[31r] My dear Lady Lovelace

You are right about
the writing down of the
terms:

 \(\frac{z}{(2n-2)(2n-3)}\) 

 is the \(n\) th term divided
by the \((n-1)\) th and the
\( \overline{n+1}\) th divided by the 
\( n\) th is \(\frac{z}{2n(2n-1)}\) as you
make it.

If I understand you correctly
[31v] you are now satisfied about
all the rest

Suppose you try at what
term convergency begins in the
following series

 \(1+\frac{x}{2.4}+\frac{x^2}{2.4.6.8}+\frac{x^3}{2.4.6.8.10.12}\) 

 \(+\cdots\cdots\) 

 when \(x=100,000\) 

With remembrances to
Lord Lovelace
                   I remain
                          Yours truly
                                ADeMorgan

 69 G.S.
Thursday

[32r] You will see the alteration 
I have made in your paper
If you do not see it
clearly, write again for
the sort of point con-
tained in it is one
of importance.