Two quick changes before diving into action:
First of all, we can factor the 100 out of the sum:
[tex] \sum_{k=1}^\infty 100(-0.9)^{k-1} = 100\sum_{k=1}^\infty (-0.9)^{k-1} [/tex]
Secondly, we can see that the index k starts from 1, but the exponent is k-1. This means that when k=1, the exponent is actually 0. When k=2, the exponent is actually 1, and so on. So, we can rewrite the sum as
[tex] 100\sum_{k=1}^\infty (-0.9)^{k-1} = 100\sum_{k=0}^\infty (-0.9)^k[/tex]
Now, let's focus on the sum. First of all, it converges, because every sum like
[tex] \sum_{k=0}^\infty a^k [/tex]
converges if and only if [tex] |a|<1 [/tex], which is the case because
[tex] |-0.9| = 0.9 < 1 [/tex]
Also, in this case we have
[tex] \sum_{k=0}^\infty a^k = \cfrac{1}{1-a} [/tex]
so in your case you have
[tex] \sum_{k=0}^\infty (-0.9)^k = \cfrac{1}{1-(-0.9)} = \cfrac{1}{1.9} [/tex]
and let's not forget the 100 we factored at the beginning!
[tex] 100\sum_{k=0}^\infty (-0.9)^k= 100\cdot\cfrac{1}{1.9} = \cfrac{100}{1.9} [/tex]
