Bounding the sum by an integral

This article is incomplete. More examples are needed and more complicated situations should be analyzed.

Quick description

Suppose we want to estimate a sum

$\sum_{a<n<b} a_n$

where the sequence $\{a_n\}$ is a monotone sequence of non negative real numbers and $a<b$ are real numbers. Later we can let for example $b\rightarrow +\infty$ and likewise $a\rightarrow -\infty$ but let us keep $a,b$ finite for the moment. In this case is a standard tactic to look for a real function $f$ with the same kind of monotonicity as the sequence $\{a_n\}$ , such that $f(n)=a_n$ for all the $n$ we are interested in. Then we have

$\sum _{a<n<b} a_n \simeq \int_a ^b f(x) dx .$

Usually the choice of the function should be obvious by looking at the sequence $\{a_n\}$ . If for example the sequence $\{a_n\}$ is explicitly given then the first obvious choice would be to replace the discrete parameter $n$ with a continuous variable $x$ and look at the resulting function $f(x)$ .

Prerequisites

calculus

Example 1

Let us look at the partial sums of the harmonic series

$\sum_{k=1} ^n \frac{1}{k}.$

The sequence $a_k=\frac{1}{k}, k=1,2,\ldots$ is a strictly decreasing sequence of positive numbers. Taking $f(x)=\frac{1}{x}$ we recover the well known estimate

$\sum_{k=1} ^n \frac{1}{k}\simeq \int_1 ^n \frac{dx}{x}= \log n$

Example 2

One can use the same technique to prove that for a positive real number $p$ , the over-harmonic series

$\sum_{k=1} ^\infty \frac{1}{n^p},$

converge exactly when $p>1$ and we have the estimate

$\sum_{k=1} ^\infty \frac{1}{n^p}\simeq \frac{1}{p-1},$

whenever $p>1$ .

General discussion

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