How do you determine whether the sequence #a_n=n!-10^n# converges, if so how do you find the limit?

Answer 1

the sequence #{a_n}# diverges

We have a sequence defined by:

# a_n = n! -10^n #
Our first observation is that for large #n# then #n!# grows much faster than any exponential so our intuition tells us that the sequence #{a_n}# diverges.
We can demonstrate this using Stirling's Approximation, which states that for large #n# then:
# n! ~ sqrt(2pin)(n/e)^n #
From which we get approximation for #a_n# given by:
# a_n = sqrt(2pin)(n/e)^n -10^n #

And clearly, we have for the dominant term, that:

# n^n " >> " 10^n #
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Answer 2

To determine convergence of the sequence (a_n = n! - 10^n), consider the behavior of the terms as (n) approaches infinity. Notice that (10^n) grows much faster than (n!), meaning (10^n) dominates the growth of the sequence. Hence, (a_n) approaches negative infinity as (n) tends to infinity. Therefore, the sequence diverges.

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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.

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