determine whether the infinite series converges or diverges? give a reason for your answer
A) the sum of (4/3)^n for n=0 and (Above the sum theres an infinty im not sure how to phrase that)
B) the sum of (n^n/n!) for n=0 and (Above the sum theres an infinty im not sure how to phrase that)
A) the sum of (4/3)^n for n=0 and (Above the sum theres an infinty im not sure how to phrase that)
B) the sum of (n^n/n!) for n=0 and (Above the sum theres an infinty im not sure how to phrase that)
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∞
∑ (4/3)ⁿ
n=0
This is a geometric series with a radius of convergence (4/3) > 1, thus it diverges. For the first few terms of n, you'll be adding
1 + (4/3) + (16/9) + (64/27) + ...
Notice that these numbers continue to grow, meaning that your partial sums will grow indefinitely as n goes to infinity.
——————————————————————
∞
∑ (nⁿ / n!)
n=0
This series begs for the ratio test. an = (nⁿ / n!) while an+1 = { (n+1)^n+1 } / { (n+1)! }
lim n→∞ | an+1 / an |
lim n→∞ [ { (n+1)^n+1 } / { (n+1)! } ]*[ n! / nⁿ ]
lim n→∞ (n+1)ⁿ / nⁿ
lim n→∞ ( n+1 / n )ⁿ
lim n→∞ ( 1 + 1/n )ⁿ
= e > 1, so the series diverges.
I apologize for omitting the ratio simplification, but it gets very messy-looking with Yahoo's basic typing interface.
∑ (4/3)ⁿ
n=0
This is a geometric series with a radius of convergence (4/3) > 1, thus it diverges. For the first few terms of n, you'll be adding
1 + (4/3) + (16/9) + (64/27) + ...
Notice that these numbers continue to grow, meaning that your partial sums will grow indefinitely as n goes to infinity.
——————————————————————
∞
∑ (nⁿ / n!)
n=0
This series begs for the ratio test. an = (nⁿ / n!) while an+1 = { (n+1)^n+1 } / { (n+1)! }
lim n→∞ | an+1 / an |
lim n→∞ [ { (n+1)^n+1 } / { (n+1)! } ]*[ n! / nⁿ ]
lim n→∞ (n+1)ⁿ / nⁿ
lim n→∞ ( n+1 / n )ⁿ
lim n→∞ ( 1 + 1/n )ⁿ
= e > 1, so the series diverges.
I apologize for omitting the ratio simplification, but it gets very messy-looking with Yahoo's basic typing interface.