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\begin{document}

\begin{titlepage}

  \begin{center}

    {\large Nested Radicals} \vspace{1cm} \\

    And Other Infinitely Recursive Expressions \vspace{3cm} \\

    Michael M$^C$Guffin \vspace{3cm} \\ % Michael McGuffin

    {\small {\em prepared} } \\
    % \today \vspace{2cm} \\
    July 17, 1998 \vspace{2cm} \\

    {\small {\em for} } \\
    The Pure Math Club \\
    University of Waterloo \\

  \end{center}

\end{titlepage}

\begin{center}Outline\end{center} \vspace{1.30cm}
1. Introduction \vspace{0.60cm} \\
2. Derivation of Identities \vspace{-0.60cm}
\begin{list}{}{}
   \item 2.1 Constant Term Expansions \vspace{-0.60cm}
   \item 2.2 Identity Transformations \vspace{-0.60cm}
   \item 2.3 Generation of Identities Using Recurrences \vspace{-0.60cm}
\end{list}
3. General Forms \vspace{0.60cm} \\
4. Selected Results from Literature

\pagebreak

{\large 1. Introduction} \\
Examples of Infinitely Recursive Expressions
\linebreak[2]

Series

   \[
      e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!}
          + \frac{1}{4!} + \ldots
   \]

Infinite Products

   % Wallis
   %
   \[ \pi/2 = \frac{2}{1} \times \frac{2}{3} \times \frac{4}{3}
      \times \frac{4}{5} \times \frac{6}{5} \times \cdots
   \]

Continued Fractions

   %  Euler
   %
   \[ e-1 = 1+\frac{2}{2+\frac{3}{3+\frac{4}{4+\frac{5}{5+\ldots}}}} \]

   % \[ e = 2 + \frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{1+\frac{1}{4+
   %       \ldots
   % %     \frac{1}{1+\frac{1}{1+\frac{1}{6+\ldots}}}
   % }}}}} \]

   % Lord Brouncker
   %
   \[ 4/\pi = 1+\frac{1}{2+\frac{3^2}{2+\frac{5^2}{2+\frac{7^2}{2+\ldots}}}}
   \]

   % \[ \pi/2 = 1-\frac{1}{3-\frac{2*3}{1-\frac{1*2}{3-\frac{4*5}{1-
   %    \frac{3*4}{3-\ldots}
   % }}}} \]

Infinitely Nested Radicals (or Continued Roots)

   % Kasner number
   %
   \[ K = \sqrt{1+\sqrt{2+\sqrt{3+\ldots}}} \]

Exponential Ladders (or Towers)

   \[ 2 = {(\sqrt{2})}^{{(\sqrt{2})}^{{(\sqrt{2})}^{\cdots}}} \]

Hybrid Forms

   % Can be trivially transformed into an exponential ladder.
   %
   \[ 4 = 2^{\sqrt{2^{\sqrt{2^{\sqrt{2^{\cdots}}}}}}} \]

   % This is actually a continued fraction in disguise.
   %
   \[ \frac{1}{2} = \frac{1}{
         \frac{1}{\frac{1}{\ldots}+1+\frac{1}{\ldots}}
         +1+
         \frac{1}{\frac{1}{\ldots}+1+\frac{1}{\ldots}}
      }
   \]

\pagebreak

Questions: \vspace{-1.25cm}
\begin{itemize}
\item Does the expression converge ?  Are there {\em tests},
      or necessary/sufficient conditions for convergence ?
   Examples: \vspace{-0.65cm}
      \begin{itemize}
         \item For series, \vspace{-0.65cm}
         \begin{itemize}
            \item Terms must go to zero \vspace{-0.65cm}
            \item d'Alembert-Cauchy Ratio Test, Cauchy {\em n}th Root Test,
            Integral Test, ...
         \end{itemize}
         \item For infinite products, \vspace{-0.65cm}
         \begin{itemize}
            \item Terms must go to a value in (-1,1]
         \end{itemize}
         \item For infinitely nested radicals, \vspace{-0.65cm}
         \begin{itemize}
            \item Terms can grow !  (But how fast ?)
         \end{itemize}
      \end{itemize}
\item What does the expression converge to ?
      Are there formulae or identities we can use 
      to evaluate the limit\nolinebreak ?
   Example: when $-1 < r < 1$,
      \[ \frac{a}{1-r} = a + ar + ar^2 + ar^3 + \ldots \]
\end{itemize}

