182G.] the Use of continued Fractions, S^c. it 



determined at '5963473, &c. as found by Euler and Hutton, are 

 left for the reader's exercise. They are very similar to those in 

 the following example, where the coefficients are the continued 

 products of the odd series 1, 3, 5, 7, &c. 



Example VIII. — Km = 1, r = 2, a: = 1, we have 

 S = 1 - 1 + 3 - 15 + 105- 945 + 10394 - &c. = 



I 1_ 2 3 4 19 



T+1 + T+1+T+ T+y (ts) 



20 



putting y, as before, to represent the residual portion — 

 The converging series proceeds thus : 



! I 1 1 1 1 1 1+t/ 



^ii?^ll P. ^ V + I9p + V y _ 



P r* 2' 4 ' 10 ' 26 q' Q^ Q + I9q +Qy '^ 



12 3 4 5 6 19 



1556683 0368 + 3 156467 520 .y __ 486463449 + 98619 61Q 3> __ o 

 23758664096 + 4809701440 i/ ~ 742458253 + 1503031701/ "" 



Serviceable limits to the value of y may be thus found. Say, 

 generally, ?/ = ^ '^^ ^^^^ + &c. Assuming this to be 



nearly = t — = ^ (\/ 4 m + 1 — 1), it is still more nearly = 

 nt mjj^^m(v ^m^5-i) ^ But if One of these is too great, 



] + m, 2 w + 2 o > 



the other is obviously too little. Use then the mean, or say if 



, m m + h . y m(v'4m-f3 — 1) , .^ 



yisnearly = y^— ^= ^—^ (A) 



3 00 



it is still more nearly 



"~" ' "' (B) 



1 + 1 + 1+ 



jn{ (m + 1) V 4 m + 7 — (m + 2) } 



2 m^ + 4 m — 1 



and there also are one greater and the other less than thfe 

 coiTect value of?/. When m = 20, these limits are, 



(A) = H»^^' = 3-95631. (B) = '^^ ^^' ^/J " ^^ = 3-95619 



Accounting .-.y = 3-9563,we findS = ^^l^i^^l^^ = -65562071 1 



y = 3-9562 S = ,^^^ = -656620704 



The latter is, therefore, the value of the series, very probably 

 corfect in every figure ; certainly in all but the last. 



W.G.H, 



E^ 



