Can this sum be defined? $\lim_{N\to \infty}\sum_{n=1}^\frac{N}{4} \frac{1}{(N-2n+1)}$ The alternating harmonic series is 1-1/2+1/3-1/4+1/5-1/6+1/7-1/8+... I am trying to see if I can eliminate the last N/4 positive terms. 
For instance, if the above were just a partial sum of the AHS (no "+..."), I would eliminate 1/5 and 1/7. When the number of terms grows to infinity, the number of terms I need to lop off has an infinite number of terms as well (N/4). Strangely enough, this series converges (and it's greater than zero) even though it belongs to the divergent series of all positive terms. But from a theoretical point of view I am a bit stumped because as N tends to infinity, I need to define a series of positive terms $\lim_{N\to \infty}\sum_{n=1}^\frac{N}{4} \frac{1}{(N-2n+1)}$ where none of the terms seem to exist because I cannot even define which would be the first term. What do you think? 
(the terms of this series are as described for any N multiple of 4, 
 A: By Riemann sums
$$\lim_{M\to +\infty}\sum_{n=1}^{M}\frac{1}{4M-2n+1}=\int_{0}^{1}\frac{dx}{4-2x}=\color{red}{\log\sqrt{2}}$$
as well as
$$ \sum_{1\leq n\leq\frac{N}{4}}\frac{1}{N-2n+1}=\log\sqrt{2}+O\left(\frac{1}{N}\right) $$
no matter what $N\pmod{4}$ is, as soon as $N$ is large enough.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
&\color{#f00}{\lim_{N \to \infty}
\sum_{n = 1}^{\left\lfloor\, N/4\, \right\rfloor}{1 \over N - 2n + 1}} =
\lim_{N \to \infty}
\sum_{n = 1}^{\left\lfloor\, N/4\, \right\rfloor}\int_{0}^{1}x^{N - 2n}
\,\,\,\dd x =
\lim_{N \to \infty}\,\,\int_{0}^{1}x^{N}
\sum_{n = 1}^{\left\lfloor\, N/4\, \right\rfloor}\pars{x^{-2}}^{n}
\,\,\,\dd x
\\[5mm] & =
\lim_{N \to \infty}\,\,\int_{0}^{1}
{x^{N - 2\left\lfloor\, N/4\, \right\rfloor}\,\,\,\,\,\, -\,\,\, x^{N} \over
1 - x^{2}}\,\,\,\dd x =
\half\,\lim_{N \to \infty}\,\,\int_{0}^{1}
{x^{N/2 - \left\lfloor\, N/4\, \right\rfloor - 1/2}\,\,\,\,\,\, -
\,\,\, x^{N/2 - 1/2} \over 1 - x}\,\,\,\dd x
\\[5mm] & =
\half\,\lim_{N \to \infty}\,\,\,\bracks{\Psi\pars{{N \over 2} + \half} -
\Psi\pars{{N \over 2} - \left\lfloor\,{N \over 4}\, \right\rfloor +
\half}\vphantom{\Huge A^{a}}}\quad\quad
\pars{~\Psi:\ Digamma\ Function~}
\end{align}

\begin{align}
&\color{#f00}{\lim_{N \to \infty}
\sum_{n = 1}^{\left\lfloor\, N/4\, \right\rfloor}{1 \over N - 2n + 1}} =
\half\,\lim_{N \to \infty}\,
\ln\pars{N/2  \over N/2 - \left\lfloor\, N/4\, \right\rfloor}
\\[5mm] = &\
-\,\half\,\lim_{N \to \infty}
\ln\pars{1 - {\left\lfloor\, N/4\, \right\rfloor \over N/2}} =
-\,\half\,\ln\pars{1 - \half} = 
\color{#f00}{\half\,\ln\pars{2}} \approx 0.3466
\end{align}


Note that
  $\ds{\Psi\pars{z + \half} \sim \ln\pars{z}  + {1 \over 24z^{2}} - {7 \over 960z^{4}} +
{31 \over 8064z^{6}} + \cdots\quad}$ as $\ds{\verts{z} \to \infty}$ with
  $\ds{\,\verts{\mrm{arg}\pars{z}} < \pi}$.

