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A convergent version ( i.e. convergent in the critical strip) of the traditional series for the Riemann Zeta is derived in the video linked at the bottom. It gives the correct numerical values (at least along the critical line, where I tried it out).

But although it works numerically, I'm wondering whether it is rigorously correct. That is, does the derivation follow the strict rules about how series are to be manipulated? (I am not an expert on those rules, so...)

For example the derivation replaces (in effect)

$$\sum_{n=1}^\infty 1/n^s$$

with

$$\sum_{n=1}^\infty \{ {1/(2n-1)}^s + 1/(2n)^s\}$$

This replacement, though invalid for a finite sum, may well be valid when summing to infinity. However, I'm not sure if the second version above can then be merged with another sum (which is what happens around 07:02 in the video). It seems to me that if we write the sums with independent indices M1 and M2, we would get an error if we then tried to set M1 = M2 = M -- at least for finite M. Given this fact, running the M up to infinity should also be problematic.

But again, it does work numerically, so this trick must be justifiable. So what is going on and how does it all work?

Link to video:

But although it works numerically, I'm wondering whether it is rigorously correct. That is, does the derivation follow the strict rules about how series are to be manipulated? (I am not an expert on those rules, so...)

For example the derivation replaces (in effect)

$$\sum_{n=1}^\infty 1/n^s$$

with

$$\sum_{n=1}^\infty \{ {1/(2n-1)}^s + 1/(2n)^s\}$$

This replacement, though invalid for a finite sum, may well be valid when summing to infinity. However, I'm not sure if the second version above can then be merged with another sum (which is what happens around 07:02 in the video). It seems to me that if we write the sums with independent indices M1 and M2, we would get an error if we then tried to set M1 = M2 = M -- at least for finite M. Given this fact, running the M up to infinity should also be problematic.

But again, it does work numerically, so this trick must be justifiable. So what is going on and how does it all work?

Link to video:

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