\pagebreak

{\large 2.1 Constant Term Expansions} \\
Assume that
   \[ \sqrt{a+b\sqrt{a+b\sqrt{a+\ldots}}} \]

converges when $a \geq 0$ and $b \geq 0$, and let $L$ be the limit.  Then
   \[ L = \sqrt{a+b\sqrt{a+b\sqrt{a+\ldots}}} \]

   \[ L = \sqrt{a+b L} \]

   \[ L^2 - b L - a = 0 \]

   \[ L = \frac{b+\sqrt{b^2+4 a}}{2} \]

Hence
   \[ \sqrt{a+b\sqrt{a+b\sqrt{a+\ldots}}} = \frac{b+\sqrt{b^2+4 a}}{2} \]

\pagebreak

Observation: when $a = 0$, we get

   \[ \sqrt{0+b\sqrt{0+b\sqrt{0+\ldots}}} = \frac{b+\sqrt{b^2+4(0)}}{2} \]

   \[ \sqrt{b \sqrt{b \sqrt{b \sqrt{\ldots}}}} = b \]

This makes sense since

\begin{eqnarray}
  \nonumber \sqrt{b \sqrt{b \sqrt{b \sqrt{\ldots}}}}
    & = & \sqrt{b} \sqrt{\sqrt{b}} \sqrt{\sqrt{\sqrt{b}}} \ldots \\
  \nonumber
    & = & b^{\frac{1}{2}} b^{\frac{1}{4}} b^{\frac{1}{8}} \ldots \\
  \nonumber
    & = & b^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots} \\
  \nonumber
    & = & b^1
\end{eqnarray}

\pagebreak

Similarly, assume that

   \[ a + \frac{b}{a + \frac{b}{a + \frac{b}{a + \ldots}}} \]

converges when $a > 0$ and $b \geq 0$, and let $L$ be the limit.  Then

   \[ L = a + \frac{b}{a + \frac{b}{a + \frac{b}{a + \ldots}}} \]

   \[ L = a + \frac{b}{L} \]

   \[ L^2 - a L - b = 0 \]

   \[ L = \frac{a+\sqrt{a^2+4 b}}{2} \]

Hence

   \[a+\frac{b}{a+\frac{b}{a+\frac{b}{a+\ldots}}}=\frac{a+\sqrt{a^2+4 b}}{2}\]

But as we saw earlier,

   \[ \sqrt{a+b\sqrt{a+b\sqrt{a+\ldots}}} = \frac{b+\sqrt{b^2+4 a}}{2} \]

Therefore,

   \[ \sqrt{a+b\sqrt{a+b\sqrt{a+\ldots}}} =
      b+\frac{a}{b+\frac{a}{b+\frac{a}{b+\ldots}}} =
      \frac{b+\sqrt{b^2+4 a}}{2}
   \]

In addition, setting $a = b = 1$, we get

   \[ \sqrt{1+\sqrt{1+\sqrt{1+\ldots}}} =
      1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}} =
      \frac{1+\sqrt{5}}{2}
   \]

which is equal to the golden ratio $\phi$.

\pagebreak

Now assume that
   \[ \sqrt[n]{a+b\sqrt[n]{a+b\sqrt[n]{a+\ldots}}} \]

converges when $a \geq 0$ and $b \geq 0$, and let $L$ be the limit.  Then
   \[ L = \sqrt[n]{a+b\sqrt[n]{a+b\sqrt[n]{a+\ldots}}} \]
   \[ L = \sqrt[n]{a+b L} \]
   \[ L^n - b L - a = 0 \]

Let $\alpha = a/L^n$ and $\beta = b/L^{n-1}$.  Then $a = \alpha L^n$,
$b = \beta L^{n-1}$ and
   \[ L^n - \beta L^n - \alpha L^n = 0 \]
   \[ 1 - \beta - \alpha = 0 \]
   \[ \beta = 1 - \alpha \]
yielding
   \[ L = \sqrt[n]{\alpha L^n+\beta L^{n-1}\sqrt[n]{
      \alpha L^n+\beta L^{n-1}\sqrt[n]{\alpha L^n+\ldots}
   }} \]

