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Estimating the size of a partial sum of the harmonic series

We know that is always less than Euler's constant, , and that is always larger than . We can use these inequalities to bound the size of the sum of the first n terms in the harmonic series:

(2.4.1.1)

When combined with an accurate value of , this gives very good bounds on the size of the partial sum of the harmonic series. Euler calculated the value of to 16-digit accuracy. To 24 digits, it is

What if we need even tighter bounds on the size of the partial harmonic series? We start by looking at the difference between ln k and ln (k+1), using the power series expansion for ln (1+x) (see equation (2.4.3) ):

(2.4.1.2)

As expected, the difference between them is a little smaller than 1/k. But we are not done with this equation. If we take value of k from 1 to n and add the left sides together, we get a phenomenon that is known as telescoping:

The right sides add up to our partial harmonic series plus an error term:

Equating these sums, we see that

(2.4.1.3)

We know that approaches as n approaches infinity, and so we have an exact expression for in terms of infinite series:

(2.4.1.4)

where is the zeta function.

Click here to learn more about the zeta function.

We can now see find the exact difference between the partial sum of the harmonic series and ln (n+1) + :

(2.4.1.5)

Each of these infinite sums is smaller than the previous one, so if we truncate the right side after subtracting one of them, we get something that is a little smaller than the partial sum of the harmonic series. If we truncate the right side after adding one of them, we get something that is a little larger than the partial sum of the harmonic series.

We can use improper integrals to bound these sums:

(2.4.1.6)

Putting this all together gives us bounds on the partial sum of the harmonic series that differ by only a little more than :

(2.4.1.7)

 



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