# Convergent fraction for constant $e$?

I've just learned about e. I am very much the novice and my problem is that while trying to calculate the convergent fractions for e. For instance:

$${2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{2}{3+\cfrac{3}{4}}}}}$$

I end up with 144/53?

I was wondering are there specific steps that I'm missing? For me I've been starting at the end of the continued fraction and working my way left. For instance:

$\frac{3}{1} + \frac{3}{4}$

And get 15/4 and then:

$\frac{2}{1} / \frac{15}{4}$

Until I finish with 144/53, which I'm not seeing this anywhere as one of the first few convergents of e.

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where is this list of convergent fractions? Typically we force the 'numerators' to be $1$ in standard form, which most lists I've seen use as well. I assume this is why you're getting different answers. –  Robert Mastragostino Jul 12 '12 at 2:40
@RobertMastragostino thanks for the edit, I had just got the correct syntax on when it said you had already corrected it. I'm going off wolfram for one, 2, 3 , 8/3, 11/4, 19/7,.... –  tijko Jul 12 '12 at 2:41

You’re using a generalized continued fraction; the convergents that you normally see listed are those for the standard continued fraction expansion of $e$, i.e., the one with $1$ for each numerator:

$$e=[2;1,2,1,1,4,1,1,6,1,1,8,\dots]\;.$$

This can also be written

$$[1;0,1,1,2,1,1,4,1,1,6,1,1,8,\dots]$$

to emphasize the pattern even more strongly.

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This is what I had original started with but, somehow veered off course. With this, I would take the integer of the divided fraction for the period and the decimals as the next divisor of 1? I will need to use the generalized continued fraction if I want the numerators to be integers other than 1? –  tijko Jul 12 '12 at 2:54
@tijko: Yes, as in this calculation of the first few figures in the continued fraction representation of $\pi$. –  Brian M. Scott Jul 12 '12 at 2:58
To get the fraction form we take the first number 2 over unity, then the next fraction would be the first period times numerator and denominator; adding unity to the numerator? –  tijko Jul 12 '12 at 3:09