I tried to prove this sum by myself, but I couldn't. $1 + 4 + 9 + 16 + ... = 0$ First, I know this sums are a bit problematic, as we can't just $'='$ an infinite sum, but I would like to see the algorithm of proving the series above. Thanks.
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One way is to use the analytical continuation of the $\zeta$-function, then use the fact that $\zeta(-2k)=0$ for every $k\in \mathbb{N}$. – MrYouMath Jun 14 '16 at 19:52
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You are asking for why $\zeta(-2)=0$. See this related question. Also related is this one where people ask why $\zeta(-1)=-\frac{1}{12}$ if you are confused by the zeta function in the first place. – JMoravitz Jun 14 '16 at 19:52
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BTW it was not Ramanujan to first derive this expression, it was the master of us all Leonhard Euler himself :). – MrYouMath Jun 14 '16 at 19:59
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Isn't the sum $n(n+1)(2n+1)\over6$ – Colbi Jun 14 '16 at 20:00
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The related alternating series $1-4+9-16+\dotsb“=”0$ is easier. – Akiva Weinberger Jun 14 '16 at 20:01
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@Colbi that is the sequence of partial sums $1+4+9+16+\dots+n^2$. The question asker is however asking why $\zeta(-2)=0$ where $\zeta(s)$ is the analytic continuation of the function $\zeta(s)=\sum\limits_{n=1}^\infty n^{-s}$. Read more here. – JMoravitz Jun 14 '16 at 20:05
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This is how Euler would have derived it using non-rigorous methods. The cool part about this non-rigorous method is that it leads to the same result as the rigorous "proof" by analyitical continuation for the $\zeta(s)$
Start with the geometric series:
$$\frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots$$
Differentiate and multiply by $x$:
$$\frac{x}{(1-x)^2}=x+2x^2+3x^3+4x^4+\cdots .$$
Differentiate once again:
$$\frac{1-x^2}{(1-x)^4}=1+2^2x+3^2x^2+4^2x^2+\cdots .$$
Note that I will use the congruent sign just to distinguish this from the normal equal sign.
Now plug in $x=-1$ and don't care about convergence :D. $$0=1^2-2^2+3^2-4^2+\cdots \cong \eta(-2) .$$
No try to establish an relationship to $$\zeta(-2)\cong1^2+2^2+3^2+4^2+\cdots.$$
This can be achieved by subtracting $\zeta(-2)$ with $\eta(-2)$. Or in general $\zeta(s)-\eta(s)=2^{1-s}\zeta(s)$ or $$\eta(s)=(1-2^{1-s})\zeta(s)$$
Note that this representation is only valid for $s \neq 1$.
From $\eta(-2)$ it follows that $\zeta(-2)=0$.
MrYouMath
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1$s=1$ is a removable singularity of the RHS, and the equality is true in the sense $\eta(1) = \lim_{ s \to 1} (1-2^{1-s}) \zeta(s)$ – reuns Jun 14 '16 at 21:03