Question 929347
The initial term is {{{a[0] = 5}}}
The common ration is 

{{{5*r=-5/3}}}

{{{r=-(5/3)/5}}}

{{{r=-5/15}}}

{{{r=-1/3}}}

Note: that the common ratio is {{{r = -1/3}}}, and that
since {{{abs(r) < 1}}} we know that the infinite geometric series {{{converges}}}

This is {{{sum((-5(-1/3)^n),n=1,infinity) =a[0]*(r/(1-r))

-5*((-1/3)/(1-(-1/3))) = -5*((-1/3)/(4/3)) = 5/4}}}

so the series converges and has a sum of {{{5/4}}}.