296 
ANALYSIS. NACHLASS. 
Ax (2) 8 '(6) (12) g _ (2) 10 _ . (6)‘ 8 r # x-| 
(1){3)(4)* (l) 4 (4)* (3)*(12) 4 L ^ 
A v r/.M (3)(12) a Wlr##N1 
— 407 L(1)J ( 4 )(6) S ( 2 ) a (6) 8 L 
[S. 45] 
m 1 
[8.] 
Durch Induction gefunden: 
(1 +2#+ 2a> 4 +2^ + ...) + (1 +2^ 3 +2^ 12 -+-2a? :i7 + ...) 
= 2 (1 + ») (1 + x s ) (1 + er 5 ) (1 + tf 7 )... X (1 - **) (1 - oß 1 ) (1 - X' 1 ] (1 - * ,B ). 
X (1 “ ^ 12 ) (1 — X U ) (1 — * 36 ) • • • 
= x (1 - *>) (1 - *') (1 - »") (1 - *“) . . • 
( [l-\-2x-\-2x i -\-2x 9 -\— •) — (1 -f- 2a? 3 -f- 2# 12 4- 2# 27 -j - • • •) 
| = 2«(1 +ic)(l+^ 3 )(l+a? 5 )(l + ^ 7 )-*- X(l-^)(l-^ ,l )(l-^ l3 )(l-^ 
[8] { x(l— ^ 12 )(1 — x 2i )... 
2x[xxJ[x'*] ^ ^ ^ < > # 
[iCj {X J 
[9.] 
Durch Induction gefunden: 
£ — ££ = ^3 
jl -f 2aj-|- 2ä; 4 + 2^ ü H j -h (® — ee ) 1 1 + 2a? 3 + 2ß u 4“ 2^ 7 4 j 
= — 2se (1 — &x) (1 — £££ 3 )(l — SXf) (1 —££^ 7 ) • . . (l 4“ # 3 ) (l + (1 + %'*) ( l + * * * 
X (1 — & 4 ) (1 — ^ 8 ) (i — £ 12 ) • • • 
Dieselbe Summe wäre demnach 
= — 2ee 1 
ex 
eex 3 4- eex c ' + ££ 10 — x 15 — x''i~ ex 28 + eex : b — etc. j 
• (1 + x 3 ) (1 4- x°) (1 4- x ,5 )... 
4-eex*~ — etc. j (1 4-x 3 ) (1 4-x?) (l 4*x u ) etc. 
eex* -\-x s -\-ex —ex 
x 
[*) Hier ist (n) = \x n ] gesetzt.] 
[**) In der Handschrift fehlt links vom Gleichheitszeichen der Faktor (1).]