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 June 7th, 2014, 05:11 AM #1 Senior Member     Joined: Apr 2014 From: zagreb, croatia Posts: 234 Thanks: 33 Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology Averages of arithmetical functions Prove If x is greater or equal to 1, and a > 0, a not equal to 1, we have $\sum_{n June 7th, 2014, 05:16 AM #2 Senior Member Joined: Apr 2014 From: zagreb, croatia Posts: 234 Thanks: 33 Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology Quote:  Originally Posted by raul21 Prove If x is greater or equal to 1, and a > 0, a not equal to 1, we have$\sum_{n
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum_{n<x} \sigma_a {n} = \frac_{\zeta{a+1}}{a+1}x^{a+1}+O{x^b}$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.

Last edited by raul21; June 7th, 2014 at 05:23 AM.

June 7th, 2014, 05:23 AM   #3
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Quote:
 Originally Posted by raul21 Prove If x is greater or equal to 1, and a > 0, a not equal to 1, we have $\sum_{n Prove If x is greater or equal to 1, and a > 0, a not equal to 1, we have$\sum_{n<x} \sigma_a {n} = \frac_{\zeta{a+1}}{a+1}x^{a+1}+O{x^b}$, where b = max{1, a} and$\sigma_a (n)$is divisor function. June 7th, 2014, 07:03 AM #4 Senior Member Joined: Apr 2014 From: zagreb, croatia Posts: 234 Thanks: 33 Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology Quote:  Originally Posted by raul21 Prove If x is greater or equal to 1, and a > 0, a not equal to 1, we have$\sum_{n
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum${n<x} $\sigma_a{n} = \frac_{\zeta{a+1}}{a+1}x^{a+1}+O{x^b}$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.

June 7th, 2014, 09:43 AM   #5
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Quote:
 Originally Posted by raul21 Prove If x is greater or equal to 1, and a > 0, a not equal to 1, we have $\sum${n
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum${n<=x} $\sigma_a (n) = \frac_{\zeta (a+1)}{a+1}x^{a+1}+O(x^b)$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.

June 7th, 2014, 09:46 AM   #6
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Joined: Apr 2014
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Quote:
 Originally Posted by raul21 Prove If x is greater or equal to 1, and a > 0, a not equal to 1, we have $\sum${n<=x} $\sigma_a (n) = \frac_{\zeta (a+1)}{a+1}x^{a+1}+O(x^b)$, where b = max{1, a} and $\sigma_a (n)$ is divisor function.
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum$ (n<=x) $\sigma_a (n) = \frac_{\zeta (a+1)} {a+1} x^(a+1) + O (x^b)$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.

June 7th, 2014, 09:51 AM   #7
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From: zagreb, croatia

Posts: 234
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Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology
Quote:
 Originally Posted by raul21 Prove If x is greater or equal to 1, and a > 0, a not equal to 1, we have $\sum$ (n<=x) $\sigma_a (n) = \frac_{\zeta (a+1)} {a+1} x^(a+1) + O (x^b)$, where b = max{1, a} and $\sigma_a (n)$ is divisor function.
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum$ (n<=x) $\sigma_a (n)$ = $\frac{\zeta (a+1)} {a+1} x^(a+1) + O (x^b)$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.

 June 10th, 2014, 01:41 PM #8 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Well, that's a pretty stinky estimate in general but if you want it ... $$\sum_{n \leq x} \sigma_a (n) = \sigma_{n \leq x} \sum_{d|n} d^a = \sum_{k \leq x} \sum_{d \leq x/k} d^a$$ Now using this teensy little twiddling in the index, you can recognize the first sum as a geometric series and evaluating it gives something like $$\frac{x^a}{1+a}\sum_{k \leq x} \frac1{k^{1+a}}$$ with an error of $\ll x^a$. Approximating the partial-zeta for large enough $x$ gives you the desired. Balarka . Thanks from raul21

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