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

We have:

The series:

By direct comparison we can then conclude that:

we can see that:

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

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.

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

- How do you use the limit comparison test for #sum ((sqrt(n+1)) / (n^2 + 1))# as n goes to infinity?
- How do you apply the ratio test to determine if #Sigma (4^n(n!)^2)/((2n)!)# from #n=[1,oo)# is convergent to divergent?
- Find the limit of the sequence an=2^n/(2n-1)?
- How do you determine if the series the converges conditionally, absolutely or diverges given #Sigma (-1)^(n+1)/(n+1)^2# from #[1,oo)#?
- How do you show whether the improper integral #int lim (lnx) / x dx# converges or diverges from 1 to infinity?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7