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When $a=1$ the sum is given by $ H_{n} $ and we have : $$H_{n}=log(n)+γ+\frac{1}{2n}-\frac{1}{12n^2}+\frac{1}{120n^4} \hspace{0.5cm}.\hspace{.1cm}.\hspace{.1cm}. $$ Does any representation of a similar type exist for arbitrary positive $a$ ?

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  • $\begingroup$ assuming, for instance, $a=1$ - why do you let the series begin with $1/2$ intead with $1/1$ as usual? Is this a typo or intentional? $\endgroup$ Commented Oct 1, 2016 at 4:33
  • $\begingroup$ I'm sorry Gottfried i have corrected the typo. $\endgroup$
    – alex
    Commented Oct 1, 2016 at 5:54

3 Answers 3

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$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 1}^{n}{1 \over 1 + ak} & = {1 \over a}\sum_{k = 0}^{n - 1}{1 \over k + 1 + 1/a} = {1 \over a}\sum_{k = 0}^{\infty} \pars{{1 \over k + 1 + 1/a} - {1 \over k + n + 1 + 1/a}} \\[5mm] & = {1 \over a}\pars{H_{n + 1/a} - H_{1/a}} \end{align}

Now you can use yor asymptotic formula as pleased.

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As Felix Marin answered,$$\sum_{k = 1}^{n}{1 \over 1 + ak}=\frac{H_{n+\frac{1}{a}}-H_{\frac{1}{a}}}{a}$$ Now, using the asymptotics $$H_m=\gamma +\log \left({m}\right)+\frac{1}{2 m}-\frac{1}{12 m^2}+\frac{1}{120 m^4}+O\left(\frac{1}{m^5}\right)$$ $$H_{n+\frac{1}{a}}=\gamma +\log \left({n+\frac{1}{a}}\right)+\frac{1}{2 (n+\frac{1}{a})}-\frac{1}{12 (n+\frac{1}{a})^2}+\frac{1}{120(n+\frac{1}{a} )^4}+O\left(\frac{1}{n^5}\right)$$ Now, developing each term as Taylor series $$\log \left({n+\frac{1}{a}}\right)=\log \left({n}\right)+\frac{1}{a n}-\frac{1}{2 a^2 n^2}+\frac{1}{3 a^3 n^3}-\frac{1}{4 a^4 n^4}+O\left(\frac{1}{n^5}\right)$$ $$\frac{1}{ (n+\frac{1}{a})}=\frac{1}{n}-\frac{1}{a n^2}+\frac{1}{a^2 n^3}-\frac{1}{a^3 n^4}+O\left(\frac{1}{n^5}\right)$$ $$\frac{1}{ (n+\frac{1}{a})^2}=\frac{1}{n^2}-\frac{2}{a n^3}+\frac{3}{a^2 n^4}+O\left(\frac{1}{n^5}\right)$$ $$\frac{1}{ (n+\frac{1}{a})^4}=\frac{1}{n^4}+O\left(\frac{1}{n^5}\right) $$and replacing, you should get

$$\sum_{k = 1}^{n}{1 \over 1 + ak}=\frac{\gamma-H_{\frac{1}{a}} }{a}+\frac{\log \left({n}\right) }{a}+\frac{a+2}{2 a^2 }\frac 1{n}-\frac{a^2+6 a+6}{12 a^3 }\frac 1{n^2}+\frac{a^2+3 a+2}{6 a^4 }\frac 1{n^3}+\frac{a^4-30 a^2-60 a-30}{120 a^5 }\frac 1{n^4}+O\left(\frac{1}{n^5}\right)$$

Take care that, for $a=1$ $$\sum_{k = 1}^{n}{1 \over 1 + k}=H_{n+1}-1=\gamma -1+\log \left({n}\right)+\frac{3}{2 n}-\frac{13}{12 n^2}+\frac{1}{n^3}-\frac{119}{120 n^4}+O\left(\frac{1}{n^5}\right)$$ and not $H_n$ (for which you gave the correct expansion) because of the shift of the index.

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  • $\begingroup$ How do you know that the asymptotic formula holds for non-integer n? $\endgroup$ Commented Oct 1, 2016 at 6:00
  • $\begingroup$ I do not see non-integer $n$ in the problem. I "accepted" the asymptotics for $H_m$ and continued. Would you accept if, instead, I used $$\sum_{k = 1}^{n}{1 \over 1 + ak}=\frac{\psi ^{(0)}\left(n+\frac{1}{a}+1\right)-\psi ^{(0)}\left(1+\frac{1}{a}\right)}{a}$$ $\endgroup$ Commented Oct 1, 2016 at 6:07
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$$\sum_{k=1}^{n}\frac{1}{1+ak} = \frac{1}{a}\sum_{k=1}^{n}\frac{1}{k+\frac{1}{a}}=\frac{H_n}{a}-\frac{1}{a^2}\sum_{k=1}^{n}\frac{1}{k\left(k+\frac{1}{a}\right)}=\frac{1}{a}H_n+O\left(\frac{1}{a^2}\right). $$

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