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How do we change [elementary way] $$\eta(s)=\sum_{n=0}^{\infty}{(-1)^n\over (n+1)^s}$$ into

$$\zeta(s)={1\over 1-2^{1-s}}\sum_{n=0}^{\infty}{1\over 2^{n+1}}\sum_{k=0}^{n}(-1)^k{n\choose k}(k+1)^{-s}$$

My try:

$${1\over 2}+{1\over 2^2}+{1\over2^3}+{1\over2^4}+\cdots=1$$

$${1\over 2^2}+{2\over 2^3}+{3\over2^4}+{4\over2^5}+\cdots=1$$

$${1\over 2^3}+{3\over 2^4}+{6\over2^5}+{10\over2^6}+\cdots=1$$

$${1\over 2^4}+{4\over 2^5}+{10\over2^6}+{20\over2^7}+\cdots=1$$

And so on ...

we expanded the eta function

$$\eta(s)=1-{1\over 2^s}+{1\over3^s}-{1\over4^s}+\cdots$$

$${1\over 1^s}\left({1\over 2}+{1\over 2^2}+{1\over2^3}+{1\over2^4}+\cdots\right)={1\over1^s}$$

$${1\over 2^s}\left({1\over 2^2}+{2\over 2^3}+{3\over2^4}+{4\over2^5}+\cdots\right)={1\over 2^s}$$

$${1\over3^s}\left({1\over 2^3}+{3\over 2^4}+{6\over2^5}+{10\over2^6}+\cdots\right)={1\over 3^s}$$

$${1\over 4^s}\left({1\over 2^4}+{4\over 2^5}+{10\over2^6}+{20\over2^7}+\cdots\right)={1\over 4^s}$$ collected the terms and we have $$\eta(s)={1\over2}\left(1\over1^s\right)+{1\over2^2}\left({1\over1^s}-{1\over2^s}\right)+{1\over2^3}\left({1\over1^s}-{2\over2^s}+{1\over3^s}\right)+\cdots$$

put in the sum notation

$$\eta(s)=\sum_{n=0}^{\infty}{1\over 2^{n+1}}\sum_{k=0}^{n}(-1)^k{n\choose k}(k+1)^{-s}$$

where

$$\zeta(s)={1\over 1-2^{1-s}}\sum_{n=0}^{\infty}{(-1)^n\over (n+1)^s}={1\over 1-2^{1-s}}\eta(s)$$

so we have

$$\zeta(s)={1\over 1-2^{1-s}}\sum_{n=0}^{\infty}{1\over 2^{n+1}}\sum_{k=0}^{n}(-1)^k{n\choose k}(k+1)^{-s}$$

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1 Answers1

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To be more specific, you just took what is known as either the Euler transform or the Binomial transform. It is basically what you did, but more general.

The Euler summation formula also happens to be basically the same.

$$_{{E_{y}}}\,\sum _{{j=0}}^{\infty }a_{j}:=\sum _{{i=0}}^{\infty }{\frac {1}{(1+y)^{{i+1}}}}\sum _{{j=0}}^{i}{i \choose j}y^{{j+1}}a_{j}$$

An Euler sum is equal to the original sum if it exists, and is useful for accelerating convergence and sometimes calculations of divergent series.

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