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I'm looking at Eqn. 13 here:

$$\sum (1/(2n-1)^s - 1/(2n)^s + 2/(2n)^s)$$

They then simplify this to obtain the zeta function:

$$\sum 1/n^s = \zeta(s)$$

Can you help with the simplication? All I can write is

$$1/(2n-1)^s + 1/(2n)^s$$

I don't know how to simplify this further. I tried Geogebra and Wolfram Alfa they both give different answers but they don't simplify to the zeta function.

EDIT: Wolfram alpha screen shot:

enter image description here

enter image description here

zeynel
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  • Hint: every integer is either of the form $2n-1$ or of the form $2n$. – Greg Martin Feb 01 '24 at 16:42
  • Try again Wolfram Alpha, but this time type it carefully. – jjagmath Feb 01 '24 at 16:56
  • @GregMartin Ok. When I plug in the values of $n$ to $1/(2n-1) + 1/(2n)^s)$ I get the zeta sum. Thanks. But I still cannot see how I can simplify it symbolically to $1/n^s$ – zeynel Feb 01 '24 at 17:10
  • @jjagmath I added the screen shot from Wolfram Alpha. What am I doing wrong? – zeynel Feb 01 '24 at 17:14
  • I do not see any attempt to sum something in the Mma input or output.... – R. J. Mathar Feb 01 '24 at 17:59
  • @R.J.Mathar Are you referring to Wolfram Alpha screen shot (I don't know what Mma means)? In that case, I was just trying to simplify the expression $1/(2n-1)^s + 1/(2n)^s$ to obtain $1/n^s$. I was not trying to sum. Now I see that, when I plug in the values of $n$ into $1/(2n-1)^s + 1/(2n)^s$ I get $1/n^s$ but I still don't know how to simply it symbolically. – zeynel Feb 01 '24 at 18:22
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    @zeynel You didn't input any sum. – jjagmath Feb 01 '24 at 18:23
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    Well, for $n\ge 1$ and $s\ge 1$ definitely $1/(2n-1)^s+1/(2n^s)\neq 1/n^s$, the two terms are different. For example for $n=s=1$ we have $1/(2n-1)^s+1/(2n)^s=1+1/2=3/2$ and $1/n^s=1$. You need to add summation to compute Riemann $\zeta$-values. – R. J. Mathar Feb 01 '24 at 18:34
  • @jjagmath Ok, added the sum :) But now it gives $\sum (1/(2n-1) + 1/(2n)^s) = \zeta(s)$ So there is no way to simplify $(1/(2n-1) + 1/(2n)^s)$ to $1/n^s$ symbolically? If so, Eqn. 13 in the linked paper confused me. It looks like they simplified $(1/(2n-1) + 1/(2n)^s)$ to $1/n^s$ . – zeynel Feb 01 '24 at 18:38
  • @R.J.Mathar Thanks. I added the summation. So, I assume that we don't try to simplify $1/(2−1)+1/(2)^$ to $1/^$. – zeynel Feb 01 '24 at 19:00
  • By the way, I thank everyone who commented. This was very helpful to me. I'm sure it will be helpful to others as well. So, I don't see any reason why this should be downvoted. – zeynel Feb 01 '24 at 20:14

1 Answers1

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$$ \sum_{n\ge 1} [\frac{1}{(2n-1)^s} -\frac{1}{(2n)^s}+\frac{2}{(2n)^s}] = \sum_{n\ge 1} [\frac{1}{(2n-1)^s} +\frac{1}{(2n)^s}] $$ $$ = \sum_{n\ge 1} \frac{1}{(2n-1)^s} +\sum_{n\ge 1} \frac{1}{(2n)^s} $$ $$ = \sum_{n=1,3,5,7,\ldots odd} \frac{1}{n^s} +\sum_{n=2,4,6,8,10 \ldots even} \frac{1}{n^s} $$ $$ = \sum_{n=1,2,3,4,5,} \frac{1}{n^s} = \zeta(s). $$

R. J. Mathar
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