Question

Test the series for convergence or divergene Σ infinite n=1 (-1) ^n 2^n/n!

Test the series for convergence or divergene Σ infinite n=1 (-1) ^n 2^n/n!

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
1. Test the series below for convergence using the Root Test. ∞∑n=1 (2n^2 / 9n+3)^n The...
1. Test the series below for convergence using the Root Test. ∞∑n=1 (2n^2 / 9n+3)^n The limit of the root test simplifies to limn→∞|f(n)|limn→∞|f(n)| where f(n)= 2. Test the series below for convergence using the Root Test. ∞∑n=1 (4n+4 / 5n+3)^n The limit of the root test simplifies to limn→∞|f(n)| where f(n)=   The limit is:
1. To test this series for convergence ∞∑n=1 n /√n^3+1 You could use the Limit Comparison...
1. To test this series for convergence ∞∑n=1 n /√n^3+1 You could use the Limit Comparison Test, comparing it to the series ∞∑n=1 1 /n^p where p= 2. Test the series below for convergence using the Ratio Test. ∞∑n=1 n^5/0.5^n The limit of the ratio test simplifies to lim n→∞|f(n)| where f(n)=    The limit is:
1. Test the series below for convergence using the Root Test. ∞∑n=1 (4n/10n+1)^n The limit of...
1. Test the series below for convergence using the Root Test. ∞∑n=1 (4n/10n+1)^n The limit of the root test simplifies to lim n→∞ |f(n)| where f(n)= The limit is:     Based on this, the series Diverges Converges 2. We want to use the Alternating Series Test to determine if the series: ∞∑k=4 (−1)^k+2 k^2/√k^5+3 converges or diverges. We can conclude that: The Alternating Series Test does not apply because the terms of the series do not alternate. The Alternating Series Test...
Test the series for convergence or divergence. ∞ en n2 n = 1 convergent or divergent    
Test the series for convergence or divergence. ∞ en n2 n = 1 convergent or divergent    
For the next two series, (1) find the interval of convergence and (2) study convergence at...
For the next two series, (1) find the interval of convergence and (2) study convergence at the end points of the interval if any. Also, (3) indicate for what values of x the series converges absolutely, conditionally, or not at all. You must indicate the test you use and show the interval of convergence both analytically and graphically and summarize your results on the picture. ∑∞ n=1 ((−1)^n−1)/ (n^1/4)) *x^n
Sigma(1, infinity) (2^n x^n)/(n!) : Radius of convergence of this series?
Sigma(1, infinity) (2^n x^n)/(n!) : Radius of convergence of this series?
Determine the convergence/divergence of the following series using the integral test: a.) ∑= (1)/n(In(n))^2 (Upper limit...
Determine the convergence/divergence of the following series using the integral test: a.) ∑= (1)/n(In(n))^2 (Upper limit of sigma is ∞ ,and the lower limit of sigma is n=2) b.) ∑ (n-4)/(n^2-2n+1) (Upper limit of sigma is ∞ and the lower limit of the sigma is n=2 c.)∑ (n)/(n^2+1) (Upper limit of sigma ∞ and the lower limit sigma is n=1) d.) ∑ e^-n^2 (Upper limit of sigma ∞ and the lower limit sigma is n=1)
find the radius of convergence and interval of convergence of the series ∑ n=1 ~ ∞...
find the radius of convergence and interval of convergence of the series ∑ n=1 ~ ∞ (3^n)((x+4)^n) / √n Please solve this problem with detailed process of solving.
Test the series for convergence using the Alternating Series Test: X∞ m=2 (−1)^m/ (m 2^m). If...
Test the series for convergence using the Alternating Series Test: X∞ m=2 (−1)^m/ (m 2^m). If convergent, determine whether this series converges absolutely or conditionally
Test the series for convergence or divergence. ∞ (−1)n 8n − 5 9n + 5 n...
Test the series for convergence or divergence. ∞ (−1)n 8n − 5 9n + 5 n = 1 Step 1 To decide whether ∞ (−1)n 8n − 5 9n + 5 n = 1 converges, we must find lim n → ∞ 8n − 5 9n + 5 . The highest power of n in the fraction is 1    1 . Step 2 Dividing numerator and denominator by n gives us lim n → ∞ 8n − 5 9n +...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT