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The sum approaches infinity as more terms are added.
It converges if ∣r∣<1|r| < 1∣r∣<1, with SnS_nSn approaching S∞S_\inftyS∞.
S∞=a1−rS_\infty = \frac{a}{1 - r}S∞=1−ra, for ∣r∣<1|r| < 1∣r∣<1.
The sum increases without bound as nnn approaches infinity.
S∞=6/(1−1/3)=9S_\infty = 6 / (1 - 1/3) = 9S∞=6/(1−1/3)=9.
S∞S_\inftyS∞ equals the first term divided by (1−r)(1 - r)(1−r), where r<1r < 1r<1.
Sn=a(1−rn)/(1−r)S_n = a(1 - r^n) / (1 - r)Sn=a(1−rn)/(1−r).
Set a/(1−r)=4aa / (1 - r) = 4aa/(1−r)=4a, leading to r=3/4r = 3/4r=3/4.
Set a∗r2=9a*r^2 = 9a∗r2=9 and solve for aaa with rrr known.
Sn=a(1−rn)/(1−r)S_n = a(1 - r^n) / (1 - r)Sn=a(1−rn)/(1−r) for ∣r∣<1|r| < 1∣r∣<1
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