Exercise 2 from Terry Tao's blog on Euler-Maclaurin, Bernouilli numbers, and the zeta function In the blog post The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation, Terry Tao looks at the commonly-cranked 'absurd' formulae
$$\begin{align}
\sum_{n \geq 1} 1 &= -1/2 \tag{1} \\
\sum_{n \geq 1} n &= -1/12 \tag{2},
\end{align}$$
where of course I do not intend these to be taken literally. These are the correct values if we interpret the left sums as $\zeta(0)$ and $\zeta(-1)$, extended through analytic continuation. But one of the big points of Terry Tao's blog post is to show that one can approach $(1)$ and $(2)$ from a completely real-variable method involving smoothed sums.
Let $\eta: \mathbb{R}^+ \longrightarrow \mathbb{R}$ be a smooth, compactly supported, bounded cutoff function which is $1$ in a neighborhood of $0$. Corresponding to $(1)$ and $(2)$ are
$$\begin{align}
\sum_{n \geq 1} 1 \cdot \eta(n/N) &= -\frac{1}{2} + C_{\eta, 0} N + O\big(\frac{1}{N}\big) \tag{3} \\
\sum_{n \geq 1} n \cdot \eta(n/N) &= - \frac{1}{12} + C_{\eta, 1} N^2 + O\big(\frac{1}{N}\big), \tag{4}
\end{align}$$
where
$$ \zeta_{\eta, j} = \int_0^\infty x^j \eta(x) dx.$$
In the blog post, Exercise 2 concerns resolving an 'apparent inconsistency' in $(1)$ and $(2)$. Adding $(1)$ and $(2)$ (formally) shows that $\sum_{n \geq 1} (1+n) = -7/12$. Subtracting the integer $1$ from $(2)$ shows (formally) that $\sum_{n \geq 2} n = \sum_{n \geq 1} (1+n) = -13/12$. Working with the smoothed sums, adding $(3)$ and $(4)$ shows that
$$ \sum_{n \geq 1} (1+n) \eta(n/N) = -\frac{7}{12} +  C_{\eta, 1} N^2 + C_{\eta, 0}N + O\big( \frac{1}{N} \big). \tag{5}$$
Subtracting $1$ (or rather $\eta(1/N)$, which is $1 + O(1/N)$ from a Taylor expansion) from $(4)$ shows that
$$ \sum_{n \geq 2} n \eta(n/N) = \sum_{n \geq 1} (1+n) \eta\big( \frac{n+1}{N} \big) = -\frac{13}{12} + C_{\eta, 1} N^2 + O\big( \frac{1}{N} \big). \tag{6}$$
We see that the difference between $(5)$ and $(6)$ is entirely in the smoothing function $\eta(n/N)$ vs $\eta( \frac{n+1}{N} )$, and this is not apparent in the "formal" manipulation leading to the apparent inconsistency.
My Question

In his post, Terry Tao puts as Exercise 2 that one can use $(3)$ and the Taylor expansion for $\eta(\frac{n+1}{N})$ to derive $(6)$ from $(5)$. (Thus these are not only consistent but essentially equivalent). But I don't see how to do this.


Initial thoughts towards a solution
There are two ways that seem natural to expand $\eta$ in a Taylor series. We could use an expansion centered at $0$, leading to expressions of the form
$$ \eta\left( \frac{n+1}{N} \right) = 1 + \eta'(0) \left( \frac{n+1}{N} \right) + \frac{\eta''(c)}{2} \left( \frac{n+1}{N} \right)^2, \qquad c \in (0, \tfrac{n+1}{N})$$
or perhaps several expansions centered at $n/N$, leading to expressions of the form
$$ \eta \left( \frac{n+1}{N} \right) = \eta \left( \frac{n}{N} \right) + \eta' \left( \frac{n}{N} \right) \frac{1}{N} + \eta''(c) \frac{1}{N^2}, \qquad c \in (\tfrac{n}{N}, \tfrac{n+1}{N}).$$
The expansions centered at $n/N$ have many advantages. In the earlier Exercise 1 in the blog post, using Taylor expansions centered at $n/N$ led to an easy solution, so I suspect that this is the way to proceed. Further, as $\eta$ is compactly supported, we have $\eta''(c) = 0$ for large $c$.
But using this Taylor expansion in the series in $(6)$, we find
$$ \sum_{n \geq 1} (1+n) \eta \left( \frac{n+1}{N} \right) = \sum_{n \geq 1} (1+n) \Big[ \eta \left( \frac{n}{N} \right) + \eta' \left( \frac{n}{N} \right) \frac{1}{N} + \eta''(c_n) \frac{1}{N^2}\Big].$$
The first terms
$$ \sum_{n \geq 1} (1+n) \eta \left( \frac{n}{N} \right)$$
are understood from $(3)$ and $(4)$ above. The error term summands are each $O(1/N)$, leading to an error term of size $O(1)$. (This indicates that one should use probably use one more term in the Taylor expansion, but that's not what I find to be the obstacle).
The secondary term is what I find confusing. We want to understand
$$ \sum_{n \geq 1} \frac{(1+n)}{N} \eta' \left( \frac{n}{N} \right),$$
but how are we to do this? I suspect there is either not much more to this line of thought, or a different line of thought is necessary.
 A: The key to the solution is that one can compute or relate $C_{\eta', j}$ to $C_{\eta, j}$ through integration by parts. For example,
$$ C_{\eta', 0} = \lim_{X \to \infty} \int_0^X \eta'(x) dx = \lim_{X \to \infty} \eta(X) - \eta(0) = - 1.$$
Similarly, you compute $C_{\eta', 1} = - C_{\eta, 0}$ and $C_{\eta'', 1} = 1$.
We must also revisit the equality
$$ \sum_{n \geq 1} (1 + n) \eta(n/N) = -\frac{7}{12} + C_{\eta, 0} N + C_{\eta, 1} N^2 + O(1/N).$$
Note that this holds for any smooth cutoff function with $\eta(0) = 1$. More generally, we have
$$ \sum_{n \geq 1} (1 + n) \eta(n/N) = -\frac{7}{12} \eta(0) + C_{\eta, 0} N + C_{\eta, 1} N^2 + O(1/N).$$
We will apply this equality repeatedly, but with $\eta, \eta'$, and $\eta''$ in place of $\eta$.
Using a Taylor approximation (centered at $n/N$, as in your second sense), we find that
$$\begin{align}
 &\sum_{n \geq 1} (1 + n) \eta( \tfrac{n+1}{N} ) \\
&= \sum_{n \geq 1} (1+n) \eta(n/N) + \sum_{n \geq 1} \frac{1+n}{N} \eta'(n/N) + \sum_{n \geq 1} \frac{1+n}{2N^2} \eta''(n/N) \\
&\quad + \sum_{n \geq 1} \frac{1+n}{6N^3} \eta'''(c_n) \\
&= \left( -\frac{7}{12} + C_{\eta, 0}N + C_{\eta, 1}N^2 + O(1/N)\right) + (C_{\eta', 0} + C_{\eta', 1} N +O(1/N)) \\
&\quad + \frac{C_{\eta'', 1}}{2} + O(1/N).
\end{align}$$
In each simplification, I've only kept the terms that are not dominated by the $O(1/N)$ error term. Plugging in that $C_{\eta', 0} = -1, C_{\eta', 1} = -C_{\eta, 0}$, and $C_{\eta'', 1} = 1$ completes the exercise. $\spadesuit$
