Question 1197684
<br>
The statement of the problem is incomplete; we don't know the starting value of k.<br>
The formula for the infinite sum of a geometric sequence is<br><pre>
        first term
--------------------------
 1 minus the common ratio</pre>
The common ratio is clear: (-6/7).<br>
But the first term depends on the starting value of k.<br>
If the starting value is k=1, then the first term is {{{(-6/7)^0=1}}}, and the infinite sum is {{{1/(1-(-6/7))=1/(1+6/7)=1/(13/7)=7/13}}}<br>
But if the starting value is k=0, then the first term is {{{(-6/7)^(-1)=-7/6}}}, and the first term is {{{(-7/6)/(1-(-6/7))=(-7/6)/(1+6/7)=(-7/6)/(13/7)=(-7/6)(7/13)=-49/78}}}<br>
And there is no reason the starting value couldn't be any other (positive or negative) integer.<br>
ANSWER (maybe): 7/13<br>