Answer :
Answer:
a) The Series (n=1→[infinity] Σ [(-3)ⁿ n²] / n!, absolutely converges!
b) The Series (n=1→[infinity] Σ [(-2)²ⁿ n²] / nⁿ, also, absolutely converges!
Step-by-step explanation:
I'll be using the ratio test.
The ratio test involves taking a ratio of (n+1)th to nth term as n --> ∞
Let the absolute value of this ratio as n tends to infinity be L.
The interpretation of the results of the ratio test:
If the absolute value of the limit of this ratio is less than 1, then the series absolutely converges. That is,
L < 1, the series absolutely converges
If the absolute value of the limit of this ratio, as n tends to infinity, is greater than 1 or tends to infinity, then, the series diverges. That is,
L > 1 or L --> ∞, the series diverges.
If the absolute value of the limit of this ratio, as n tends to infinity, is equal to 1, the test is inconclusive.
L = 1, the test is inconclusive
For this question.
(a) (n=1→[infinity] Σ [(-3)ⁿ n²] / n!
The nth term of the series is given as
[(-3)ⁿ n²] / n!
The (n+1)th term is expressed by replacing n with (n+1). The (n+1)th term is
(-3)ⁿ⁺¹ (n+1)² / (n+1)!
The ratio (n+1)th term divided by the nth term is given by
[(-3)ⁿ⁺¹ (n+1)²/(n+1)!] ÷ [(-3)ⁿ(n²)/n!]
This is evaluated on the first page of the attached image to this solution.
On evaluating the limit of the ratio as n tends to infinity, L = 0
L = 0 < 1, from the list of the interpretations of the ratio test, it is evident that this series absolutely converges.
(b) (n=1→[infinity] Σ [(-2)²ⁿ n²] / nⁿ
The nth term of this series is
[(-2)²ⁿ n²] / nⁿ
The (n+1)th term of this series is
[(-2)²ⁿ⁺² (n+1)²]/(n+1)ⁿ⁺¹
The limit of the ratio, as n tends to infinity, is evaluated on the second and third pages of the attached image.
On evaluating the limit of the ratio as n tends to infinity, L = 0
L = 0 < 1, from the list of the interpretations of the ratio test, it is evident that this series absolutely converges too!
Hope this Helps!!!
Answer:
a) converges absolutely
b) converges absolutely
Step-by-step explanation:
Solution:-
(a) (n=1→[infinity] Σ [(-3)ⁿ n²] / n!
- The given series can be inspected to be an alternating series. With the general form:
Σ [(-3)ⁿ n²] / n! = -3 + 9*(4) / 2 - 27(9)/6 + ..... - [(-3)ⁿ n²] / n!
- We see that each term has an oscillating sign ( + and - ).
- First we will check for convergence.
From definition we have that a series (n=1→[infinity] Σ an exhibits "absolute convergence" if (n=1→[infinity] Σ |an| converges.
(n=1→[infinity] Σ | (-3)ⁿ n²] / n! | = (n=1→[infinity] Σ | (3)ⁿ n²] / n! |
- We will use Ratio Test to test absolute convergence of (n=1→[infinity] Σ |an| :
[tex]\lim_{n \to \infty} | \frac{a_n_+_1}{a_n} |\\\\\lim_{n \to \infty} | \frac{(3)^n^+^1 (n+1)^2 / ( n + 1 )!}{(3)^n (n)^2 / ( n )!} | \\\\\lim_{n \to \infty} | \frac{(3) (n+1)^2 }{(n)^2 ( n + 1 )} | = \lim_{n \to \infty} | \frac{(3) (n+1) }{(n)^2 } | = 0[/tex]
- From results of (Ratio Test) we have that :
[tex]\lim_{n \to \infty} | \frac{a_n_+_1}{a_n} | = p\\\\[/tex]
- If ( 0 ≤ p < 1 ) , then (n=1→[infinity] Σ |an| converges.
- Hence, from definition if (n=1→[infinity] Σ |an| converges then it implies "absolute convergence" for (n=1→[infinity] Σ an.
b) (n=1→[infinity] Σ [(-2)²ⁿ n²] / nⁿ
- We can re-write the given series as:
(n=1→[infinity] Σ [(-2)²ⁿ n²] / nⁿ = (n=1→[infinity] Σ [(2)²ⁿ n²] / nⁿ
- From definition we have that a series (n=1→[infinity] Σ an exhibits "absolute convergence" if (n=1→[infinity] Σ |an| converges.
(n=1→[infinity] Σ | [(2)²ⁿ n²] / nⁿ |
- We will use Ratio Test to test absolute convergence of (n=1→[infinity] Σ |an| :
[tex]\lim_{n \to \infty} | \frac{a_n_+_1}{a_n} |\\\\\lim_{n \to \infty} | \frac{(2)^2^n^+^2 (n+1)^2 / ( n + 1 )^n^+^1}{(2)^2^n (n)^2 / n^n } | \\\\\lim_{n \to \infty} | \frac{(4) (n+1)^2*(n)^n }{(n)^2 ( n + 1 )*(n+1)^n} | = lim_{n \to \infty} | \frac{(4) (n+1)*(n)^n }{(n)^2 (n+1)^n} | \\\\= 4* \lim_{n \to \infty} |\frac{(n+1)}{(n)^2} |* \lim_{n \to \infty} | ( \frac{n}{n+1})^n| \\\\ = 4* 0 * \lim_{n \to \infty} | ( \frac{1}{1+1/n})^n| = 4* 0 * 1 = 0[/tex]
- From results of (Ratio Test) we have that :
[tex]\lim_{n \to \infty} | \frac{a_n_+_1}{a_n} | = p\\\\[/tex]
- If ( 0 ≤ p < 1 ) , then (n=1→[infinity] Σ |an| converges.
- Hence, from definition if (n=1→[infinity] Σ |an| converges then it implies "absolute convergence" for (n=1→[infinity] Σ an.