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Bounding the sum by an integral

Note iconIncomplete 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).



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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