = 1/(n!)#

#sum_(n=0)^oo1/(n!) = e = 2.71828...#

Explanation:

There are a couple of issues to address, here. First to answer the exact question asked:

#lim_(x->oo)1/(n!) = 1/(n!)#

As there is no #x# term in #1/(n!)#, a limit involving #x# does not change the value.

From the additional information, however, the intended question seems to be:
"What is the limit as #n# approaches infinity of #sum_(k=0)^n1/(k!)#?"

This is actually one way of calculating the well-known irrational constant #e=2.71828...#
In fact, the power series representation of the function #e^x# is given by

#e^x = sum_(k=0)^oox^k/(k!)#

Substituting #x=1# into the above sum gives us

#e = e^1 = sum_(k=0)^oo1^k/(k!) = sum_(k=0)^oo1/(k!)#

matching the above claim.

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