Find the limit of the sequence of partial sums whose general term is a sub n equals 100 to the nth power over n factorial . (4 points)
A) 10
B) 1
C) Does not exist
D) 0
If Sn = ∑an, then recall that
e^k = 1 + k/1! + k^2/2! + ...
I don't see e^100 up there, so I guess maybe you mean
Sn = 100^n/n!
In that case, Sn=0, since factorials grow faster than powers.
To find the limit of the sequence of partial sums, we need to sum up the terms of the sequence as n approaches infinity.
The general term of the sequence is given by a sub n = (100^n) / n!.
To find the limit, we can rewrite the sequence of partial sums as follows:
S sub n = a sub 1 + a sub 2 + a sub 3 + ... + a sub n
Now, let's simplify each term of the sequence of partial sums:
a sub 1 = (100^1) / 1!
a sub 2 = (100^2) / 2!
a sub 3 = (100^3) / 3!
...
a sub n = (100^n) / n!
We can see that each term of the sequence involves a power of 100 divided by a factorial. As n approaches infinity, the power of 100 will dominate the factorial in the denominator. Since factorial grows much faster than exponential functions, the denominator will become larger at a much faster rate compared to the numerator. Hence, as n approaches infinity, each term of the sequence will tend to 0.
Therefore, the answer is D) 0.