\pagebreak

{\large 2.2 Identity Transformations} \\

Pushing terms through radicals,
{\small
\begin{eqnarray}
   \nonumber \frac{b+\sqrt{b^2+4 a}}{2}
   & = &
      \sqrt{
         a+b\sqrt{
            a+b\sqrt{
               a+b\sqrt{
                  a+\ldots
               }
            }
         }
      } \\
   \nonumber
   & = &
      \sqrt{
         a+\sqrt{
            a b^2+b^3\sqrt{
               a+b\sqrt{
                  a+\ldots
               }
            }
         }
      } \\
   \nonumber
   & = &
      \sqrt{
         a+\sqrt{
            a b^2+\sqrt{
               a b^6+b^7\sqrt{
                  a+\ldots
               }
            }
         }
      } \\
   \nonumber
   & = &
      \sqrt{
         a+\sqrt{
            a b^2+\sqrt{
               a b^6+\sqrt{
                  a b^{14}+\ldots
               }
            }
         }
      } \\
   \nonumber
   & = &
      \sqrt{
         \left(\frac{a}{b^2}\right)b^2+\sqrt{
            \left(\frac{a}{b^2}\right)b^4+\sqrt{
               \left(\frac{a}{b^2}\right)b^8+\sqrt{
                  \left(\frac{a}{b^2}\right) b^{16} +\ldots
               }  
            }  
         }  
      }
\end{eqnarray}
}

Set $\alpha = a/b^2$.  Then
{\small
\begin{eqnarray}
   \nonumber
      \sqrt{
         \alpha b^2+\sqrt{
            \alpha b^4+\sqrt{
               \alpha b^8+\sqrt{
                  \alpha b^{16} +\ldots
               }
            }
         }
      }
   & = & \frac{b+\sqrt{b^2+4 a}}{2} \\
   \nonumber
   & = & \frac{b+\sqrt{b^2+4 \alpha b^2}}{2} \\
   \nonumber
   & = & \frac{b}{2}\left(1+\sqrt{1+4 \alpha}\right)
\end{eqnarray}
}
\pagebreak

Setting $\alpha = 2$, $b = 1/2$,
   \[ \sqrt{\frac{2}{2^2}+\sqrt{
      \frac{2}{2^4}+\sqrt{\frac{2}{2^8}+\sqrt{\frac{2}{2^{16}}+\ldots}}
   }} = 1\]

   \[ \sqrt{\frac{2}{2^1}+\sqrt{
      \frac{2}{2^2}+\sqrt{\frac{2}{2^4}+\sqrt{\frac{2}{2^8}+\ldots}}
   }} = \sqrt{2} \]

This can be rewritten as
   \[ 2^{1-2^{-1}} =
      \sqrt{2^{1-2^0} +
         \sqrt{2^{1-2^1} +
            \sqrt{2^{1-2^2} + \ldots
            }
         }
      }
   \]

And generalized to
   \[ 2^{1-2^k} =
      \sqrt{2^{1-2^{k+1}} +
         \sqrt{2^{1-2^{k+2}} +
            \sqrt{2^{1-2^{k+3}} + \ldots
            }
         }
      }
   \]

Letting $k \rightarrow -\infty$,
   \[ 2 =
      \sqrt{\ldots+
         \sqrt{2^{1-2^{-1}} +
            \sqrt{2^{1-2^0} +
               \sqrt{2^{1-2^1} + \ldots
               }
            }
         }
      }
   \]

\pagebreak

Transformations for "pushing" terms through radicals:

   \[ \sqrt{a_0+b_0\sqrt{a_1+b_1\sqrt{a_2+b_2\sqrt{a_3+\ldots}}}} \]
   \[ = \sqrt{a_0+\sqrt{a_1 b_0^2+\sqrt{a_2 b_1^2 b_0^4+\sqrt{
         a_3 b_2^2 b_1^4 b_0^8+\ldots
      }}}}
   \]

\vspace{2cm}

   \[\sqrt[n]{a_0+b_0\sqrt[n]{a_1+b_1\sqrt[n]{a_2+b_2\sqrt[n]{a_3+\ldots}}}}\]
   \[ = \sqrt[n]{a_0+\sqrt[n]{a_1 b_0^n+\sqrt[n]{a_2 b_1^n b_0^{n^2}+\sqrt[n]{
         a_3 b_2^n b_1^{n^2} b_0^{n^3}+\ldots
      }}}}
   \]

\pagebreak

{\large 2.3 Generation of Identities Using Recurrences} \\

Srinivasa Ramanujan (1887-1920)
%
% pronounced Shrinivasa Rah-MAH-na-jun
%             ^
%
% Amazing intuition, self-taught, often unable to
% explain his results or reasoning, was known to
% awaken in the middle of the night after dreaming
% up equations.  His notebooks are filled with
% thousands of theorems, corollaries, and examples.
% After 25 years of steady work, the task of cataloging
% and proving all his results is still not complete.

