How can I convert the old tex file below into a pdf? I would very much like to know as I have a few other files that I would like to convert. Thanks.
PS: I selected the shortest example I had to hand.
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\centerline{\magtwotenrm Some reciprocal summation identities with
applications to the}
\centerline{\magtwotenrm Fibonacci and Lucas numbers}
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\centerline{\magonetenit Derek Jennings}
\centerline{\magonetenrm Faculty of Mathematics, University of Southampton,}
\centerline{\magonetenrm Southampton, Hampshire, SO17 1BJ, England.}
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\centerline{\magtwotenrm 1: Introduction}
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\noindent In this arcticle we use theta functions and simple telescoping
of series to produce some reciprocal summation results for the Fibonacci
and Lucas numbers. The two results that we prove are the following:
\gap
{\noindent\bf Theorem 1}
$${\left( \sum_{n=1}^\infty {1 \over F_{2n-1}} \right)}^2 -
\sum_{n=1}^\infty {1 \over F_{2n-1}^2} =
2\sum_{n=1}^\infty {1 \over F_{4n-2}^2}$$
\gap
{\noindent\bf Theorem 2}
$$\sum_{n=1}^\infty {\left( {2^n \over L_{2^n} } \right)}^2 = {4 \over 5}$$
\gap
\noindent $F_n$ and $L_n$ are the Fibonacci and Lucas numbers respectively,
satisfying the usual recurrence $U_{n+1}=U_n+U_{n-1}$ where $F_0=0$, $F_1=1$,
$L_0=2$ and $L_1=1$.
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\centerline{\magtwotenrm 2: Proof of Theorem 1}
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\noindent By simple series rearrangement we have
$$\sum_{n=1}^\infty {q^{2n-1} \over (1+q^{2n-1})^2} =
\sum_{n=1}^\infty (-1)^{n-1}{ nq^n \over 1-q^{2n}}.$$
\noindent Replacing $q$ by $q^2$ in the above then splitting the summation
over the odd and even numbers we obtain
$$\eqalign{
\sum_{n=1}^\infty {q^{4n-2} \over (1+q^{4n-2})^2} &=
\sum_{n=1}^\infty (-1)^{n-1}{ nq^{2n} \over 1-q^{4n}} \cr
&= \sum_{n=1}^\infty (2n-1){q^{4n-2} \over 1-q^{8n-4}} -
2\sum_{n=1}^\infty {nq^{4n} \over 1-q^{8n}}. \cr }\eqno(1)$$
\noindent Now a theorem originally due to Guass gives
$$\sum_{n=1}^\infty (2n-1){q^{8n-4} \over 1-q^{16n-8}} =
{ \left\{ {1 \over 2} \theta_2(q^4) \right\} }^4$$
\noindent where
$$\theta_2(q) = \sum_{-\infty}^\infty q^{(n+1/2)^2} =
2q^{1/4} \prod_{n=1}^\infty(1-q^{2n})(1+q^{2n})^2.$$
\noindent Therefore
$$\sum_{n=1}^\infty (2n-1){q^{4n-2} \over 1-q^{8n-4}} =
{\left( \sum_{n=1}^\infty {q^{2n-1} \over 1+q^{4n-2} } \right)}^2,\eqno(2)$$
\noindent using
$$\theta_2(q^2)^2 = 4 \sum_{n=1}^\infty {q^{2n-1} \over 1+q^{4n-2} }.$$
\noindent The last equality follows directly from equating the coefficient
of $x$ in (2) of {\bf [1]}.
\noindent Also, by simple series rearrangement we have
$$\sum_{n=1}^\infty { nq^n \over 1-q^{2n}} =
\sum_{n=1}^\infty {q^{2n-1} \over (1-q^{2n-1})^2},$$
\noindent which on replacing $q$ by $q^4$ gives
$$\sum_{n=1}^\infty { nq^{4n} \over 1-q^{8n}} =
\sum_{n=1}^\infty {q^{8n-4} \over (1-q^{8n-4})^2}.\eqno(3)$$
\noindent Using (2) and (3) to substitute for the summations in the RHS of
(1) gives
$$\sum_{n=1}^\infty {q^{4n-2} \over (1+q^{4n-2})^2}=
{\left( \sum_{n=1}^\infty {q^{2n-1} \over 1+q^{4n-2} } \right)}^2 -2
\sum_{n=1}^\infty {q^{8n-4} \over (1-q^{8n-4})^2}.\eqno(4)$$
\noindent Setting $q=(1-\sqrt 5)/2$ in (4), and using the Binet form
$F_n = (\alpha^n - \beta^n)/(\alpha -\beta)$ of the Fibonacci numbers
($\alpha = (1+\sqrt 5)/2$, $\beta = (1-\sqrt 5)/2$) we obtain Theorem 1.
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\centerline{\magtwotenrm 3: Proof of Theorem 2}
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\noindent Our starting point is the identity
$${q \over (1+q)^2} + {4q^2 \over (1-q^2)^2 } = {q \over (1-q)^2}.$$
\noindent This telescopes to
$${q \over (1+q)^2} + {4q^2 \over (1+q^2)^2 } + {16q^4 \over (1-q^4)^2}
= {q \over (1-q)^2}$$
\noindent and continuing the expansion process we arrive at
$$\sum_{n=0}^\infty {2^{2n} q^{2^n} \over (1+q^{2^n})^2} =
{q \over (1-q)^2}.$$
\noindent Now we put $q=(1-\sqrt 5)/2$ in the above identity and use the
Binet form of $L_n = \alpha^n + \beta^n$ of the Lucas numbers ($\alpha$,
$\beta$ as in section 2) to obtain Theorem 2.
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\centerline{\magonetenrm References}
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\item{[1]} D.Jennings {\sl An identity for the Fibonacci and Lucas numbers}
Glasgow Math. J. {\bf 35} (1993) 381--384.
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