How do you use the direct comparison test to determine if #Sigma 1/(n!)# from #[0,oo)# is convergent or divergent?

Answer 1

#sum_(n=0) 1/(n!) = e#

We have:

#a_n = 1/(n!) = 1/(1*2*3*...*n) < 1/(2*2*...*2) = (1/2)^(n-1)#

The series:

#sum_(n=0)^oo (1/2)^(n-1)#
can be expressed as the sum of a geometric series of ratio #1/2# and as #1/2 < 1# such series is convergent:
#sum_(n=0)^oo (1/2)^(n-1)= 2 sum_(n=0)^oo (1/2)^n = 2/(1-1/2) =4#

By direct comparison we can then conclude that:

#sum_(n=0) 1/(n!)#
is also convergent, ant that its sum is less than #4#. In fact if we recall that the MacLaurin series of the exponential function is:
#e^x = sum_(n=0)^oo x^n/(n!)#

we can see that:

#sum_(n=0) 1/(n!) = e^1 = e#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To use the direct comparison test to determine if the series ( \sum \frac{1}{n!} ) from ( n = 0 ) to ( \infty ) is convergent or divergent, we compare it with a known convergent or divergent series.

Since ( n! ) grows faster than any exponential function, we can compare ( \frac{1}{n!} ) with ( \frac{1}{n^n} ).

Now, consider the series ( \sum \frac{1}{n^n} ). This series is known to converge by the ratio test or the root test.

So, for all ( n \geq 1 ), ( \frac{1}{n!} \leq \frac{1}{n^n} ).

Since ( \sum \frac{1}{n^n} ) converges, by the direct comparison test, ( \sum \frac{1}{n!} ) converges as well.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7