Expanded form of $\sum _{k=0}^n \frac{1}{k + a}$

Question

I have a linear combination of the following sum: $\sum _{k=0}^n \frac{1}{k + a}$, $a$ being a real number.

Let's say I have:

$H(n) = \sum _{k=0}^n \frac{1}{k + 2} + \sum _{k=0}^n \frac{1}{k + 3} - 2 \sum _{k=0}^n \frac{1}{k + 4} $

Here is my problem: the series $\sum _{k} \frac{1}{k + a}$ diverges by comparison with the harmonic series.

However, I know H(n) converges, but I cannot show why and what is the limit. Can I expand $\sum _{k=0}^n \frac{1}{k + a}$ in a more specific expression in that purpose? Thanks.

Answer 1

Hint: write

$$\sum_{k=0}^n\frac{1}{k+a}=\sum_{k=a}^{n+a}\frac{1}{k}$$

Take out the terms that don't match in the sums so that both the limits of the sums match as well as the term you sum over (with the above, that is $\tfrac1k$). Then you have only loose terms left and you can subtract the sums, leading to something that converges.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note $\ds{\sum_{k = 0}^{n}{1 \over k + a} = \sum_{k = a}^{n + a}{1 \over k} = \sum_{k = 1}^{n + a}{1 \over k} - \sum_{k = 1}^{a - 1}{1 \over k} = H_{n + a} - H_{a - 1}}$.

Then \begin{align} \mrm{H}\pars{n} & = \pars{H_{n + 2} - H_{1}} + \pars{H_{n + 3} - H_{2}} - 2\pars{H_{n + 4} - H_{3}} \\[5mm] & = H_{n + 2} + H_{n + 3} - 2H_{n + 4} - 1 - {3 \over 2} + 2\,{11 \over 6} = {7 \over 6} + H_{n + 2} - H_{n + 3} - {2 \over n + 4} \\[5mm] & = {7 \over 6} - {1 \over n + 3} - {2 \over n + 4} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, \bbx{7 \over 6} \end{align}