Dirichlet’s Test
Let \((a_n)\) and \((b_n)\) be sequences. Suppose \(a_n\) is non-negative and is decreasing to \(0\) that is,
\[
a_n\ge 0,\qquad a_{n+1}\le a_n,\qquad \lim_{n\to\infty}a_n=0,
\]and the partial sums of \((b_n)\) are bounded so that
\[
B: = \sup_{N\ge 1} \left| \sum_{n=1}^{N} b_n \right| <\infty.
\]
Then \(\sum_{n=1}^{\infty}a_nb_n\) converges.
Proof
Let \(B_N=\sum_{n=1}^{N}b_n\) and define \(B_0=0\). Then
$$
\begin{align*}
B_n &= b_1+b_2+\cdots+b_n \\
B_{n-1} &= b_1+b_2+\cdots+b_{n-1}.
\end{align*}
$$
Subtracting gives \(b_n=B_n-B_{n-1}\). Consider a finite tail of the series \(\sum_{n=M}^{N}a_nb_n\).
Using \(b_n=B_n-B_{n-1}\). Re-write it as follows
$$
\begin{align*}
\sum_{n=M}^{N}a_nb_n &= \sum_{n=M}^{N}a_n(B_n-B_{n-1}) \\
&= \sum_{n=M}^{N}a_nB_n - \sum_{n=M}^{N}a_nB_{n-1}. \tag{1}
\end{align*}
$$
Now focus on the second sum above \(\sum_{n=M}^{N}a_nB_{n-1}\).
$$
\begin{align*}
k=n-1.
\end{align*}
$$
Then \(n = k+1\). The endpoints also change:
$$
\begin{align*}
n=M\implies k=M-1,
\qquad
n=N\implies k=N-1.
\end{align*}
$$
Therefore,
$$
\begin{align*}
\sum_{n=M}^{N}a_nB_{n-1} = \sum_{k=M-1}^{N-1}a_{k+1}B_k.
\end{align*}
$$
Since \(k\) is just a placeholder. We can rename it back to \(n\) and also plug it in (1)
$$
\begin{align*}
\sum_{n=M}^{N}a_nb_n = \sum_{n=M}^{N}a_nB_n - \sum_{n=M-1}^{N-1}a_{n+1}B_n.
\end{align*}
$$
Now observe that
$$
\begin{align*}
\sum_{n=M}^{N} a_n b_n
&= \sum_{n=M}^{N} a_nB_n
-\sum_{n=M-1}^{N-1}a_{n+1}B_n \\[4pt]
&= \left(
\sum_{n=M}^{N-1}a_nB_n+a_NB_N
\right)
-
\left(
a_MB_{M-1}
+\sum_{n=M}^{N-1}a_{n+1}B_n
\right) \\[4pt]
&= a_NB_N-a_MB_{M-1}
+\sum_{n=M}^{N-1}a_nB_n
-\sum_{n=M}^{N-1}a_{n+1}B_n \\[4pt]
&= a_NB_N-a_MB_{M-1}
+\sum_{n=M}^{N-1}
\left(a_nB_n-a_{n+1}B_n\right) \\[4pt]
&= a_NB_N-a_MB_{M-1}
+\sum_{n=M}^{N-1}
(a_n-a_{n+1})B_n.
\end{align*}
$$
We know the partial sums \(B_n\) are bounded. Let
$$
\begin{align*}
C=\sup_{n\ge 1}|B_n|<\infty.
\end{align*}
$$
Therefore, \(|B_n|\le C\) for every \(n\). Now apply the triangle inequality to our formula. Also recall that \(a_n\) is decreasing and non-negative. Hence
$$
\begin{align*}
\left| \sum_{n=M}^{N}a_nb_n \right|
&= \left| a_NB_N-a_MB_{M-1} +\sum_{n=M}^{N-1}(a_n-a_{n+1})B_n \right| \\[4pt]
&\le a_N|B_N| +a_M|B_{M-1}| +\sum_{n=M}^{N-1} (a_n-a_{n+1})|B_n| \\
&\le
a_N|B_N|
+a_M|B_{M-1}|
+\sum_{n=M}^{N-1}(a_n-a_{n+1})|B_n| \\[4pt]
&\le
Ca_N+Ca_M
+C\sum_{n=M}^{N-1}(a_n-a_{n+1}) \\[4pt]
&=
C\left(
a_N+a_M+
\sum_{n=M}^{N-1}(a_n-a_{n+1})
\right).
