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Math  /  Calculus

QuestionDetermine the Convergence or divergence of these series: (1) ∑n=0∞(3+i)2n(2n)!\sum_{n=0}^{\infty} \frac{(3+i)^{2 n}}{(2 n)!} (2) ∑n=0∞n(i2)n\sum_{n=0}^{\infty} n\left(\frac{i}{2}\right)^{n} (3) ∑n=1∞in2n\sum_{n=1}^{\infty} \frac{i^{n}}{2^{n}}

Studdy Solution

STEP 1

1. We are dealing with infinite series, and we need to determine whether each series converges or diverges.
2. We will use standard tests for convergence such as the Ratio Test, Root Test, and properties of geometric series.
3. Complex numbers are involved, so we may need to consider their magnitudes.

STEP 2

1. Analyze the series ∑n=0∞(3+i)2n(2n)!\sum_{n=0}^{\infty} \frac{(3+i)^{2 n}}{(2 n)!} using the Ratio Test.
2. Analyze the series ∑n=0∞n(i2)n\sum_{n=0}^{\infty} n\left(\frac{i}{2}\right)^{n} using the Ratio Test.
3. Analyze the series ∑n=1∞in2n\sum_{n=1}^{\infty} \frac{i^{n}}{2^{n}} as a geometric series.

STEP 3

For the series ∑n=0∞(3+i)2n(2n)!\sum_{n=0}^{\infty} \frac{(3+i)^{2 n}}{(2 n)!}, apply the Ratio Test:
Consider the general term an=(3+i)2n(2n)! a_n = \frac{(3+i)^{2n}}{(2n)!} .
Compute the ratio:
∣an+1an∣=∣(3+i)2(n+1)(2(n+1))!⋅(2n)!(3+i)2n∣\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(3+i)^{2(n+1)}}{(2(n+1))!} \cdot \frac{(2n)!}{(3+i)^{2n}} \right|
Simplify the expression:
=∣(3+i)2(2n+2)(2n+1)∣= \left| \frac{(3+i)^2}{(2n+2)(2n+1)} \right|
Calculate (3+i)2(3+i)^2:
(3+i)2=9+6i+i2=9+6i−1=8+6i(3+i)^2 = 9 + 6i + i^2 = 9 + 6i - 1 = 8 + 6i
Find the magnitude:
∣8+6i∣=82+62=64+36=100=10|8 + 6i| = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10
Thus, the ratio becomes:
∣10(2n+2)(2n+1)∣\left| \frac{10}{(2n+2)(2n+1)} \right|
As n→∞ n \to \infty , the ratio approaches 0. Since the limit is less than 1, the series converges by the Ratio Test.

STEP 4

For the series ∑n=0∞n(i2)n\sum_{n=0}^{\infty} n\left(\frac{i}{2}\right)^{n}, apply the Ratio Test:
Consider the general term bn=n(i2)n b_n = n\left(\frac{i}{2}\right)^n .
Compute the ratio:
∣bn+1bn∣=∣(n+1)(i2)n+1n(i2)n∣\left| \frac{b_{n+1}}{b_n} \right| = \left| \frac{(n+1)\left(\frac{i}{2}\right)^{n+1}}{n\left(\frac{i}{2}\right)^n} \right|
Simplify the expression:
=∣n+1n⋅i2∣=∣n+1n∣⋅∣i2∣= \left| \frac{n+1}{n} \cdot \frac{i}{2} \right| = \left| \frac{n+1}{n} \right| \cdot \left| \frac{i}{2} \right|
=(1+1n)⋅12= \left(1 + \frac{1}{n}\right) \cdot \frac{1}{2}
As n→∞ n \to \infty , the ratio approaches 12\frac{1}{2}. Since the limit is less than 1, the series converges by the Ratio Test.

STEP 5

For the series ∑n=1∞in2n\sum_{n=1}^{\infty} \frac{i^{n}}{2^{n}}, recognize it as a geometric series:
The general term is cn=(i2)n c_n = \left(\frac{i}{2}\right)^n .
The series is geometric with common ratio r=i2 r = \frac{i}{2} .
Calculate the magnitude of the common ratio:
∣r∣=∣i2∣=12|r| = \left| \frac{i}{2} \right| = \frac{1}{2}
Since ∣r∣<1|r| < 1, the series converges.
The series (1), (2), and (3) all converge.

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