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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 $\sum_{n<x} \sigma_a (n) = \frac_{\zeta(a+1)}{a+1}x^{a+1}+O(x^b)$, where b = max{1, a} and $\sigma_a (n)$ is divisor function. Last edited by raul21; June 7th, 2014 at 05:15 AM. 
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:
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  
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:
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:
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  
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:
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  
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:
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  
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:
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 ... $$\sum_{n \leq x} \sigma_a (n) = \sigma_{n \leq x} \sum_{dn} d^a = \sum_{k \leq x} \sum_{d \leq x/k} d^a$$ 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 partialzeta for large enough $x$ gives you the desired. Balarka . 

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