Question

In each problem, determine if the series converges absolutely, converges conditionally, or diverges. Show work and state explicitly which test or tests you are using (Ratio, Root, Alternating Series, etc.).

(a) (n=1→[infinity] Σ [(-3)ⁿ n²] / n!
(b) (n=1→[infinity] Σ [(-2)²ⁿ n²] / nⁿ

Answer 1

a) converges absolutely

b) converges absolutely

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| :

`\lim_(n \to \infty) | (a_n_+_1)/(a_n) |\\\\\lim_(n \to \infty) | ((3)^n^+^1 (n+1)^2 / ( n + 1 )!)/((3)^n (n)^2 / ( n )!) | \\\\\lim_(n \to \infty) | ((3) (n+1)^2 )/((n)^2 ( n + 1 )) | = \lim_(n \to \infty) | ((3) (n+1) )/((n)^2 ) | = 0`

- From results of (Ratio Test) we have that :

`\lim_(n \to \infty) | (a_n_+_1)/(a_n) | = p\\\\`

• 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| :

`\lim_(n \to \infty) | (a_n_+_1)/(a_n) |\\\\\lim_(n \to \infty) | ((2)^2^n^+^2 (n+1)^2 / ( n + 1 )^n^+^1)/((2)^2^n (n)^2 / n^n ) | \\\\\lim_(n \to \infty) | ((4) (n+1)^2*(n)^n )/((n)^2 ( n + 1 )*(n+1)^n) | = lim_(n \to \infty) | ((4) (n+1)*(n)^n )/((n)^2 (n+1)^n) | \\\\= 4* \lim_(n \to \infty) |((n+1))/((n)^2) |* \lim_(n \to \infty) | ( (n)/(n+1))^n| \\\\ = 4* 0 * \lim_(n \to \infty) | ( (1)/(1+1/n))^n| = 4* 0 * 1 = 0`

- From results of (Ratio Test) we have that :

`\lim_(n \to \infty) | (a_n_+_1)/(a_n) | = p\\\\`

• 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.

Answer 2

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!

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!!!