\end{align*}
$$
Now observe that
$$
\begin{align*}
\sum_{n=M}^{N-1}(a_n-a_{n+1}) &=(a_M-a_{M+1}) +(a_{M+1}-a_{M+2})+\cdots +(a_{N-1}-a_N) = a_M-a_N.
\end{align*}
$$
Hence
$$
\begin{align*}
\left| \sum_{n=M}^{N}a_nb_n \right| &\le C\left(a_N+a_M+a_M-a_N\right) =2Ca_M.
\end{align*}
$$
Since \(2C\) is a fixed constant and since \(a_M\to 0\), then as \(M\to\infty\), we have \(2Ca_M\to 0\). Hence by definition, for all \(\varepsilon>0\), there exists an index \(M_0\) such that whenever \(M\ge M_0\),
$$
\begin{align*}
|2Ca_M-0| &< \varepsilon \\
2Ca_M &< \varepsilon \quad \text{(since $2Ca_m \geq 0$)} \tag{2}
\end{align*}
$$
Now choose any two integers \(M\) and \(N\) satisfying \(N\ge M\ge M_0\). By (2)
$$
\begin{align*}
\left|\sum_{n=M}^{N}a_nb_n\right|\le 2Ca_M <\varepsilon.
\end{align*}
$$
Therefore, by the Cauchy criterion for series \(\sum_{n=1}^{\infty}a_nb_n\) converges. \(\blacksquare\)
Example 1
Let \(b_n=(-1)^n\). Dirichlet’s test asks whether the partial sums of \(b_n\) are bounded. Define
$$
\begin{align*}
B_N=\sum_{n=1}^{N}(-1)^n.
\end{align*}
$$
Let’s calculate the first few:
$$
\begin{align*}
B_1 &= -1 \\
B_2 &= -1+1=0 \\
B_3 &= -1+1-1=-1 \\
B_4 &= -1+1-1+1=0.
\end{align*}
$$
Thus,
$$
\begin{align*}
B_N =
\begin{cases}
-1, & N\text{ odd},\\[4pt]
0, & N\text{ even}.
\end{cases}
\end{align*}
$$
Therefore, \(|B_N|\le1\). Eequivalently,
$$
\begin{align*}
\sup_{N\ge1}
\left| \sum_{n=1}^{N}(-1)^n \right| =1<\infty.
\end{align*}
$$
So \(b_n=(-1)^n\) satisfies the bounded-partial-sums condition in Dirichlet’s test.
Consequently, for any sequence \(a_n\searrow0\),
$$
\begin{align*}
\sum_{n=1}^{\infty}a_n(-1)^n
\end{align*}
$$
converges.
Example 2
Recall the sawtooth example from the previous lecture where
$$
\begin{align*}
f(x)=x,\qquad -\pi< x < \pi,
\end{align*}
$$
then its Fourier series is
$$
\begin{align*}
f(x)\sim \sum_{n=1}^{\infty} \frac{2(-1)^{n-1}}{n}\sin(nx).
\end{align*}
$$
Let
$$
\begin{align*}
a_n=\frac{2}{n},
\qquad
b_n(x)=(-1)^{n-1}\sin(nx),
\end{align*}
$$
Note that \(a_n\) is non-negative, decreasing and converges to zero.
Define
$$
\begin{align*}
B_M(x)=\sum_{n=1}^{M}b_n(x) = \sum_{n=1}^{M}(-1)^{n-1}\sin(nx).
