Kural:|1a|<1ise1+1a+1a2+1a3+.......+1an+....=aa−1
∞∑n=1nan(a∈R>1)=1a+2a2+3a3+.......+nan+....
S=1+1a+2a2+3a3+.......+nan+....
⋯
S1=1a+1a2+1a3+.......+1an+....
S2=1a2+1a3+.......+1an+....
S3=1a3+.......+1an+....
bunları toplarsak istenen ifadeyi buluruz(S1+S2+S3+......=∞∑n=1nan)
−−−−−−−−−−−−−−−−−−−−−−−−−−−
S1=1a+1a2+1a3+.......+1an+....=1a.(aa−1)
S2=1a2+1a3+.......+1an+....=1a2.(aa−1)
S3=1a3+.......+1an+....=1a3.(aa−1)
⋮
bunları toplayıp , paranteze alalım
∞∑n=1nan(a∈R>1)=S1+S2+S3+....=1a.(aa−1)+1a2.(aa−1)+.....
=
1a.(aa−1)[1+1a+1a2+1a3+......]⏟(aa−1)
⟺
∞∑n=1nan(a∈R>1)=1a(aa−1)2