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I would like to give other representation for zeta function using

fundemental arithmitic I have got this: $\zeta(s-1)=\sum_{n=1}^{\infty}\frac{\gcd(n,n)}{{\operatorname{lcm}(n,n)}^{s}}$ where $\gcd(n,n)$ is the greatest common divisor between $(n,n)$ and $\operatorname{lcm}$ is the least common multiple of $(n,n)$.

My question here is:

Is this true: $$\zeta(s-1)=\sum_{n=1}^{\infty}\frac{\gcd(n,n)}{{\operatorname{lcm}(n,n)}^{s}}?$$

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Yes, since $\gcd{(n,n)}=n$ and $\operatorname{lcm}{(n,n)}=n$, so the sum is of $n/n^s = n^{-(s-1)}$.

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  • $\begingroup$ thanks, sorry for this question, i was affraid from gcd(\infty,infty) and lcm(\infty,infty) $\endgroup$ – zeraoulia rafik Feb 3 '16 at 14:28
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    $\begingroup$ That's not in the sum. $\endgroup$ – Chappers Feb 3 '16 at 14:29
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You know that $\gcd(n,n)=n$ and $\operatorname{lcm}(n,n)=\dfrac{|n\cdot n|}{\gcd(n,n)}=n$ so your original question reduces to the definition of the Zeta function:

$$\zeta(s-1)=\sum_{n=1}^{\infty}\dfrac{1}{n^{s-1}}$$

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  • $\begingroup$ $n$ is not a variable, you sum over $n$. You can't do $\zeta(n-1)$, it makes no sense. $\endgroup$ – vrugtehagel Feb 3 '16 at 14:31
  • $\begingroup$ relax, it was just a typo! $\endgroup$ – Jose Lopez Garcia Feb 3 '16 at 14:31
  • $\begingroup$ I know, I was just pointing it out :) $\endgroup$ – vrugtehagel Feb 3 '16 at 14:32
  • $\begingroup$ Yea, I edited it, thanks! $\endgroup$ – Jose Lopez Garcia Feb 3 '16 at 14:32

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