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∑
n
=
0
∞
1
3
n
+
1
−
1
3
n
+
2
=
\sum_{n=0}^{\infty} \frac{1}{3^{n+1}} - \frac{1}{3^{n+2}} =
n
=
0
∑
∞
3
n
+
1
1
−
3
n
+
2
1
=
Mathematics
Problem
∑
n
=
0
∞
1
3
n
+
1
−
1
3
n
+
2
=
\sum_{n=0}^{\infty} \frac{1}{3^{n+1}} - \frac{1}{3^{n+2}} =
n
=
0
∑
∞
3
n
+
1
1
−
3
n
+
2
1
=
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