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This question may appear to be a silly one for experts.

From long back I have been observing all kinds of series but every-series contain a reciprocal part, I mean the " one over something " , is there any interesting reason for it ? .

In general if we consider the Dirichlet's L-series its given by $$L(s,\chi)=\sum^{\infty}_{n=1}\frac{\large \chi (n)}{\large n^s}$$ , so I know that its a generalization of Riemann Zeta function, and also know that these series are useful in proving the theorem of primes in arithmetic progression.

But I observed that every such " series " contain the reciprocal term ( I was referring to $\large \frac{1}{n^s}$ ) . As far as I know the Dirichlet theorem for arithmetic progression of primes implies that $\large \frac{1}{3}+\frac{1}{7}+\frac{1}{11}+.........$ is a divergent series, but I was wondering that $\large 3+7+11+......$ is also a divergent series and still diverge faster than the former one.

So why is the reason of using the " one over term in the series ", like why cant the series be stated as $$L(s,\chi)=\sum^{\infty}_{n=1}\large \chi (n)\large n^s$$ instead of $$L(s,\chi)=\sum^{\infty}_{n=1}\frac{\large \chi (n)}{\large n^s}.$$

What is the significance of such reciprocal term ?

And also if one takes the Hasse-weil L-function of a curve its defined as $$L_p(s)= 1/P(\large p^{-s})$$ where $P(T)=\begin{cases} \large1-a_pT+pT^2 & \text{ If its good reduction } , \\ 1-T & \text{If reduction is split multiplicative } , \\ 1+T & \text { If reduction is non-split multiplicative },\\1 & \text{ If its additive } .\end{cases}$

Here also we can notice that we took a reciprocal of the term $P(\large p^{-s})$ instead of directly writing it.

And actual $L$-function is $L(s,E)=\prod_p L_p(s)$.

So finally to put it in short why does one always encounters these reciprocals when talking about series and other functions in Analytical Number Theory ?

And also I heard that Hasse-Weil L-function carries the information about local points, so there must be some interesting reason behind considering the reciprocals when one fabricate any such functions.

If the answer is " Its a result of random attempt like one needs to take sum of reciprocals to obtain something then please explain me the use of taking reciprocals "

Thanks a lot.

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1 Answer 1

up vote 6 down vote accepted

As far as I can tell this is a matter of historical convention. Replacing $1/n^s$ by $n^s$ is just a matter of replacing $s$ by $-s$. But the primary interest in the $\zeta$ function (or, at least, the one that motivated its introduction) is that the series for $\zeta(s)$ converges for $s > 1$, and diverges precisely at $s = 1$. Adopting the alternative convention (i.e. replacing $s$ by $-s$ in the notation), one would say that the series for $\zeta(s)$ converges for $s < -1$, and diverges precisely when $s = -1$.

Given the choice between considering a function of $s >1$ or $ s < - 1$, it doesn't seem so surprising that the former choice was made. Positive real numbers are psychologically simpler than negative ones.

Incidentally, the fact that $\sum_n n$ also diverges is misleading; the Riemann $\zeta$ function does not blow up at $-1$, but rather has the value $-1/12$. Also, if you work through Dirichlet's proof, or even Euler's proof of infinitude of primes using the divergence of $\zeta(1)$, you will see that it does not work if consider instead the series $\sum_n n$, rather than $\sum_n 1/n$.

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Wow !, simply fantastic thats it !! , I must copy paste the word "fantastic" as every answer you give is by default fantastic. Thanks a ton Prof.Emerton @Matt E –  Iyengar Dec 17 '11 at 13:32

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