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Real Analysis
book

Real Analysis

by V. Karunakaran
May 2024
Intermediate to advanced content levelIntermediate to advanced
585 pages
15h 38m
English
Pearson India
Content preview from Real Analysis
“real: chapter_13” 2011/5/22 23:35 page 26 #26
13-26 Real Analysis
The inequality
n f (x)<
j
2
n+1
implies 2
n+1
n < j or that 2
n+1
n j 1, which is equivalent to
s
n
(x) s
n+1
(x). Since x is arbitrary the proof of (i) is complete.
Proof of (ii): We can assume f (x)< and proceed (see the obser-
vation in case 1 above). Choose an integer m such that f (x)<m. For
n m, s
n
(x) =
i1
2
n
for a unique integer i with
i1
2
n
f (x)<
i
2
n
.It
now follows that s
n
(x) f (x)<s
n
(x) +
1
2
n
or that
|s
n
(x) f (x)|=f (x) s
n
(x)<
1
2
n
0
as n →∞. Hence s
n
(x) f (x) as n →∞pointwise.
Proof of (iii): If f (x)<M (x
R), then the above argument is valid
for all x if n M (note ...
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Publisher Resources

ISBN: 9781299447561Publisher Website