Question

Find Sum of power series ,then answer the questions

series from 2 to infinity of [ 3 ^(n+1) . {z^n+2)} ] indicate the convergence nhd (neighborhood).

is series from 2 to infinity of [ 3 ^ {(3n+4)/2} /n!) convergent if yes compute its sum

Answer 1

Answer on Question #49960 – Math – Complex Analysis

1) Given:

$$\sum_{n=1}^{\infty} z^{n+2} 3^{n+1}$$

Solution:

$$\sum_{n=1}^{\infty} z^{n+2} 3^{n+1} = \frac{1}{3} \sum_{n=1}^{\infty} z^{n+2} 3^{n+2} = \frac{1}{3} \sum_{n=3}^{\infty} z^n 3^n$$ $$c_n = 3^n$$

$R$ is a radius of convergence

$U_R$ is the convergence neighborhood

R = \frac{1}{\lim_{n \to \infty} \sqrt[n]{|3^n|} = \frac{1}{\lim_{n \to \infty} \sqrt[n]{3^n}} = \frac{1}{3}

Then the convergence neighborhood is

$$U_R = \left\{ z : |z| < \frac{1}{3} \right\}$$

Answer:

$$R = \frac{1}{3}$$ $$U_R = \left\{ z : |z| < \frac{1}{3} \right\}$$

2) Given:

$$\sum_{n=2}^{\infty} \frac{3^{\frac{3n+4}{2}}}{n!}$$

Solution:

$$a_n = \frac{9 \cdot 3^{\frac{3n}{2}}}{n!} \qquad \frac{a_{n+1}}{a_n} = \frac{9 \cdot 3^{\frac{3n+3}{2}}}{(n+1) \cdot n!} \cdot \frac{n!}{9 \cdot 3^{\frac{3n}{2}}} = \frac{\sqrt{27}}{n+1} < 1 \quad \text{when } n \to \infty, \text{ so the series is convergent.}$$

The sum of series is

$$\sum_{n=2}^{\infty} \frac{3^{\frac{3n+4}{2}}}{n!} = 3^2 \sum_{n=2}^{\infty} \frac{(3^{3/2})^n}{n!} = 9 \sum_{n=2}^{\infty} \frac{(\sqrt{27})^n}{n!} = 9 \left( \sum_{n=0}^{\infty} \frac{(\sqrt{27})^n}{n!} - 1 - \sqrt{27} \right) = 9 \left( e^{\sqrt{27}} - 1 - \sqrt{27} \right).$$

We used Maclaurin series of function $e^i$ .

Answer: $S = 9\left(e^{\sqrt{27}} - 1 - \sqrt{27}\right)$

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