\end{align*}
$$
Our task now is to show that the partial sum made out of \(b_n\)’s is actually bounded in order to use the Dirichlet’s test. In other words, we need to show that, for each fixed \(x\), there is some finite number \(C_x\) such that
$$
\begin{align*}
\left|
\sum_{n=1}^{M}(-1)^{n-1}\sin(nx)
\right|
\le C_x
\quad\text{for every }M.
\end{align*}
$$
Equivalently,
$$
\begin{align*}
\sup_{M\ge1} \left| \sum_{n=1}^{M}(-1)^{n-1}\sin(nx) \right| <\infty.
\end{align*}
$$
The bound \(C_x\) may depend on \(x\), but it must not depend on \(M\). That will prove pointwise convergence by Dirichlet’s test. Now, use the following identity
$$
\begin{align*}
(-1)^{n-1}\sin(nx)=\sin\bigl(n(\pi-x)\bigr).
\end{align*}
$$
To verify it, observe that
$$
\begin{align*}
\sin\bigl(n(\pi-x)\bigr)
&=\sin(n\pi-nx)\\
&=\sin(n\pi)\cos(nx)-\cos(n\pi)\sin(nx)\\
&=0-(-1)^n\sin(nx)\\
&=(-1)^{n-1}\sin(nx).
\end{align*}
$$
Therefore,
$$
\begin{align*}
B_M(x) &=\sum_{n=1}^{M}(-1)^{n-1}\sin(nx)\\
&=\sum_{n=1}^{M}\sin\bigl(n(\pi-x)\bigr).
\end{align*}
$$
If we let \(y=\pi-x\). Then
$$
\begin{align*}
B_M(x)=\sum_{n=1}^{M}\sin(ny).
\end{align*}
$$
Recall Euler’s formula \(e^{iny}=\cos(ny)+i\sin(ny)\). Hence \(\operatorname{Im}\left(e^{iny}\right)=\sin(ny)\). So we can rewrite the partial sum as
$$
\begin{align*}
B_M(x)
&=\sum_{n=1}^{M}\sin(ny)\\
&=\sum_{n=1}^{M}\operatorname{Im}\left(e^{iny}\right)\\
&=\operatorname{Im}\left(\sum_{n=1}^{M}e^{iny}\right).
\end{align*}
$$
This is a geometric series since if we let \(r=e^{iy}\), then
$$
\begin{align*}
\sum_{n=1}^{M}e^{iny} = \sum_{n=1}^{M}r^n = r+r^2+\cdots+r^M = \frac{r(1-r^M)}{1-r}.
\end{align*}
$$
Assuming \(e^{iy}\neq1\), we obtain
$$
\begin{align*}
\sum_{n=1}^{M}e^{iny} &= \frac{e^{iy}\left(1-e^{iMy}\right)}{1-e^{iy}}.
\end{align*}
$$
Therefore,
$$
\begin{align*}
B_M(x) = \operatorname{Im}\left(
\frac{e^{iy}\left(1-e^{iMy}\right)}
{1-e^{iy}}
\right),
\qquad y=\pi-x.
\end{align*}
$$
The condition \(e^{iy}\neq1\) implies \(y\notin2\pi\mathbb Z\). Now recall that for every complex number \(z\),
$$
\begin{align*}
|\operatorname{Im}(z)|\le |z|.
\end{align*}
$$
Therefore,
$$
\begin{align*}
|B_M(x)|
&=
\left|
\operatorname{Im}\left(
\frac{e^{iy}(1-e^{iMy})}{1-e^{iy}}
\right)
\right|\\[4pt]
&\le
\left|
\frac{e^{iy}(1-e^{iMy})}{1-e^{iy}}
\right|\\[4pt]
&=
\frac{|e^{iy}|\,|1-e^{iMy}|}
{|1-e^{iy}|}.
\end{align*}
$$
Recall that our goal is to bound the term \(|B_M(x)|\). The bound can depend on \(x\) but it should depend on \(M\). Since there is an \(M\) term in the numerator, we need to get rid of it. Note first that \(e^{iy}\) lies on the unit circle, \(|e^{iy}|=1\). Thus,
$$
\begin{align*}
|B_M(x)| \le \frac{|1-e^{iMy}|}{|1-e^{iy}|}.
