Question

any body can explain this step?
cauchy condensation test (rudins book)
Question is when n <2^k then why terms are a(2^k+1)
see the attachment

Answer 1

this is geometric series

Answer 2

its basically expanding the expression

Answer 3

\[a _{2^{k+1}-1}\]

Answer 4

is just the next term after \[a _{2^k}\]

Answer 5

t1=a1
t2=a2+a3
t3= a3+a4+a5
t4=a4+a5+a6+a7
1
1
1
1
t2k

Answer 6

Could you share the previous page of your attachment ?

Answer 7

@ziawasim

Answer 8

yes

Answer 9

Thank you :)

First notice that the given sequence \(\{a_n\}\) is "decreasing". Next lets try and "over estimate" the sum of the terms in this series.

Answer 10

@ayeshaafzal221 i want to know about when we are taking terms 2^k then how can we go for 2^k+1

Answer 11

Group the terms in the series as below :

\(a_1\)
\(a_2 + a_3\)
\(a_4+a_5+a_6+a_7\)
\(a_8+a_9+\cdots + a_{15}\)
\(a_{16}+a_{17} + \cdots + a_{31}\)
\(\vdots \)

Answer 12

Since the sequence \(\{a_n\}\) is decreasing, would you agree that we can over estimate the sums in each of the above groups as below :

\(a_1\)
\(2a_2\)
\(4a_4\)
\(8a_8\)
\(16a_{16}\)
\(\vdots\)

?

Answer 13

In other words, we have :

\(a_1\ge a_1\)
\(2a_2\ge a_2 + a_3\)
\(4a_4\ge a_4+a_5+a_6+a_7\)
\(8a_8\ge a_8+a_9+\cdots + a_{15}\)
\(16a_{16}\ge a_{16}+a_{17} + \cdots + a_{31}\)
\(\vdots \)

Answer 14

Let me know once you digest above. Rest of the proof is trivial..

Answer 15

yes both are same nature sequences
but i m want to know when n<2^k but but our terms are 2^k+1

Answer 16

I think this short video addresses your question
https://d3c33hcgiwev3.cloudfront.net/cauchy-condensation.9784ac54df87615f7bc738ba81d6e1d5/full/360p/index.mp4?Expires=1451347200&Signature=D5L63Tgr-~rHeQSD1KcUmccLlmiqjfDG-8sb0TteK0J7IoS38haEyd-1ijCEhRGOiX85HJr0nlw7GzU3vlsRZJzWZ5LeafolaU4aR9QqmZYfJ086vR1JZr8mhEZD-CaKapILNs6QcIayqijeGjsgSjGKau8aouF7DPLizvj0wRs_&Key-Pair-Id=APKAJLTNE6QMUY6HBC5A

Answer 17

let suppose k=4 then n will be less then \[n<2^k\]\[n<2^4\] <16
but we are taking terms 2^k+1 means \[2^5=32\] my question not about the grouping , my question numbers of term of the sequence
\[S_n= a_1+a_2+......+a_n<a_1+a_2+......+a_2^k\] but here
\[S_n a_1+a_2+......+a _{2^k})+...... +a _{2^k+1})\]

Answer 18

how S_n is equal to the terms a_{2^k+1}

Answer 19

I think I have got what you're asking... yeah the explanation in your notes is a bit confusing...
let me read it again and see..

Answer 20

your issue is with the number of terms in the last group right ?

Answer 21

yes this is my issue

Answer 22

Do below steps look fine to you ?

suppose \(n\lt 2^k\) :
\[S_n= a_1+a_2+a_3+a_4+\cdots +a_n\\~\\
=(a_1)+(a_2+a_3)+(a_4+\cdots + a_7) + \cdots + (a_{2^{k-1}} +\cdots + a_{2^k-1})\\~\\
\le a_1+2a_2+4a_4+\cdots + 2^{k-1}a_{2^{k-1}}\]

Answer 23

yes

Answer 24

suppose \(n\lt 2^k\) :
\[S_n= a_1+a_2+a_3+a_4+\cdots +a_n\\~\\
=(a_1)+(a_2+a_3)+(a_4+\cdots + a_7) + \cdots + (a_{2^{k-1}} +\cdots + a_{2^k-1})\\~\\
\le a_1+2a_2+4a_4+\cdots + 2^{k-1}a_{2^{k-1}}\\~\\
\le a_1+2a_2+4a_4+\cdots + 2^{k-1}a_{2^{k-1}}\color{red}{+2^ka_{2^k}}\\~\\
\]

Answer 25

that last line above still troubles you is it ?

Answer 26

The key idea here is that all the terms of given sequence are "positive"; therefore adding more terms only results in a greater sum.

Answer 27

*nonnegative

Answer 28

For example,

If your height is less than 6 feet,
then your height is also less than (6 + 1) feet

Answer 29

just this was the idea behind it oohhh

Answer 30

thank you

Answer 31

Yes :) yw !