sequences and series - Sum of all odd numbers to infinity

Starting from the idea that $$\sum_{n=1}^\infty n = -\frac{1}{12}$$
It's fairly natural to ask about the series of odd numbers $$\sum_{n=1}^{\infty} (2n - 1)$$

I worked this out in two different ways, and get two different answers. By my first method
$$\sum_{n=1}^{\infty} (2n - 1) + 2\bigg( \sum_{n=1}^\infty n \bigg) = \sum_{n=1}^\infty n$$
$$\therefore ~\sum_{n=1}^{\infty} (2n - 1) = - \sum_{n=1}^\infty n = \frac{1}{12}$$
But then by the second
$$\sum_{n=1}^{\infty} (2n - 1) - \sum_{n=1}^\infty n = \sum_{n=1}^\infty n$$
$$\therefore ~\sum_{n=1}^{\infty} (2n - 1) = 2 \sum_{n=1}^\infty n = -
\frac{1}{6}$$
Is there any reason to prefer one of these answers over the other? Or is the sum over all odd numbers simply undefined? In which case, was there a way to tell that in advance?

I'm also curious if this extends to other series of a similar form

$$\sum_{n=1}^{\infty} (an + b)$$
Are such series undefined whenever $b \neq 0$?

Answer

With the usual caveat that $$ \sum_{n=1}^\infty n \ne -\frac{1}{12}$$
we can do a similar zeta function regularization for the sum of odd integers. We start with the fact that $$ \sum_{n = 1}^\infty \frac{1}{(2n-1)^s} =(1-2^{-s})\zeta(s)$$ for $\Re(s) > 1$ and then analytically continue to $s=-1$ to get $$ \sum_{n=1}^\infty(2n+1) "=" (1-2)\zeta(-1) = \frac{1}{12}$$

Edit

Zeta function regularization and Ramanujan get the same answer here. As for why your first method gets the "right answer" and the second doesn't, note that the first is argued by the exact same formal steps used to derive $$ \sum_{n=1}^\infty\frac{1}{(2n-1)^s} = (1-2^{-s})\zeta(s)$$ while the second uses both linearity and index shifting which are generally not preserved by the regularization methods.