\end{align*}
$$
Moreover, using the triangle inequality,
$$
\begin{align*}
|z-w|\le |z|+|w|.
\end{align*}
$$
Therefore,
$$
\begin{align*}
|1-e^{iMy}|
&\le |1|+|e^{iMy}| =1+1 =2.
\end{align*}
$$
Substituting this into our bound gives
$$
\begin{align*}
|B_M(x)| &\le \frac{|1-e^{iMy}|}{|1-e^{iy}|}\\[4pt]
&\le \frac{2}{|1-e^{iy}|}.
\end{align*}
$$
Thus,For a fixed \(y\) with \(e^{iy}\neq1\), we found
$$
\begin{align*}
|B_M(x)|\le\frac{2}{|1-e^{iy}|}.
\end{align*}
$$
The right-hand side may depend on \(y\), but it does not depend on \(M\). Therefore,
$$
\begin{align*}
\sup_{M\ge1}|B_M(x)|
\le
\frac{2}{|1-e^{iy}|}
<\infty.
\end{align*}
$$
Now, let’s handle the case when \(e^{iy} = 1\). This happens when \(y = 2\pi t\) where \(t\) is an integer. Recall that \(y = \pi - x\). Hence, \(x = \pi - 2\pi t\). If we plug in this value in the original series, then
$$
\begin{align*} d
B_M(x) &=\sum_{n=1}^{M}\sin\bigl(n(\pi-x)\bigr) \\
&=\sum_{n=1}^{M}\sin\bigl(n(2\pi t\bigr)) = 0.
\end{align*}
$$
So \(B_M(x)=0\) for every \(M\). Thus the partial sums are bounded there too. We have now verified both Dirichlet conditions:
$$
\begin{align*}
a_n=\frac2n\searrow0
\end{align*}
$$
and
$$
\begin{align*}
\sup_{M\ge1}
\left|
\sum_{n=1}^{M}b_n(x)
\right|<\infty.
\end{align*}
$$
Therefore,
$$
\begin{align*}
\sum_{n=1}^{\infty}
\frac{2(-1)^{n-1}}{n}\sin(nx)
\end{align*}
$$
converges for each fixed \(x\).
Does the Sawtooth Series Converge Uniformly?
In the Dirichlet test above we let \(a_n = 2/n\) and \(b_n(x)=(-1)^{n-1}\sin(nx)\), then we showed that the partial sum made out of \(b_n\)’s is bounded. Hence the Dirichlet’s test confirms the series \(f(x)\) converges pointwise.
But now we are now asking a stronger question:
Does the series converge uniformly in \(x\)?
For pointwise convergence, the bound may depend on the chosen point:
$$
\begin{align*}
|B_M(x)|\le C_x.
\end{align*}
$$
We fix \(x\), and then \(C_x\) is just some finite number. For uniform convergence on a set \(E\), we need one constant that works simultaneously for every \(x\in E\):
$$
\begin{align*}
|B_M(x)|\le C
\qquad
\text{for every }M\text{ and every }x\in E.
\end{align*}
$$
In our \(y\)-coordinates, the bound was
$$
\begin{align*}
|B_M(y)| \le \frac{2}{|1-e^{iy}|}.
\end{align*}
$$
This becomes very large as \(y\to0\) or \(y\to2\pi\). So it is not a uniform bound on all of \([0,2\pi]\) Hence if we stay away from those problem points and restrict to
$$
\begin{align*}
y\in[\delta,2\pi-\delta],
\end{align*}
$$
then the denominator stays away from zero. We can obtain one constant that works for every \(y\) in that interval. Returning to the original variable \(x\), this means uniform convergence on
$$
\begin{align*}
[-\pi+\delta,\pi-\delta]
\end{align*}
$$
away from the sawtooth’s jump points.
References
- SFSU Fourier Analysis by Professor Chun-Kit Lai
- Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi