Calculus BC: Series


The Integral Test

Suppose we have a function f (x) , defined for all x≥1 , which is positive and decreasing. This function defines a sequence {f (n)} and a series

f (n) = f (1) + f (2) + ...    

Considering the following figure, we see that

f (2)≤ f (x)dx    

since a rectangle with height f (1) and width 1 is contained within the region below the graph of f from 0 to 1 .

Figure %: The Function Contains the Rectangular Regions

Similarly,

f (3)≤ f (x)dx    

and so on. Thus we have

f (1) + f (2) + ... + f (n)≤f (1) + f (x)dx    

But the left side of this inequality is simply the n th partial sum for the series under consideration. If

f (x)dx    

is defined, then the partial sums are bounded, so the series converges.

This logic goes the other way too. As the figure below demonstrates,

f (1)≥ f (x)dx    

and so on.

Figure %: The Rectangular Regions Contain the Function

Thus

f (1) + f (2) + ... + f (n)≥ f (x)dx    

If does not exist, then the integral becomes arbitrarily large for large n , as do the partial sums for the series. Therefore the series f (n) does not converge. We summarize the results of this section in the following statement.

If f is a positive, decreasing function defined for x≥1 , then f (n) converges if and only if the limit

f (x)dx    

exists. If so, denote this limit by L . Then the value to which the sum converges satisfies the following inequality:

L f (n)≤f (1) + L    

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