## Folios 74-76: AAL to ADM

Ockham

Monday. 4^{th} Jan^{y} ['1841' added by later reader]

Dear M^{r} De Morgan. We have

had company ever since I

last wrote to you, so I have

been at a Stand-still, &

only yesterday was able to

read over with attention your

replies. I am reluctant to

trouble you a__gain__ with remarks

on the Series \(1+\frac{x^2}{2}+\frac{x^4}{2.3.4}+\) &c,

for it seems as if I was

determined to plague you

about it. However I feel I

must do so. Your added**[74v] ** remarks of last time, about

\( B\) &c, are quite clear in

themselves, but I felt at

once that they did not

meet my difficulty which

was that as long as \(\frac{5}{4}\)

(which is gr__eater__ than \(1\) ) is to

be added to \(\frac{1}{2}\sqrt{1000,000+\frac{1}{4}}\),

it matters not whether for

\( \sqrt{1000,000+\frac{1}{4}}\) we substitute

the w__hole number__ next above

it or ''the intermediate fraction''

alluded to by you in the line

I have marked [mark a bit like her '\( \sqrt{}\) '], but we

never can bring out \(n=\) to

anything l__ess__ than \(502\), whence

\( n+1\) the required term must**[75r] ** be \(503\) . This, after reading

over & over, remains in my

mind a most obstinate fact,

and I believe I have found

out the real source of the

discrepancy between the result

at the bottom of the first

page & the top of the Second

one. I am presumptuous

enough to think there is

certainly an error in your

writing out, in the line I

have marked X, & it is one

which is very likely to have

occurred in writing ['it' crossed out] in

a hurry. __ __

The \((n+1)^{\rm th}\)^{ }term divided by

the *\(n\)*^{th} is I believe n__ot__**[75v] ** \(\frac{Z}{(2n-2)(2n-3)}\), but \(\frac{Z}{2n(2n-1)}\),

and I have re-written & now

enclose the rest of the demonstration

(exactly like yours) with this

correction. The result comes

out as I expected, owing to

\( \frac{1}{4}\) taking the place of \(\frac{5}{4}\),

& everything appears to me

consistent. the \(*n\)*^{th} term

divided by the \((n-1)^\text{th}\) term

would be, (as you have

written) \(\frac{Z}{(2n-2)(2n-3)}\), & this

correction would do instead of

the others, & be perhaps ['a' inserted] more

simple mode of making it,

as y__our__ demonstration would**[76r] ** then remain correct, the \(__n__\)^{th}

term being in that case the

required unknown one instead

of the \((n+1)^\text{th}\) . __ __

I am afraid all this is a

little complicated to explain

in letters, & perhaps I have

still not succeeded very

perfectly in doing so; but

I feel it now all very

clear in my own mind,

& am only anxious to receive

c__onfirmation__ as to my being

right, both as satisfactory

to me in the present instance,

& as tending to give me**[76v] ** confidence in future in

my own conclusions, or, (if

I am in this case puzzle-

-headed), a due diffidence

of them. __ __

I therefore beg your indulgence

for being so teasing. __ __

Believe me

Yours most truly

A. A. Lovelace

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