Estimating the size of a partial sum of the
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:
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
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
We know that approaches as n approaches
infinity, and so we have an exact expression for in
terms of infinite series:
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) + :
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:
Putting this all together gives us bounds on the partial sum
of the harmonic series that differ by only a little more than :