Aleah Booth

2022-07-17

Inequality involving sum with parameter in denominator (leading to harmonic number problem)

Given integers $a$, $b$ and $c$, I am trying to find $n\le b-1$ such that this inequality holds :

$\sum _{i=0}^{n}\frac{a}{b-i}\le c\le \sum _{i=0}^{n+1}\frac{a}{b-i}$

That is to say : the largest $n$ up to which the sum is the closest to $c$

Since :

$\sum _{i=0}^{n}\frac{a}{b-i}=a\cdot (\sum _{i=0}^{b}\frac{1}{i}-\sum _{i=0}^{b-n-1}\frac{1}{i})$

And $\sum _{i=0}^{k}\frac{1}{i}={H}_{k}$ where ${H}_{k}$ is the $k$-th harmonic number,

We can rewrite the inequality into :

$a({H}_{b}-{H}_{b-n-1})\le c\le a({H}_{b}-{H}_{b-n-2})$

How to find a tight lower and upper bound for $n$, such that this inequality holds ?

Given integers $a$, $b$ and $c$, I am trying to find $n\le b-1$ such that this inequality holds :

$\sum _{i=0}^{n}\frac{a}{b-i}\le c\le \sum _{i=0}^{n+1}\frac{a}{b-i}$

That is to say : the largest $n$ up to which the sum is the closest to $c$

Since :

$\sum _{i=0}^{n}\frac{a}{b-i}=a\cdot (\sum _{i=0}^{b}\frac{1}{i}-\sum _{i=0}^{b-n-1}\frac{1}{i})$

And $\sum _{i=0}^{k}\frac{1}{i}={H}_{k}$ where ${H}_{k}$ is the $k$-th harmonic number,

We can rewrite the inequality into :

$a({H}_{b}-{H}_{b-n-1})\le c\le a({H}_{b}-{H}_{b-n-2})$

How to find a tight lower and upper bound for $n$, such that this inequality holds ?

agyalapi60

Beginner2022-07-18Added 17 answers

Let us rewrite the inequality as

${S}_{n}\u2a7d\frac{c}{a}\u2a7d{S}_{n+1}\phantom{\rule{thinmathspace}{0ex}},$

where ${S}_{n}=\frac{1}{b}\sum _{i=0}^{n}f(\frac{i}{b})$ and $f$ is defined by $x\mapsto \frac{1}{1-x}$ for $x$ in $[0,1[$. Thus, one can view ${S}_{n}$ as the rectangle method of numerical integration applied to $f$. Since $f$ is strictly increasing, one has

$\frac{1}{b}f({\textstyle \frac{i}{b}})<{\int}_{i/b}^{(i+1)/b}f(x)\phantom{\rule{thinmathspace}{0ex}}dx<\frac{1}{b}f({\textstyle \frac{i+1}{b}})\phantom{\rule{thinmathspace}{0ex}}.$

By summation over $i$, one obtains for all $n$

${S}_{n}<\underset{{I}_{n+1}}{\underset{\u23df}{{\int}_{0}^{(n+1)/b}f(x)\phantom{\rule{thinmathspace}{0ex}}dx}}<{S}_{n+1}-{S}_{0}<{S}_{n+1}\phantom{\rule{thinmathspace}{0ex}},$

where the integral is given by ${I}_{n+1}=-\mathrm{ln}(1-{\textstyle \frac{n+1}{b}})$. Let us examine two cases:

if $c/a$ is larger than ${S}_{b-1}$, then no such $n$ can be found. Taking $n=b-1$ insures ${S}_{n}\u2a7dc/a$ only.

else, we can find $n$ such that the first inequality is satisfied. If ${S}_{n}\u2a7dc/a<{I}_{n+1}$, then ${I}_{n}<c/a<{I}_{n+1}$. Therefore,

$n=\lfloor b(1-\mathrm{exp}(-\frac{c}{a}))\rfloor ,$

where $\lfloor \cdot \rfloor $ denotes the floor function. Otherwise, ${I}_{n+1}\u2a7dc/a\u2a7d{S}_{n+1}$, and

$n=\lfloor b(1-\mathrm{exp}(-\frac{c}{a}))\rfloor -1\phantom{\rule{thinmathspace}{0ex}}.$

${S}_{n}\u2a7d\frac{c}{a}\u2a7d{S}_{n+1}\phantom{\rule{thinmathspace}{0ex}},$

where ${S}_{n}=\frac{1}{b}\sum _{i=0}^{n}f(\frac{i}{b})$ and $f$ is defined by $x\mapsto \frac{1}{1-x}$ for $x$ in $[0,1[$. Thus, one can view ${S}_{n}$ as the rectangle method of numerical integration applied to $f$. Since $f$ is strictly increasing, one has

$\frac{1}{b}f({\textstyle \frac{i}{b}})<{\int}_{i/b}^{(i+1)/b}f(x)\phantom{\rule{thinmathspace}{0ex}}dx<\frac{1}{b}f({\textstyle \frac{i+1}{b}})\phantom{\rule{thinmathspace}{0ex}}.$

By summation over $i$, one obtains for all $n$

${S}_{n}<\underset{{I}_{n+1}}{\underset{\u23df}{{\int}_{0}^{(n+1)/b}f(x)\phantom{\rule{thinmathspace}{0ex}}dx}}<{S}_{n+1}-{S}_{0}<{S}_{n+1}\phantom{\rule{thinmathspace}{0ex}},$

where the integral is given by ${I}_{n+1}=-\mathrm{ln}(1-{\textstyle \frac{n+1}{b}})$. Let us examine two cases:

if $c/a$ is larger than ${S}_{b-1}$, then no such $n$ can be found. Taking $n=b-1$ insures ${S}_{n}\u2a7dc/a$ only.

else, we can find $n$ such that the first inequality is satisfied. If ${S}_{n}\u2a7dc/a<{I}_{n+1}$, then ${I}_{n}<c/a<{I}_{n+1}$. Therefore,

$n=\lfloor b(1-\mathrm{exp}(-\frac{c}{a}))\rfloor ,$

where $\lfloor \cdot \rfloor $ denotes the floor function. Otherwise, ${I}_{n+1}\u2a7dc/a\u2a7d{S}_{n+1}$, and

$n=\lfloor b(1-\mathrm{exp}(-\frac{c}{a}))\rfloor -1\phantom{\rule{thinmathspace}{0ex}}.$

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