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VE> 0 3Ng s.t. Vk?Ng I (-1* (2k) (2k )+1 Proot:0 Given E> 0 Choose NE= (-1)? -||< E any O For any k> Ng Solve u for k Justify alown wards Finish this Classwork! (final -1|< E 14(서2)

VE> 0 3Ng s.t. Vk?Ng I (-1* (2k) (2k )+1 Proot:0 Given E> 0 Choose NE= (-1)? -||< E any O For any k> Ng Solve u for k Justify alown wards Finish this Classwork! (final -1|< E 14(서2)

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.2: Arithmetic Sequences
Problem 52E
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Please solve homework question in the following format like in two columns: 

 

•statment          |             •justifications 

 

Note: My previous homework was cancelled due to not following format.

 

i want to give you thumbs up. So please follow attached format

VE> O 3 Ng s.t. Vk>Ng I- (-1)?k (zk) 3, -||< E Proof:O Given E>o Choose NE=I 1+(서2) any O For any kzNg 'solve up for k Justify down wards Fnish tais Classword.' 1+(서2) Part 1 HW3 finish classwork above final) (-1)** (zk) - ||< E
Transcribed Image Text:VE> O 3 Ng s.t. Vk>Ng I- (-1)?k (zk) 3, -||< E Proof:O Given E>o Choose NE=I 1+(서2) any O For any kzNg 'solve up for k Justify down wards Fnish tais Classword.' 1+(서2) Part 1 HW3 finish classwork above final) (-1)** (zk) - ||< E
Please solve my homework question like this (statement and justifications) columns 1) Given any ɛ > 0, choose N, = = +1 1) OK because ɛ > 0 %3D 2) For any n > N, we have <n 2) by a+1 > a 3) 8 – E < En 3) no switch when mult by ɛ > 0 4) 8 < en + ɛ 4) adding ɛ to both sides 5) 8 < e(n+ 1) 5) algebra (factoring) 6) <E 6) no switch when mult by 1/(n+1) > 0 because n+1 > 0. n+1 7) E 7) defn of abs value 8) - < e Sn+8 8) algebra n+1 n+1 9) - < 8(n+1) n+1 8n 9) algebra n+1 - 8| < e. 10) algebra QED n+1
Transcribed Image Text:Please solve my homework question like this (statement and justifications) columns 1) Given any ɛ > 0, choose N, = = +1 1) OK because ɛ > 0 %3D 2) For any n > N, we have <n 2) by a+1 > a 3) 8 – E < En 3) no switch when mult by ɛ > 0 4) 8 < en + ɛ 4) adding ɛ to both sides 5) 8 < e(n+ 1) 5) algebra (factoring) 6) <E 6) no switch when mult by 1/(n+1) > 0 because n+1 > 0. n+1 7) E 7) defn of abs value 8) - < e Sn+8 8) algebra n+1 n+1 9) - < 8(n+1) n+1 8n 9) algebra n+1 - 8| < e. 10) algebra QED n+1
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