Prove that $\zeta(-1)=\zeta(-13)$. Basically what the title says. 
I saw this through another Math Platform but did not get any response to it. The original question was to find distinct integers of $ x $ and $y$ such that $ \zeta(x) = \zeta(y) $.
I'm not too familiar with zeta functions but I'm sure that there must be a simple explanation for this.
Can anyone help me? Thank you.
 A: The general formula for the Riemann zeta function at negative integers, taken from the Wikipedia entry, is
$$\zeta(-n)=-{B_{n+1}\over n+1},$$
where $B_{n+1}$ is a Bernoulli number.  As it happens, $B_2=1/6$ and $B_{14}=7/6$, so 
$$\zeta(-1)=\zeta(-13)=-\frac{1}{12}.$$
As far as I know, the equality can be chalked up to coincidence, and any proof will require a fair amount of analytic infrastructure, such as the functional equation
$$\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s).$$
A: Continuing the functions described in this answer and this answer, which are differences of finite sums of reciprocals of powers of integers and their Euler-Maclaurin Sum Formula approximations, set
$$
\begin{align}
\zeta_n(z)
&=\sum_{k=1}^n\small\frac1{k^z}-\frac1{1-z}n^{1-z}-\frac12n^{-z}+\frac{z}{12}n^{-1-z}-\frac{z(z+1)(z+2)}{120\cdot3!}n^{-3-z}\\
&\small+\frac{z(z+1)(z+2)(z+3)(z+4)}{252\cdot5!}n^{-5-z}\\
&\small-\frac{z(z+1)(z+2)(z+3)(z+4)(z+5)(z+6)}{240\cdot7!}n^{-7-z}\\
&\small+\frac{z(z+1)(z+2)(z+3)(z+4)(z+5)(z+6)(z+7)(z+8)}{132\cdot9!}n^{-9-z}\\
&-\tfrac{691z(z+1)(z+2)(z+3)(z+4)(z+5)(z+6)(z+7)(z+8)(z+9)(z+10)}{32760\cdot11!}n^{-11-z}\\
&+\tfrac{z(z+1)(z+2)(z+3)(z+4)(z+5)(z+6)(z+7)(z+8)(z+9)(z+10)(z+11)(z+12)}{12\cdot13!}n^{-13-z}
\end{align}
$$
then for $\mathrm{Re}(z)\gt-15$,
$$
\lim_{n\to\infty}\zeta_n(z)=\zeta(z)
$$
For $z=-1$,
$$
\zeta_n(-1)=\underbrace{\sum_{k=1}^nk-\left(\frac{n^2}2+\frac{n}2\right)}_{\text{these must cancel}}\underbrace{\vphantom{\sum_{k=1}^n}\ -\frac1{12}\ \ }_{\zeta(-1)}\underbrace{\vphantom{\sum_{k=1}^n}+O\left(n^{-2}\right)}_{\text{vanishes}}
$$
For $z=-13$,
$$
\zeta_n(-13)=\underbrace{\sum_{k=1}^nk^{13}-P_{13}(n)}_{\text{these must cancel}}\underbrace{\vphantom{\sum_{k=1}^n}\ -\frac1{12}\ \ }_{\zeta(-13)}
$$
The parts that "must cancel" must cancel because, when $z$ is a non-positive integer, they are the difference of two polynomials with no constant terms. If they did not match, there would be no way for the limit to exist. Other than those terms are the constant term and the terms that vanish as $n\to\infty$. Thus, the constant term is $\zeta(z)$.
As Barry Cipra says, I don't know that there is any reason that these are the same other than coincidence.
A: $$\zeta(1-2k)=-\frac{B_{2k}}{2k}:k\gt 0$$
$$\displaystyle \sum_{k=0}^{n-1}k^{m}=B_{m}n+A_{2}n^{2}+...+A_{m+1}n^{m+1}$$
$$B_{m}=\displaystyle \frac{-1}{m+1}\sum_{k=0}^{m-1}\binom{m+1}{k}B_{k}$$
$$B_{0}=1:B_{1}=-\frac{1}{2},B_{2}=\frac{1}{6},B_{14}=\frac{7}{6}$$
$$k=1,\zeta(-1)=-\frac{B_{2}}{2\times 1}=-\frac{\frac{1}{6}}{2}=-\frac{1}{12}$$
$$k=7,\zeta(-13)=-\frac{B_{14}}{2\times 7}=-\frac{\frac{7}{6}}{2\times 7}=-\frac{1}{12}$$
