summation by part and dirichlet test

hantian

June 27, 2024

it is often-pactised to introduce integral by parts via the product rule. intuitively

duv = u dv + v du,

this is a perfect starting point. however students tends to overlook the other side of this result, that is, the implication of summation.

indeed, the theorem in concern can be proven in a pure integral form:

uv dx =u dv

discretise the above,

i=1nu i(vi+1 vi) = i=1nu ivi+1 i=1nu ivi = i=2nv i(ui1 ui) v1u1 + vn+1un. (1)

write (1) back into integral form, that is

uv v du = uv uv dx.

1 dirichlet test of convergence

equation (1) by itself is interesting, as it offers an alternative way to calculate infinite series, this was first introduced by abel1 as a discrete analog to integral by parts.

however, an important question that pursuing the formula is the convergence of the summation itself. now with abel’s summation, comes the following dirichlet test2

the summation anbn converges if series bn is absolute convergent |bn| < M and series an decreases to zero lim an = 0.

the result is only natural with the abel’s summation. consider the partial sum

SN = i=1Na ibi = i=1Na i(Bi Bi1) = anBn a1B1 i=2n1B i(ai ai1).

obviously, if the conditions are satisfied

lim N|SN| < lim NaNBN a1B1 + M(a1 aN1) = |M b1|a1.

it is also possible to extend this test to continuous functions, as to test the convergence of improper integrals.

integral fg dx converges if f dx is uniformly bounded and none-negative function g monotonically decreasing.

both dirichlet’s test and abel’s test which is delibrately ommitted in this post have important significance in real analysis, that is, when an and bn are functions of x.