maths

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!!

infinity since it is infinite, just like 1+2+4+8+16...