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## Main Question or Discussion Point

using an exponential regulator [tex] exp(-\epsilon n) [/tex] the sum

[tex] 1+2+3+4+5+6+7+............= -1/12+ 1/\epsilon^{2} [/tex]

and for Casimir effect [tex] 1+8+27+64+125+............= -1/120+ 1/\epsilon^{4} [/tex]

can i simply remove in the calculations of divergent series 1+2+3+4+5.. and similar the epsilon terms imposing renormalization conditions ??

how about for the rest of sums [tex] 1+2^{m}+3^{m}+..........= \zeta (-m) + 1/\epsilon ^{m+1} [/tex]

if i introducte a power regulator [tex] n^{-s} [/tex] in the limit s-->0+ i would get

[tex] \zeta(s-m)=\zeta(-m) [/tex] but i am not sure, why this work

for example in the definition of a functional determinant (in differential geommetry )

[tex] \prod_{i} \lambda_{i} [/tex] apparently there is no divergent term proportional to [tex] log(\epsilon) [/tex] as one would expect since the product is divergent

[tex] 1+2+3+4+5+6+7+............= -1/12+ 1/\epsilon^{2} [/tex]

and for Casimir effect [tex] 1+8+27+64+125+............= -1/120+ 1/\epsilon^{4} [/tex]

can i simply remove in the calculations of divergent series 1+2+3+4+5.. and similar the epsilon terms imposing renormalization conditions ??

how about for the rest of sums [tex] 1+2^{m}+3^{m}+..........= \zeta (-m) + 1/\epsilon ^{m+1} [/tex]

if i introducte a power regulator [tex] n^{-s} [/tex] in the limit s-->0+ i would get

[tex] \zeta(s-m)=\zeta(-m) [/tex] but i am not sure, why this work

for example in the definition of a functional determinant (in differential geommetry )

[tex] \prod_{i} \lambda_{i} [/tex] apparently there is no divergent term proportional to [tex] log(\epsilon) [/tex] as one would expect since the product is divergent