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PROBLEM 5 (Attach additional pages as needed for detailed work) Fill in the blank cells. Justify your results in detail. Infinite Series Radius of Convergence Center of Convergence $\sum_{n=0}^{\infty} \frac{n!(2n)!(n+1)!}{9^n(4n)!(n+2)!}(z-2+i2)^n$ $\sum_{n=0}^{\infty} (\frac{z-2}{2-i})^{2n}$ $\sum_{n=0}^{\infty} \frac{(4n+1)!n!}{10^n \times (2n)!(2n)^{3n}} z^n$ Hints $(n)! = n(n-1)(n-2) \times ... \times 1.$ $(2n)! = 2n(2n-1)(2n-2) \times ... \times 1.$ $(n+2)! = (n+2)(n+1)(n)!$ $(2(n+1))! = (2n+2)! = (2n+2)(2n+1)(2n)!$ Let $B = (\frac{m+1}{m})^{4m}$ then $B_m = (1+\frac{1}{m})^{4m}$ and $\lim_{m \to \infty} B = e^4$

Name: problem 5 attach additional pages as needed for detailed work fill in the blank cells justify your results in detail infinite series radius of convergence center of convergence sumn0infty f 03725
Uploaded: 2023-11-27T22:13:33-08:00
Duration: 5 min 12 s
Channel: Adi S
Description: problem 5 attach additional pages as needed for detailed work fill in the blank cells justify your results in detail infinite series radius of convergence center of convergence sumn0infty f 03725

          PROBLEM 5 (Attach additional pages as needed for detailed work)
Fill in the blank cells. Justify your results in detail.
Infinite Series
Radius of
Convergence
Center of
Convergence
$\sum_{n=0}^{\infty} \frac{n!(2n)!(n+1)!}{9^n(4n)!(n+2)!}(z-2+i2)^n$
$\sum_{n=0}^{\infty} (\frac{z-2}{2-i})^{2n}$
$\sum_{n=0}^{\infty} \frac{(4n+1)!n!}{10^n \times (2n)!(2n)^{3n}} z^n$
Hints
$(n)! = n(n-1)(n-2) \times ... \times 1.$
$(2n)! = 2n(2n-1)(2n-2) \times ... \times 1.$
$(n+2)! = (n+2)(n+1)(n)!$
$(2(n+1))! = (2n+2)! = (2n+2)(2n+1)(2n)!$
Let $B = (\frac{m+1}{m})^{4m}$ then $B_m = (1+\frac{1}{m})^{4m}$ and $\lim_{m \to \infty} B = e^4$

problem 5 attach additional pages as needed for detailed work fill in the blank cells justify your results in detail infinite series radius of convergence center of convergence sumn0infty f 03725

Added by Cheryl M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

11/27/2023

Step 1

We'll use the ratio test to determine the radius of convergence and the center is directly observable from the series. Show more…

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PROBLEM 5 (Attach additional pages as needed for detailed work) Fill in the blank cells. Justify your results in detail. Infinite Series Radius of Convergence Center of Convergence $\sum_{n=0}^{\infty} \frac{n!(2n)!(n+1)!}{9^n(4n)!(n+2)!}(z-2+i2)^n$ $\sum_{n=0}^{\infty} (\frac{z-2}{2-i})^{2n}$ $\sum_{n=0}^{\infty} \frac{(4n+1)!n!}{10^n \times (2n)!(2n)^{3n}} z^n$ Hints $(n)! = n(n-1)(n-2) \times ... \times 1.$ $(2n)! = 2n(2n-1)(2n-2) \times ... \times 1.$ $(n+2)! = (n+2)(n+1)(n)!$ $(2(n+1))! = (2n+2)! = (2n+2)(2n+1)(2n)!$ Let $B = (\frac{m+1}{m})^{4m}$ then $B_m = (1+\frac{1}{m})^{4m}$ and $\lim_{m \to \infty} B = e^4$