Suppose we have: f(x) = Σ n=0 4n (n+1) xn n! |x| < ∞ Determine and simplify expressions for: (a) f(x)dx (b) f'(x) EC) Suppose we have 4" (n+1) X^ n 1x1 <0 n=0 n! f(x) = 2 4" (n+1) Determine and simplify expressions for: (f(x) dx Ratio Test an Lim 11700 ant tim n>x 4" (^+ ((n+1)+1) (h+i)! 4" (n+1) un n! 'X' n+l X 4+ ((n+1)+1) A! x Tim no∞ 4" 5+1 * (n+1) (n+1)/** 4 ((n+1)+1)/x/ Iim n->∞ (n+1)(n+1) Tim 4((1+1)+1) (n+1)2 1x Tim 15 4n+8 Cn+1) ① f(x).
Suppose we have: f(x) = Σ n=0 4n (n+1) xn n! |x| < ∞ Determine and simplify expressions for: (a) f(x)dx (b) f'(x) EC) Suppose we have 4" (n+1) X^ n 1x1 <0 n=0 n! f(x) = 2 4" (n+1) Determine and simplify expressions for: (f(x) dx Ratio Test an Lim 11700 ant tim n>x 4" (^+ ((n+1)+1) (h+i)! 4" (n+1) un n! 'X' n+l X 4+ ((n+1)+1) A! x Tim no∞ 4" 5+1 * (n+1) (n+1)/** 4 ((n+1)+1)/x/ Iim n->∞ (n+1)(n+1) Tim 4((1+1)+1) (n+1)2 1x Tim 15 4n+8 Cn+1) ① f(x).
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.4: Definition Of Function
Problem 51E
Question
Please check this for me
![Suppose we have:
f(x) = Σ
n=0
4n (n+1)
xn
n!
|x| < ∞
Determine and simplify expressions for:
(a) f(x)dx
(b) f'(x)
EC) Suppose
we have
4" (n+1) X^
n
1x1 <0
n=0
n!
f(x) = 2 4" (n+1)
Determine and simplify expressions for:
(f(x) dx
Ratio Test
an
Lim
11700
ant
tim
n>x
4"
(^+ ((n+1)+1)
(h+i)!
4" (n+1) un
n!
'X'
n+l
X
4+ ((n+1)+1) A! x
Tim
no∞
4"
5+1
* (n+1) (n+1)/**
4 ((n+1)+1)/x/
Iim
n->∞
(n+1)(n+1)
Tim
4((1+1)+1)
(n+1)2
1x
Tim
15
4n+8
Cn+1)
① f(x).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a0802a2-9dd1-43c0-9b3f-3c3accbadbd6%2F252a0330-e3d6-4908-988d-481ce3eb129b%2Fnojjdpi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Suppose we have:
f(x) = Σ
n=0
4n (n+1)
xn
n!
|x| < ∞
Determine and simplify expressions for:
(a) f(x)dx
(b) f'(x)
EC) Suppose
we have
4" (n+1) X^
n
1x1 <0
n=0
n!
f(x) = 2 4" (n+1)
Determine and simplify expressions for:
(f(x) dx
Ratio Test
an
Lim
11700
ant
tim
n>x
4"
(^+ ((n+1)+1)
(h+i)!
4" (n+1) un
n!
'X'
n+l
X
4+ ((n+1)+1) A! x
Tim
no∞
4"
5+1
* (n+1) (n+1)/**
4 ((n+1)+1)/x/
Iim
n->∞
(n+1)(n+1)
Tim
4((1+1)+1)
(n+1)2
1x
Tim
15
4n+8
Cn+1)
① f(x).
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