Question

HELP!!
http://prntscr.com/4f5b3i

Answer 1

i have no idea where to start this >.<

Answer 2

d is the devisors of n and \( \sigma (n)\) is sum of devisors function ?

Answer 3

yes

Answer 4

this explains the notation clearly :
http://prntscr.com/4f5efd

Answer 5

ok , start from here
\(\sum_{d|n} 1/d=1/d_1+1/d_2+1/d_3 +..+1/d_k\)
idk if thiss leads any where
\( 1/d_1+1/d_2+1/d_3 +..+1/d_k \\ = d_2d_3..dk+d_1d_3..dk+...+d_1d_2..dk+d_2d_3..d_{k}/ d_1d_2d_3 ...d_k \)

Answer 6

Consider how \[
n\times \frac 1d_i = \frac nd_i = d_i'
\]Where \(d_i'd_i = n\)

Answer 7

oh

Answer 8

\[\large n = d_i * \dfrac{n}{d_i}\]

Answer 9

I'm guessing \[
d_i' = d_{k - i} \quad k = \tau(n)
\]When you order the divisors.

Answer 10

cool ! it works now

Answer 11

we only need to multiply by n/n >.<

Answer 12

Yeah, multiply by \(n/n\) seems to work.

Answer 13

The only subtle part is to show that there is a one to one correspondence to divisors and quotient.

Answer 14

you want to compute ? \[ \dfrac{1}{n}\sum \limits_{d|n} n/d\]

Answer 15

@rational \[
f(d) = \frac nd
\]In this case, \(f\) is a one to one function form the set of divisors to the set of divisors.

Answer 16

\[\dfrac{1}{n}\sum \limits_{d|n} n/d = \dfrac{1}{n}\sum \limits_{d|n} d = \dfrac{\sigma(n)}{n} \]


sweet :3

Answer 17

it is sweet :o
i almost forgot how these funtions are awesome

Answer 18

yes, as \(d\) ranges over all the divisors of \(n\), so does \(\dfrac{n}{d}\),
so both sums evaluate to same nujmber

Answer 19

Or we could have simply substituted the dummy variable \(d = \dfrac{n}{d}\) to make it look more plausible :

\[\large \sum \limits_{d|n} 1/d = \sum \limits_{\frac{n}{d}|n} d/n = \dfrac{1}{n} \sum \limits_{\frac{n}{d}|n} d = \dfrac{\sigma(n)}{n}\]

Answer 20

both work

Answer 21

:o

Answer 22

thank you both :)

Answer 23

thank you for refreshing >.<

Answer 24

http://prntscr.com/4f5pyo

i have worked 9th question, need help with 10th question

Answer 25

any ideas on how to start #10a ?

Answer 26

there is a function u(n) mmm we could use

Answer 27

brb , i think i worked this before

Answer 28

ok

Answer 29

ok hope this make since
\(\large \frac{\sigma(n)}{n}=\sum_{d|n} 1/d >1 \\\large \therefore \frac{n}{\sigma(n)}<1 \)


next :-
\( \large \frac{n}{\sigma(n)}= n \prod_{i=1}^r \frac{p_i-1}{p_i^{k_i+1}-1} \)

Answer 30

yes <1 is clear

Answer 31

\( \large \frac{n}{\sigma(n)}= n \prod_{i=1}^r \frac{p_i-1}{p_i^{k_i+1}-1}\\\large = (p_1^{k_{1}}.p_2^{k_{2}}...p_r^{k_{r}})\prod_{i=1}^r \frac{p_i-1}{p_i^{k_i+1}-1} \\\large = \prod_{i=1}^r \frac{p_i^{k_i}(p_i-1)}{p_i^{k_i+1}-1} \)

Answer 32

@ganeshie8 wow !
finally ! >.<

Answer 33

\( \large \prod_{i=1}^r \frac{p_i^{k_i}(p_i-1)}{p_i^{k_i+1}-1}> \prod_{i=1}^r \frac{p_i^{k_i}(p_i-1)}{p_i^{k_i+1} } = \prod_{i=1}^r \frac{p_i^{k_i}(p_i-1)}{p^i .p_i^{k_i } }\\\large = \prod_{i=1}^r \frac{ (p_i-1)}{p_i } = \prod_{i=1}^r 1- \frac{ 1}{p_i } \)

Answer 34

just wonderful !!!!! XD

Answer 35

\[\large \begin{align} \\ \prod \limits_{1\le i \le r} \left(1- \dfrac{1}{p_i}\right) &= \prod \limits_{1\le i \le r} \left(\dfrac{p_i-1}{p_i}\right) \\~\\
&= \prod \limits_{1\le i \le r} \left(\dfrac{p_i^{k_i}(p_i-1)}{p_i^{k_i+1}}\right) \\~\\
&= n\prod \limits_{1\le i \le r} \left(\dfrac{p_i-1}{p_i^{k_i+1}}\right) \\~\\

& \lt \dfrac{n}{\sigma(n)} \\~\\

\end{align} \]

Answer 36

for part b :


\[\large \begin{align} \\

\dfrac{\sigma(n!)}{n!} &= \sum \limits_{d|n!}\dfrac{1}{d} \\~\\
&\ge 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n} ~~\color{gray}{\because d \in \{1,2,3\cdots n\} | n}

\end{align} \]

Answer 37

yeah direct !
ohh u wanna c as well

Answer 38

the Hint solved it >.<

Answer 39

Not yet, im still not sure how to use the hint :S

Answer 40

posted it here
http://math.stackexchange.com/questions/906255/show-that-sum-of-divisors-of-a-composite-number-n-is-n-sqrtn

Answer 41

\(n | n\) and for composite numbers, there always exists another divisor \(\gt \sqrt{n}\)

so the sum has to be \(\large \gt n + \sqrt{n}\)

Answer 42

i see it clear in hint but i cant like type/explain it neat

Answer 43

im becoming lazy because of MSE

Answer 44

posting questions without even trying >.<

Answer 45

oh i get bored of that site >.<
good question , but hehe i cant type long answer xd
anyway liked u said
n is also divisors and there is at least d < sqrt n

Answer 46

what else u have ?