Calculation: Consider the series E 3n +3 Since, by the nth-Term test for convergence series. If lima, 0 then a, convergence. Thus, for the series,) V3n? the ith-Term will be: Compute the limit, lima, as shown below: lim - lim - lim Now, since the limit nth-Term is * * 0. So, by nth-Term test for divergence, Hence, the series diverges. V3n +3
Calculation: Consider the series E 3n +3 Since, by the nth-Term test for convergence series. If lima, 0 then a, convergence. Thus, for the series,) V3n? the ith-Term will be: Compute the limit, lima, as shown below: lim - lim - lim Now, since the limit nth-Term is * * 0. So, by nth-Term test for divergence, Hence, the series diverges. V3n +3
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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I'm confused this problem. I dont understand how it is reduced after the computing the limits. I dont understand where the n comes from in the denominator

Transcribed Image Text:Convergence Table-1.pdf - Adobe Acrobat Pro DC (32-bit)
M Modul
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E CalcCH
P Reviev
b Re X
Sy sum fr
G nth te
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| Review In Exercises 69-80, determine the convergence or divergence of the series. En = 1 o n 3 n2 + 3
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Calculation:
SUMMARY OFTESTS FOR SERIES
Combine Files
Condition(s)
Condition(s)
Comment
Consider the series Y
Test
Series
of Convergence
of Divergence
Edit PDF
n=1 V3n2 + 3
This test cannot bc used
to show convergence.
lim a, + 0
FEATURED
nth-Term
A Request Signatures
Since, by the nth-Term test for convergence series. If lima, # 0 then
an convergence.
n=1
0 < |r| < 1
| 2 1
O Fill & Sign
Geometric Series
ar"
Sum: S=
Thus, for the series,
the nth-Term will be:
E Export PDF
/3n² +3
Telescoping Series
2 (b, - b+1)
lim b, = L
Sum: S = b, - L
n=1
Σ
0 < ps 1
EI Organize Pages
an =
p-Scries
p > 1
Van?43
Compute the limit, lima, as shown below:
Ž (- 1)-'a,
0 < a, +1 s a,.
Remainder:
Send for Comments
Alternating Series
and lim a, = 0
Integral
(fis continuous,
positive, and
decreasing)
lim = lim
Remainder:
Comment
3+2
fx) dx diverges 0 < Ry <
f(x) dx converges
a, = f(n) 2 0
Scan & OCR
= lim
lim la > I or
Test is inconclusive when
Root
lim a < 1
lim la= 1.
U Protect
E oc
Test is inconclusive when
> or
A More Tools
lim
Ratio
lim
< 1
lim
= 1.
= co
Now, since the limit nth-Term is + 0.
0 < a, s b,
0 < b, s a,
Direct Comparison3
(a, b, > 0)
and
b, converges
and
b, diverges
So, by nth-Term test for divergence,
lim
= L > 0
lim
= L >0
Limit Comparison
(a, b, > 0)
Hence, the series
diverges.
b, converges
and
and
b, d
n=1 V3n2 + 3
E Chapter 9.3, Problem 74E
Chapter 9.3, Problem 76E >
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