Why $\zeta(-1)=-\frac{1}{12}$ does not mean the sum from $1$ to infinity is $-\frac{1}{12}$

Question

Since $\zeta(-1)=\frac{1}{1^{-1}}+\frac{1}{2^{-1}}+\frac{1}{3^{-1}}+\cdots=-\frac{1}{12}$, why do we still say that $\sum^\infty_{n=1}n\rightarrow+\infty$?

Answer 1

First of all, it is NOT TRUE that

$$\zeta(-1) = \frac{1}{1^{-1}} + \frac{1}{2^{-1}} + \dots$$

The Riemann Zeta function is only defined as

$$\zeta(s) = \sum_{i=1}^\infty \frac{1}{n^s}$$

for values of $s$ for which the real part of $s$ is greater than one.

For other values, we can calculate the analytic continuation of $\zeta$ to define the function on the entire complex plane, but that means that we are expanding the definition beyond what it originally was.

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You need to understand where our definitions come from:

• First, we have infinite sequences.
• From infinite sequences, we construct infinite series, and we define when these series converge.
• Then, we take a very specific family of infinite series, and we construct a function that sums them.
• Then, we expand the definition of this function to other numbers.

Now, if we take a series and ask "does it converge", we cannot say "yes because Riemann's zeta says so". We have an earlier definition, and to answer convergence, we need to use that definition.

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So, why does $$\sum_{i=0}^\infty i$$

diverge? Simply. Take the original definition of convergence of series, and the answer becomes:

Because $$\sum_{i=0}^\infty a_i$$

is defined as $$\lim_{N\to\infty} \sum_{i=0}^N a_i$$

And the sequence in the case when $a_i = i$ diverges toward $\infty.$