[EX] Series

[gugo82]

My first thread here in a while.
Enjoy!

***

Exercise:

Let \(x,y,z\) be real numbers.

1. Find a closed form for:
\[
\tag{A}
\sum_{n=0}^N (y+nz)\ x^n\; ,
\]
where \(N\in \mathbb{N}\).

2. Let \(y,z\) be held fixed. Find both the convegence set and the sum of the series:
\[
\tag{B}
\sum_{n=0}^\infty (y+nz)\ x^n
\]
w.r.t. the variable \(x\) alone.

3. Find the convergence set and analyse the mode of convergence of the series (B) w.r.t. the three variables \(x,y,z\).

[gugo82]

"gugo82":Exercise:
Let \(x,y,z\) be real numbers.
1. Find a closed form for:
\[
\tag{A}
\sum_{n=0}^N (y+nz)\ x^n\; ,
\]
where \(N\in \mathbb{N}\).

Let:
\[
S_N:=\sum_{n=0}^N (y+nz)\ x^n\; .
\]
First, assume \(x=1\). Then:
\[
\begin{split}
S_N &= \sum_{n=0}^N (y+nz)\\
&= y\ \sum_{n=0}^N 1 + z\ \sum_{n=0}^N n\\
&= y\ (N+1) + z\ \frac{N(N+1)}{2}\\
&= \frac{N+1}{2}\ (y+Nz)\; .
\end{split}
\]
Now, assume \(x\neq 1\). Then:
\[
\begin{split}
S_N &= y &+ (y+z)\ x &+ (y+2z)\ x^2 &+\cdots &+ (y+Nz)\ x^N\\
-x\ S_N &= &-y\ x &- (y+z)\ x^2 &-\cdots &- (y+(N-1)z)\ x^N - (y+Nz)\ x^{N+1}\\
\hline\\
(1-x)\ S_N &= y &+ z\ x &+z\ x^2&+\cdots &+z\ x^N - (y+Nz)\ x^{N+1}
\end{split}
\]
and a little algebra gives:
\[
\begin{split}
S_N &= y\ \frac{1-x^{N+1}}{1-x} +z\ \frac{1}{1-x}\ \left(x\ \sum_{n=0}^{N-1} x^n - N\ x^{N+1}\right) \\
&=y\ \frac{1-x^{N+1}}{1-x} + z\ \frac{1}{1-x}\ \left( x\ \frac{1-x^N}{1-x} - N\ x^{N+1}\right)\\
&=y\ \frac{1-x^{N+1}}{1-x} + z\ \left( \frac{1-x^N}{1-x} - N\ x^N\right)\ \frac{x}{1-x}\\
&= y\ \frac{1-x^{N+1}}{1-x} + z\ \frac{1-(N+1)\ x^N + x^{N+1}}{(1-x)^2}\ x\; .
\end{split}
\]
Therefore the closed form to be determined is:
\[
\tag{1}
S_N = \begin{cases} \frac{N+1}{2}\ (y+Nz) &\text{, if } x=1\\
y\ \frac{1-x^{N+1}}{1-x} + z\ \frac{1-(N+1)\ x^N + x^{N+1}}{(1-x)^2}\ x &\text{, otherwise.}
\end{cases}
\]

*

"gugo82":2. Let \(y,z\) be held fixed. Find both the convegence set and the sum of the series:
\[
\tag{B}
\sum_{n=0}^\infty (y+nz)\ x^n
\]
w.r.t. the variable \(x\) alone.

If \(y=0=z\), then series (B) converges in the whole of \(\mathbb{R}\) to the zero function \(f(x;0,0)=0\).
So, assume \(y\neq 0\) or \(z\neq 0\). Then, passing (1) to the limit, it becomes clear the series (B) converges iff \(|x|<1\) to the function:
\[
\begin{split}
f(x;y,z) &:= \frac{y}{1-x} + \frac{z\ x}{(1-x)^2} \\
&= \frac{y+(z-y)\ x}{(1-x)^2}\; .
\end{split}
\]

*

"gugo82":3. Find the convergence set and analyse the mode of convergence of the series (B) w.r.t. the three variables \(x,y,z\).

Thanks to part 2, series (B) converges in the set:
\[
D:=\underbrace{\{(x,0,0),\ x\in \mathbb{R}\}}_{=:D_0} \cup \underbrace{\{(x,y,z):\ |x|<1\}}_{=: D_1} \subseteq \mathbb{R}^3
\]
(which is the union of the \(x\)-axis \(D_0\) and of the unbounded double layer \(D_1\)) to the three variables function:
\[
s(x,y,z) := f(x;y,z) = \begin{cases}
0 &\text{, if } (x,y,z)\in D_0\\
\frac{y+(z-y)\ x}{(1-x)^2} &\text{, if } (x,y,z)\in D_1\; .
\end{cases}
\]
Let \(M_n\) be the least upper bound of \(|y+nz|\ |x|^n\) in \(D\). Then \(M_n=+\infty\) and Weierstrass test does not apply to (B) over \(D\).
On the other hand, the least upper bound \(M_n(x_0,y_0,z_0)\) of \(|y+nz|\ |x|^n\) over any region of the type:
\[
D(x_0,y_0,z_0):=D_0\cup \big( ]-x_0,x_0[\times ]-y_0,y_0[\times ]-z_0,z_0[ \big)
\]
(with \(00\)) is less than \(( y_0+nz_0)\ x_0^n\), therefore (B) converges totally (thus uniformly and absolutely) in \(D(x_0,y_0,z_0)\) by the Weierstrass test.
Thus, the convergence of (B) is total over bounded sets contained in \(D\) for every bounded set \(B\subset D\) is contained into a suitable set of the type \(D(x_0,y_0,z_0)\).

[gugo82]

Hints:

1. Let \(S_N\) be the sum to solve. Use the same trick usually used to find the sum of a geometric progression, i.e. evaluate \(xS_N\) and subtract from \(S_N\).
2. Pass (A) to the limit.
3. (B) is a series of functions of three real variables, hence Weierstrass test for total convergence should be used.