Question

show that
\[\sum\limits_{d\mid n}d\tau(d) = \sum\limits_{d\mid n}(n/d)\sigma(d)\]

Answer 1

it is given, \(n\) is of form \(p^k\)
where \(p\) is a prime

Answer 2

in other words \(n\) is power of a prime

Answer 3

As always ...explain the term first pls
So that we can also understand it

Answer 4

Here is the idenity incase if the latex has trouble displaying in main question
\[\sum\limits_{d\mid n}d\tau(d) = \sum\limits_{d\mid n}(n/d)\sigma(d)\]

Answer 5

\(\tau(n)\) gives the "number" of positive divisors of \(n\)

\(\sigma(n)\) gives the "sum" of positive divisors of \(n\)

Answer 6

the sum \(\sum\limits_{d\mid n}f(d)\) means : "Sum the values f(d) as d runs over all the positive divisors of n"

Answer 7

example :
\[\sum\limits_{d\mid 10}f(d) = f(1)+f(2)+f(5)+f(10)\]

because positive divisors of 10 are {1, 2, 5, 10}

Answer 8

Can I say LHS as
f(1)+ f(p)+ f(p^2)+f(p^3)+...............+f(p^n)

Answer 9

f(1)+f(p)+f(p^2)+...........+f(p^k)

Answer 10

LHS is \(\sum\limits_{d\mid n}d\tau(d)\)

i think we get :
\[1\tau(1) + p\tau(p) + p^2\tau(p^2) + \cdots + p^k\tau(p^k)\]

Answer 11

it may help you :-
1) http://primes.utm.edu/glossary/xpage/tau.html
2) http://2000clicks.com/mathhelp/NumberTh05DivisorsTau.aspx

Answer 12

Yes, my i understand my mistake

Answer 13

@ganeshie8

Answer 14

yes i think we may use those results...

Answer 15

So LHS would be
1+2p+3(p^2)+.........+(k+1)(p^k)

Answer 16

happy to help you :)

Answer 17

Oh nice, i think that is correct !

Answer 18

But T(1)=1
T(p)=2
As p a prime it have two +ve divisor

Answer 19

Yes but what next

Answer 20

So, according to you :
\(\sum\limits_{d\mid n}d\tau(d)=1\tau(1) + p\tau(p) + p^2\tau(p^2) + \cdots + p^k\tau(p^k)\\~\\~\\
=1*1+p*2+p^2*3+\cdots + p^k(k+1)\\~\\
=1+2p+3p^2+\cdots+(k+1)p^k

\)

Answer 21

I don't know whats next, still thinking...

Answer 22

Is it a aritho-geo-progression

Answer 23

Both AP and GP

Answer 24

yeah it may be sufficient if we could show that RHS also equals this aritho-geo-progression

Answer 25

I can solve it ...OK now starts with RHS

Answer 26

So RHS =
n S(1) + n/p S(p)+.......+n/n S(n)

Answer 27

Where S represents Sigma
And n=p^k

Answer 28

yes

\(\sigma(p^k) = 1+p+p^2+\cdots+p^k\)

Answer 29

because the positive divisors of \(p^k\) are simply \(\{1,p,p^2,\ldots,p^k\}\)

Answer 30

That looks nice!
the sum should be from i=0 to i=k right ?

Answer 31

oops sooryy
yea that should be i=k :)

Answer 32

\(\tau(p^k) = k+1\)

Answer 33

How to take summation of that that series.

Answer 34

ok i'll edit the solution
edited solution->
my work so far-
\(\sum\limits_{d\mid n}d\tau(d) = \sum\limits_{d\mid n}(n/d)\sigma(d)\)


We know that \(n=p^k\)
divisors of n will be->\(1, p, p^2, p^3...p^{k}\)

putting these values in LHS we get a series like this->
\[1 \tau(1)+p \tau(p) + p^2 \tau(p^2)....p^{k} \tau(p^{k}) \]
\[=>1 +2p+3p^2 +4p^3......(k+2)p^{k+1}\]\[=> \sum_{i=0}^{k}(i+1)p^i\]

now the RHS

\(\sum\limits_{d\mid n}(n/d)\sigma(d)\)

\(\sum\limits_{d\mid n}(p^k /d)\sigma(d)\)
putting d=\(1,~p,~p^2,~p^3...p^{k+1}\)

we get a series like this->
\(p^k,~p^{k-1}[1+p], ~p^{k-2}[1+p+p^2],.....p^0[1+p+p^2..p^k]\)

opening the brackets the series looks like this->

\(p^k,~p^{k}+p^{k-1}, ~ p^k+p^{k-1}+p^{k-2},.......,p^k+p^{k-1}..+p^{k-k}\)

taking the summation we will get p^k we will get it k+1 times
so we can write the RHS as\[\sum_{i=0}^{k}(i+1)p^i\]

\(LHS=RHS\)
hence proved

Answer 35

Can you tell me pls how to take summation of that series?

Answer 36

which series @pawanyadav ?

Answer 37

That we found in RHS