   % Rediscovered by Ramanujan, originally due to Bauer
   %
   \[
      1 - 5\left(\frac{1}{2}\right)^3 +
             9\left(\frac{1 \times 3}{2 \times 4}\right)^3
      -13\left(\frac{1\times 3\times 5}{2\times 4\times 6}\right)^3 + \ldots
      = 2/\pi
   \]

   \[ \frac{1}{1+\frac{e^{-2\pi}}{1+\frac{e^{-4\pi}}{1+
         \frac{e^{-6\pi}}{1+\ldots}
      }}}
      =
      \left(\sqrt{\frac{5+\sqrt{5}}{2}}-\frac{\sqrt{5}+1}{2}\right)e^{(2\pi/3)}
   \]

   \[
      \left(1+\frac{1}{1 \times 3}+\frac{1}{1\times 3\times 5}+\ldots\right)
      + \frac{1}{1+\frac{1}{1+\frac{2}{1+\frac{3}{1+\ldots}}}}
      = \sqrt{\frac{\pi e}{2}}
   \]

\pagebreak

Problem:

   \[ ? = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\ldots}}}} \]

Ramanujan claimed:

   \[ x+n = \sqrt{n^2+x\sqrt{n^2+(x+n)
                 \sqrt{n^2+(x+2n)\sqrt{\ldots}}
              }} \]

Setting {\em n}=1 and {\em x}=2 we find

   \[ 3 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\ldots}}}} \]

\pagebreak

Notice
   \[ [a+b] = \sqrt{b^2+a^2+2ab} = \sqrt{b^2+a[a+b+b]} \]

Expanding the square-bracketed portions,

{\small
\begin{eqnarray}
  \nonumber [x+n]
    & = & \sqrt{n^2+x[x+n+n]} \\
  \nonumber
    & = & \sqrt{n^2+x\sqrt{n^2+(x+n)[x+2n+n]}} \\
  \nonumber
    & = & \sqrt{n^2+x\sqrt{n^2+(x+n)\sqrt{n^2+(x+2n)[x+3n+n]}}} \\
  \nonumber
    & . & \\
  \nonumber
    & . & \\
  \nonumber
    & . & \\
  \nonumber
    & = & \sqrt{n^2+x\sqrt{n^2+(x+n)\sqrt{n^2+(x+2n)\sqrt{\ldots}}}}
\end{eqnarray}
}

Basic Idea: \vspace{-1.5cm}
\begin{itemize}
  \item Find a "telescoping" recurrence relation \vspace{-1.5cm}
  \item Use it to generate an infinitely recursive expression \vspace{-1.5cm}
  \item Hope that it converges (!)
\end{itemize}

\pagebreak

Consider a more familiar recurrence relation
   \[ \left[\frac{1}{k}\right] = \frac{1}{k(k+1)}+\left[\frac{1}{k+1}\right] \]

Expanding the square-bracketed portions,

{\small
\begin{eqnarray}
  \nonumber \left[ \frac{1}{n} \right]
    & = & \frac{1}{n(n+1)} + \left[ \frac{1}{n+1} \right] \\
  \nonumber
    & = & \frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}+\left[ \frac{1}{n+2} \right] \\
  \nonumber
    & . & \\
  \nonumber
    & . & \\
  \nonumber
    & . & \\
  \nonumber
    & = & \frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+2)(n+3)}+\ldots
\end{eqnarray}
}

In this case, the infinite expansion is valid.

