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Math  /  Calculus

QuestionWhich of the following series converge?
1. ∑n=1∞n(1e)n\sum_{n=1}^{\infty} n\left(\frac{1}{e}\right)^{n} II. ∑n=1∞n!100n\sum_{n=1}^{\infty} \frac{n!}{100^{n}} III. ∑n=1∞(n!)2(2n)!\sum_{n=1}^{\infty} \frac{(n!)^{2}}{(2 n)!} (A) Ionly
C land llonly (D) I and ili only

Studdy Solution

STEP 1

1. We are given three series and need to determine which ones converge.
2. A series converges if the sequence of partial sums converges to a finite limit.
3. We will use various convergence tests such as the ratio test, root test, and comparison test.

STEP 2

1. Analyze series I: ∑n=1∞n(1e)n\sum_{n=1}^{\infty} n\left(\frac{1}{e}\right)^{n}.
2. Analyze series II: ∑n=1∞n!100n\sum_{n=1}^{\infty} \frac{n!}{100^{n}}.
3. Analyze series III: ∑n=1∞(n!)2(2n)!\sum_{n=1}^{\infty} \frac{(n!)^{2}}{(2 n)!}.
4. Determine which series converge.

STEP 3

Analyze series I: ∑n=1∞n(1e)n\sum_{n=1}^{\infty} n\left(\frac{1}{e}\right)^{n}.
We will use the ratio test, which states that a series ∑an\sum a_n converges if:
lim⁡n→∞∣an+1an∣<1\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1
For this series, an=n(1e)na_n = n\left(\frac{1}{e}\right)^{n}.
Calculate the ratio:
an+1an=(n+1)(1e)n+1n(1e)n=n+1n⋅1e\frac{a_{n+1}}{a_n} = \frac{(n+1)\left(\frac{1}{e}\right)^{n+1}}{n\left(\frac{1}{e}\right)^{n}} = \frac{n+1}{n} \cdot \frac{1}{e}
Simplify:
=(1+1n)⋅1e= \left(1 + \frac{1}{n}\right) \cdot \frac{1}{e}
Take the limit as n→∞n \to \infty:
lim⁡n→∞(1+1n)⋅1e=1e<1\lim_{n \to \infty} \left(1 + \frac{1}{n}\right) \cdot \frac{1}{e} = \frac{1}{e} < 1
Since the limit is less than 1, the series converges by the ratio test.

STEP 4

Analyze series II: ∑n=1∞n!100n\sum_{n=1}^{\infty} \frac{n!}{100^{n}}.
Use the ratio test again. Let an=n!100na_n = \frac{n!}{100^n}.
Calculate the ratio:
an+1an=(n+1)!100n+1⋅100nn!=n+1100\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{100^{n+1}} \cdot \frac{100^n}{n!} = \frac{n+1}{100}
Take the limit as n→∞n \to \infty:
lim⁡n→∞n+1100=∞\lim_{n \to \infty} \frac{n+1}{100} = \infty
Since the limit is greater than 1, the series diverges by the ratio test.

STEP 5

Analyze series III: ∑n=1∞(n!)2(2n)!\sum_{n=1}^{\infty} \frac{(n!)^{2}}{(2 n)!}.
Use the ratio test. Let an=(n!)2(2n)!a_n = \frac{(n!)^2}{(2n)!}.
Calculate the ratio:
an+1an=((n+1)!)2(2(n+1))!⋅(2n)!(n!)2\frac{a_{n+1}}{a_n} = \frac{((n+1)!)^2}{(2(n+1))!} \cdot \frac{(2n)!}{(n!)^2}
Simplify:
=(n+1)2(2n+2)(2n+1)= \frac{(n+1)^2}{(2n+2)(2n+1)}
Take the limit as n→∞n \to \infty:
lim⁡n→∞(n+1)2(2n+2)(2n+1)=lim⁡n→∞n2+2n+14n2+6n+2=14\lim_{n \to \infty} \frac{(n+1)^2}{(2n+2)(2n+1)} = \lim_{n \to \infty} \frac{n^2 + 2n + 1}{4n^2 + 6n + 2} = \frac{1}{4}
Since the limit is less than 1, the series converges by the ratio test.

STEP 6

Determine which series converge:
- Series I converges. - Series II diverges. - Series III converges.
Therefore, the convergent series are I and III.
The correct answer is D: I and III only \boxed{\text{D: I and III only}} .

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