n(n+2) | ||
∑n=1∞ ln | ||
(n+1)2 |
1*3 | n(n+2) | |||
Sn=ln | +...+ln | = ln1+ln3−2ln2+...+lnn+ln(n+2)−2ln(n+1)= | ||
22 | (n+1)2 |
n+2 | ||
=ln3−ln2+ln4−ln3+...+ln(n+2)−2ln(n+1)=−ln2+ln(n+2)−ln(n+1)=ln | −ln2 | |
n+1 |
n+2 | ||
S=limn→∞ ln | −ln2 = −ln2 | |
n+1 |
n+2 | ||
Sn=ln | −ln2 poprzez indukcję | |
n+1 |
3 | 3 | |||
1. S1=ln | = ln | −ln2 | ||
4 | 2 |
n+2 | ||
2. zakładamy że Sn=ln | −ln2 | |
n+1 |
n+2 | (n+1)(n+3) | n+3 | ||||
3. Sn+1=ln | −ln2+ln | = ln | −ln2 | |||
n+1 | (n+2)2 | n+2 |
n+2 | ||
na mocy indukcji Sn=ln | −ln2 | |
n+1 |