\pagebreak

Consider the recurrence
   \[ \left[2^{1-2^k}\right] =
      \sqrt{2^{1-2^{k+1}} + \left[2^{1-2^{k+1}}\right]}
   \]

which expands into

   \[ \left[2^{1-2^k}\right] =
      \sqrt{2^{1-2^{k+1}} +
         \sqrt{2^{1-2^{k+2}} +
            \sqrt{2^{1-2^{k+3}} + \ldots
            }
         }
      }
   \]

Next, consider the recurrence

   \[ \left[1+2^{-2^{k+1}}\right] =
      \sqrt{2^{1-2^{k+1}} + \left[1+2^{-2^{k+2}}\right]}
   \]

which expands into

   \[ \left[1+2^{-2^{k+1}}\right] =
      \sqrt{2^{1-2^{k+1}} +
         \sqrt{2^{1-2^{k+2}} +
            \sqrt{2^{1-2^{k+3}} + \ldots
            }
         }
      }
   \]

How can two identities have the same right hand side but different
left hand sides ?  Answer: in the second identity, the infinite
expansion is not valid.

\pagebreak

Another example (this time of a valid expansion).  The recurrence

   \[ \left[n! + (n+1)!\right] =
      \sqrt{n!^2 + n!\left[(n+1)!+(n+2)!\right]}
   \]

expands into

{\small
   \[ \left[n! + (n+1)!\right] =
      \sqrt{n!^2 +
         n!\sqrt{(n+1)!^2 +
            (n+1)!\sqrt{(n+2)!^2 + \ldots
            }
         }
      }
   \]
}

Recalling that $ \Gamma(k+1) = k! $ for natural $k$, we can generalize to

{\small
   \[ \left[\Gamma(x) + \Gamma(x+1)\right] =
      \sqrt{\Gamma^2(x) +
         \Gamma(x)\sqrt{\Gamma^2(x+1) +
            \Gamma(x+1)\sqrt{\ldots
            }
         }
      }
   \]
}

\pagebreak

{\large 3. General Forms} \\
Consider a "continued power" of the form
  \[ a_0+b_0(a_1+b_1(a_2+b_2(a_3+\ldots)^{p_2})^{p_1})^{p_0} \]
Setting $p_j = 1$ and $b_j = 1$, we get a series
  \[ a_0+a_1+a_2+a_3+\ldots \]
Setting $p_j = 1$ and $a_j = 0$, we get an infinite product
  \[ b_0 b_1 b_2 b_3 \ldots \]
Setting $p_j = -1$, we get a continued fraction
  \[ a_0 + \frac{b_0}{a_1 + \frac{b_1}{a_2 + \frac{b_2}{a_3 + \ldots}}} \]
Setting $p_j = 1$ and $b_j = 1/c_j$, we get an ascending continued fraction
  \[ a_0+\frac{a_1+\frac{a_2+\frac{a_3+\ldots}{c_2}}{c_1}}{c_0} \]
Setting $p_j = 1/n$, we get a nested radical
  \[ a_0+b_0\sqrt[n]{a_1+b_1\sqrt[n]{a_2+b_2\sqrt[n]{a_3+\ldots}}} \]
Setting $p_j = -1/n$, we get a hybrid form
  \[ a_0 + \frac{b_0}{
      \sqrt[n]{a_1 + \frac{b_1}{
        \sqrt[n]{a_2 + \frac{b_2}{
          \sqrt[n]{a_3 + \ldots}
        }}
      }}
  } \]

Observation: series, infinite products, continued fractions
and nested radicals are all special cases of this generalized
"continued power" form ! \linebreak[2]

Question: can another general form be found for which
exponential ladders are also a special case ? \linebreak[2]
\pagebreak

We can imagine constructing the expression
  \[ a_0+b_0(a_1+b_1(a_2+b_2(a_3+\ldots)^{p_2})^{p_1})^{p_0} \]
by starting with a "seed" term and repeating the following steps:
\vspace{-1.5cm}
\begin{itemize}
  \item Raise to the exponent $p_j$ \vspace{-1.5cm}
  \item Multiply by $b_j$ \vspace{-1.5cm}
  \item Add $a_j$ \vspace{-0.75cm}
\end{itemize}
Of these 3 operations, only the first is non-commutative.
What if we change the ordering of the operands in the first
step ?  Then we would constuct an expression like

  \[ a_0+b_0 p_0^{a_1+b_1 p_1^{a_2+b_2 p_2^{a_3+\ldots}}} \]

Setting $a_j = 0$ and $b_j = 1$, we get an exponential ladder
  \[ p_0^{p_1^{p_2^{p_3^{\cdots}}}} \]

