# 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,

• The terms of this series are as described for any N multiple of 4, zero otherwise. Note that the first term would be, for n=1, $\frac{1}{(N-1)}$ and the last term, for n=N/4, $\frac{1}{(N/2+1)}$ so both are infinitely far away in the AHS. – paul hawkey Aug 30 '16 at 19:34

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.

• Thank you! although I think is ln, not log. – paul hawkey Aug 30 '16 at 21:29
• @paulhawkey Mathematicians often use $\log$ to denote the logarithm with base $e$. – Did Aug 30 '16 at 21:50

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $\ds{\,\verts{\mrm{arg}\pars{z}} < \pi}$.

• @paulhawkey You're welcome. Glad it was helpful to you. – Felix Marin Aug 30 '16 at 21:23
• As witnessed by the "miraculous" cancellations at the end, none of this détour is necessary, really. To invoke the digamma function in this context is (tiring and) to use a sledgehammer to kill a fly. – Did Aug 30 '16 at 21:49
• @Did You are quite right. However, the 'easy one' was already ( or simultaneously ) worked out by $\texttt{Jack D'Aurizio}$. – Felix Marin Aug 30 '16 at 21:54
• Since you are systematically flooding the site with these digamma-tricks and a few related ones (and with next to nothing else), @Jack's answer is not the reason you posted yours and you know it. So much for candor. – Did Aug 30 '16 at 22:02