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As the other answer has pointed out, $-1/12$ is not the only value that can obtained with analytic continuation. However, it is the unique constant term of the asymptotic expansion of the smoothed partial sums, which perhaps explains why it is the most "natural" value.

Let $\eta$ be any Schwartz function such that $\eta(0) = 1$. Then

\begin{align} \sum_{n=0}^\infty n^s \eta(n \varepsilon) &= \zeta(-s) + O(\varepsilon) + \frac{1}{\varepsilon^{s+1}} \int_0^\infty x^s \eta(x) dx \end{align}

Therefore, by choosing for any given $s$ an $\eta$ that makes the last integral zero, we get

\begin{align} \sum_{n=0}^\infty n^s &= \sum_{n=0}^\infty n^s \lim_{\varepsilon \rightarrow 0^+} \eta(n \varepsilon) \\ &\overset{!}{=} \lim_{\varepsilon \rightarrow 0^+} \sum_{n=0}^\infty n^s \eta(n \varepsilon) \\ &= \lim_{\varepsilon \rightarrow 0^+} \left( \zeta(-s) + O(\varepsilon) \right) \\ &= \zeta(-s) \end{align}

As the other answer has pointed out, $-1/12$ is not the only value that can obtained with analytic continuation. However, it is the unique constant term of the asymptotic expansion of the smoothed partial sums, which perhaps explains why it is the most "natural" value.

Let $\eta$ be any Schwartz function such that $\eta(0) = 1$. Then

\begin{align} \sum_{n=1}^\infty n^s \eta(n \varepsilon) &= \zeta(-s) + O(\varepsilon) + \frac{1}{\varepsilon^{s+1}} \int_0^\infty x^s \eta(x) dx \end{align}

Therefore, by choosing for any given $s$ an $\eta$ that makes the last integral zero, we get

\begin{align} \sum_{n=1}^\infty n^s &= \sum_{n=1}^\infty n^s \lim_{\varepsilon \rightarrow 0^+} \eta(n \varepsilon) \\ &\overset{!}{=} \lim_{\varepsilon \rightarrow 0^+} \sum_{n=1}^\infty n^s \eta(n \varepsilon) \\ &= \lim_{\varepsilon \rightarrow 0^+} \left( \zeta(-s) + O(\varepsilon) \right) \\ &= \zeta(-s) \end{align}

As the other answer has pointed out, $-1/12$ is not the only value that can obtained with analytic continuation. However, it is the unique constant term of the asymptotic expansion of the smoothed partial sums, which perhaps explains why it is the "most natural" value.

Let $\eta$ be any Schwartz function such that $\eta(0) = 1$. Then

\begin{align} \sum_{n=0}^\infty n^s \eta(n \varepsilon) &= \zeta(-s) + O(\varepsilon) + \frac{1}{\varepsilon^{s+1}} \int_0^\infty x^s \eta(x) dx \end{align}

Therefore, by choosing an $\eta$ that makes the last integral zero, we get

\begin{align} \sum_{n=0}^\infty n^s &= \sum_{n=0}^\infty n^s \lim_{\varepsilon \rightarrow 0^+} \eta(n \varepsilon) \\ &\overset{\star}{=} \lim_{\varepsilon \rightarrow 0^+} \sum_{n=0}^\infty n^s \eta(n \varepsilon) \\ &= \lim_{\varepsilon \rightarrow 0^+} \left( \zeta(-s) + O(\varepsilon) \right) \\ &= \zeta(-s) \end{align}

As the other answer has pointed out, $-1/12$ is not the only value that can obtained with analytic continuation. However, it is the unique constant term of the asymptotic expansion of the smoothed partial sums, which perhaps explains why it is the most "natural" value.

Let $\eta$ be any Schwartz function such that $\eta(0) = 1$. Then

\begin{align} \sum_{n=0}^\infty n^s \eta(n \varepsilon) &= \zeta(-s) + O(\varepsilon) + \frac{1}{\varepsilon^{s+1}} \int_0^\infty x^s \eta(x) dx \end{align}

Therefore, by choosing for any given $s$ an $\eta$ that makes the last integral zero, we get

\begin{align} \sum_{n=0}^\infty n^s &= \sum_{n=0}^\infty n^s \lim_{\varepsilon \rightarrow 0^+} \eta(n \varepsilon) \\ &\overset{!}{=} \lim_{\varepsilon \rightarrow 0^+} \sum_{n=0}^\infty n^s \eta(n \varepsilon) \\ &= \lim_{\varepsilon \rightarrow 0^+} \left( \zeta(-s) + O(\varepsilon) \right) \\ &= \zeta(-s) \end{align}

As the other answer has pointed out, $-1/12$ is not the only value that can obtained with analytic continuation. However, it is the unique constant term of the asymptotic expansion of the smoothed partial sums, which perhaps explains why it is the "most natural" value.

Let $\eta$ be any Schwartz function such that $\eta(0) = 1$. Then

\begin{align} \sum_{n=0}^\infty n^s \eta(n \varepsilon) &= \zeta(-s) + O(\varepsilon) + \frac{1}{\varepsilon^{s+1}} \int_0^\infty x^s \eta(x) dx \end{align}

Therefore, by choosing an $\eta$ that makes the last integral zero, we get

\begin{align} \sum_{n=0}^\infty n^s &= \sum_{n=0}^\infty n^s \lim_{\varepsilon \rightarrow 0^+} \eta(n \varepsilon) \\ &\overset{\star}{=} \lim_{\varepsilon \rightarrow 0^+} \sum_{n=0}^\infty n^s \eta(n \varepsilon) \\ &= \lim_{\varepsilon \rightarrow 0^+} \left( \zeta(-s) + O(\varepsilon) \right) \\ &= \zeta(-s) \end{align}

As the other answer has pointed out, $-1/12$ is not the only value that can obtained with analytic continuation. However, it is the unique constant term of the asymptotic expansion of the smoothed partial sums, which perhaps explains why it is the "most natural" value.

Let $\eta$ be any Schwartz function such that $\eta(0) = 1$. Then

\begin{align} \sum_{n=0}^\infty n^s \eta(n \varepsilon) &= \zeta(-s) + O(\varepsilon) + \frac{1}{\varepsilon^{s+1}} \int_0^\infty x^s \eta(x) dx \end{align}

Therefore, by choosing an $\eta$ that makes the last integral zero, we get

\begin{align} \sum_{n=0}^\infty n^s &= \sum_{n=0}^\infty n^s \lim_{\varepsilon \rightarrow 0^+} \eta(n \varepsilon) \\ &\overset{\star}{=} \lim_{\varepsilon \rightarrow 0^+} \sum_{n=0}^\infty n^s \eta(n \varepsilon) \\ &= \lim_{\varepsilon \rightarrow 0^+} \left( \zeta(-s) + O(\varepsilon) \right) \\ &= \zeta(-s) \end{align}