Since this process may be continued indefinitely, we obtain an infinite continued fraction by substituting, in succession, the values obtained for d1, d2, dз,

We then have

....

is the same as that for d1, so that

deda, d, da, ds = d1, dy = d= d1, etc.

=

=

Therefore the partial quotients 1, 2, 1, 6, are repeated indefinitely.

11. A Periodic Continued Fraction is an infinite continued fraction in which the partial quotients are repeated in sets of

Therefore, only the third and fourth partial quotients are repeated, and the required fraction is

1 1 1 1 1 1

1 +1 + 5 + 10 + 5 + 10 +

Application of Convergents.

12. It is often convenient to substitute for a fraction with large terms, or for a quadratic surd, a convergent with comparatively small terms, provided that convergent approximates closely enough to the true value.

By Art. 9, we should expect the third convergent to be a close approximation, since the following partial quotient, 50,

Therefore, by Art. 9, the error of the third convergent is less than

Consequently, 2 represents the true value of 35 correctly

to four decimal places.

Assume

x=3+

Number.

13. We will take as an example the result of Ex. Art. 10. 1 1 1 1 1 1 + 2 + 1 + 6 + 1 + 1 1 1 1 1+2+1+6+

then

x- 3 =

....

Since the partial quotients 1, 2, 1, 6 are repeated indefinitely in that order, the continued fraction whose first partial quotient is the first periodic number (i.e., 1) at any stage, and which is continued indefinitely, differs in no respect from the given periodic continued fraction. For example, the periodic continued fraction which follows the heavy plus sign (+), in the value of x 3 above, is the same as the entire continued fraction, which is the value of x − 3.

We may therefore substitute x 3 for the part of the continued fraction which follows that particular plus sign. We thus have

14. If the continued fraction be not periodic from the beginning, we first reduce the periodic part by itself as above, and substitute its value in the given continued fraction. The latter is then a terminating continued fraction and can be reduced to a simple fraction, whose numerator and denominator will not, however, be rational.

1 1 1 1 1 1

Ex.

x =

2+1+3 + 5 + 3 + 5 +

the periodic part commencing with the third partial quotient.

Reduce each of the following fractions to a continued fraction, find its convergents, and determine a limit to the error of the third convergent.

Reduce each of the following surds to continued fractions, find the first five convergents, and determine a limit to the error of the fourth

Reduce each of the following periodic continued fractions to a surd :

25. Express the decimal .43429 as a continued fraction, find its fifth convergent, and determine the limit to the error of this convergent.

26. Express the decimal 2.71828 as a continued fraction, find its seventh convergent, and determine a limit to the error of this convergent.

27. The true length of the equinoctial year is 365d 5h 48m 46a. Reduce the ratio 5h 48m 468: 24h, to a continued fraction, and hence show how often leap year should come.

CHAPTER XXXVII.

SUMMATION OF SERIES.

By Undetermined Coefficients.

1. When the nth term of a series is a rational, integral function of n, the sum of n terms can be found by means of undetermined coefficients. The form which the sum of n terms of an arithmetical progression assumes will suggest a method of procedure.

By Ch. XXVII., § 2, Art. 5, the sum of n terms of the A. P.

3 + 5+ 7+9+ ·· + [3 + (n − 1)2] is 2 n + n2.

...

Now observe that the sum of n terms is an integral function of n, of degree one higher than the nth term.

In applying the method of undetermined coefficients, we start with this assumption.

Ex. Find the sum of n terms of the series

1.2+2.3+ 3.4 + ··· + n (n + 1) + ....

...

Since the nth term, n(n + 1), is of the second degree, we

assume

...

1.2+2.3+3 · 4 + ··· + n(n+1)=A+Bn + Cn2 + Dn3. (1) The validity of this assumption will be proved by mathematical induction. The method of proof will at the same time determine the values of A, B, C, D.

We have now to prove that if this relation hold for the sum of n terms, it holds for the sum of n + 1 terms. Evidently the latter sum will involve n + 1 just as the sum of n terms involves n.

That is,

1.2+2 3+3 · 4 + ··· + n (n + 1) + (n + 1) (n + 2)

= A + B(n + 1) + C(n + 1)2 + D(n + 1)3.

(2)