\pagebreak

What other things can we generalize ?
\begin{itemize}
  \item Identities.  Example (constant term expansion):
    \[ L = \sqrt[n]{\alpha L^n+\beta L^{n-1}\sqrt[n]{
       \alpha L^n+\beta L^{n-1}\sqrt[n]{\alpha L^n+\ldots}
    }} \]
    becomes
    \[ L = (\alpha L^{1/p}+\beta L^{1/p-1}(
       \alpha L^{1/p}+\beta L^{1/p-1}(\alpha L^{1/p}+\ldots)^p
    )^p)^p \]
    where $\beta = 1 - \alpha$.
  \item Recurrences.  Example:
    \[ \left[2^{1-2^k}\right] =
       \sqrt{2^{1-2^{k+1}} + \left[2^{1-2^{k+1}}\right]}
    \]
    becomes
    \[ \left[2^{\frac{p^k-1}{p^{k-1}-p^k}}\right] =
       \left(
          2^{\frac{p^{k+1}-1}{p^k-p^{k+1}}} +
          \left[ 2^{\frac{p^{k+1}-1}{p^k-p^{k+1}}} \right]
       \right)^p
    \]
\pagebreak

  \item Transformations.  Example:
    \[\sqrt[n]{a_0+b_0\sqrt[n]{a_1+b_1\sqrt[n]{a_2+b_2\sqrt[n]{a_3+\ldots}}}}\]
    \[ = \sqrt[n]{a_0+\sqrt[n]{a_1 b_0^n+\sqrt[n]{a_2 b_1^n b_0^{n^2}+\sqrt[n]{
         a_3 b_2^n b_1^{n^2} b_0^{n^3}+\ldots
      }}}}
    \]
    becomes
    {\small
    \[(a_0+b_0(a_1+b_1(a_2+b_2(a_3+\ldots)^p)^p)^p)^p\]
    \[ = (a_0+(a_1 b_0^{p^-1}+(a_2 b_1^{p^-1}b_0^{p^-2}+(
         a_3 b_2^{p^-1} b_1^{p^-2} b_0^{p^-3}+\ldots
       )^p)^p)^p)^p
    \]
    }
  \item Convergence Tests.  Example: Is there a generalized
        ratio test like the one used with series ?
\end{itemize}
\pagebreak

{\large 4. Selected Results from Literature} \\
Infinite Products \\

A) If $-1 < x < 1$, then
  \[ \prod_{j=0}^\infty\left(1+x^{2^j}\right) = \frac{1}{1-x} \]
Incidentally, this identity can be generated with the recurrence
  \[ \left[\frac{1}{1-x}\right] = (1+x)\left[\frac{1}{1-x^2}\right] \]

B) If $F_n = 2^{2^n}+1 =$ the $n$th Fermat number, then
  \[ \prod_{n=0}^\infty\left(1-\frac{1}{F_n}\right) = \frac{1}{2} \]

C) If the factors of an infinite product all exceed unity by
   small amounts that form a convergent series, then the
   infinite product also conveges.

\pagebreak

Exponential Ladders \\

If 0.06599 $\approx e^{-e} \leq x \leq e^{1/e} \approx$ 1.44467, then

  \[ x^{x^{x^{x^{\cdots}}}} \]

converges to a limit $L$ such that $L^{1/L} = x$. \\

\pagebreak

Herschfeld's Convergence Theorem (restricted), published 1935.
When $x_n > 0$ and $0 < p < 1$, the expression

  \[ \lim_{k\rightarrow\infty}x_0+(x_1+(\ldots+(x_k)^p\ldots)^p)^p \]

converges if {\em and only if} $\{x_n^{p^n}\}$ is bounded. \\

Special case: $p = 1/2$.  Then

  \[ \lim_{k\rightarrow\infty}x_0+\sqrt{x_1+\sqrt{\ldots+\sqrt{x_k}}} \]

converges if and only if $\{x_n^{2^{-n}}\}$ is bounded. \\

\pagebreak

"Souped-up" ratio test (due to Dixon Jones, 1988).  When $x_n > 0$
and $p > 1$, the continued power

  \[ \lim_{k\rightarrow\infty}x_0+(x_1+(\ldots+(x_k)^p\ldots)^p)^p \]

converges if

  \[ \frac{x_{n+1}^p}{x_n} \leq \frac{(p-1)^{p-1}}{p^p} \]

for all sufficiently large $n$. \\

Observation: as $p \rightarrow 1$, we {\em almost} get back d'Alembert's
ratio test for series.

\end{document}