+33661 No, this sum has a finite limit:
$$\sum_{k=2}^{\inf}(\frac{-3}{7})^k=\frac{9}{70}$$
See the following figure;
In this respect it is more like the infinite series 1 + 1/2 + 1/4 + 1/8 etc. which has the finite sum of 2. In other words it is a geometric series, the sum to infinity of which is given by:
sum = a0/(1-r) where a0 is the first term [(-3/7)2 here] and r [= (-3/7)] is the ratio between successive terms. Because r is less than 1, successive terms get smaller, r∞→0, and the sum converges to a finite value.
$${\mathtt{sum}} = {\frac{{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)}^{{\mathtt{2}}}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)\right)}} \Rightarrow {\mathtt{sum}} = {\mathtt{0.128\: \!571\: \!428\: \!571\: \!428\: \!6}}$$
$${\frac{{\mathtt{9}}}{{\mathtt{70}}}} = {\mathtt{0.128\: \!571\: \!428\: \!571\: \!428\: \!6}}$$
Just noticed this is a very old unanswered question!!
+33661 No, this sum has a finite limit:
$$\sum_{k=2}^{\inf}(\frac{-3}{7})^k=\frac{9}{70}$$
See the following figure;
In this respect it is more like the infinite series 1 + 1/2 + 1/4 + 1/8 etc. which has the finite sum of 2. In other words it is a geometric series, the sum to infinity of which is given by:
sum = a0/(1-r) where a0 is the first term [(-3/7)2 here] and r [= (-3/7)] is the ratio between successive terms. Because r is less than 1, successive terms get smaller, r∞→0, and the sum converges to a finite value.
$${\mathtt{sum}} = {\frac{{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)}^{{\mathtt{2}}}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)\right)}} \Rightarrow {\mathtt{sum}} = {\mathtt{0.128\: \!571\: \!428\: \!571\: \!428\: \!6}}$$
$${\frac{{\mathtt{9}}}{{\mathtt{70}}}} = {\mathtt{0.128\: \!571\: \!428\: \!571\: \!428\: \!6}}$$
Just noticed this is a very old unanswered question!!
