SOLUTION: Compute the infinite sum of (-6/7)^k-1

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Question 1197684: Compute the infinite sum of (-6/7)^k-1
Found 3 solutions by greenestamps, ewatrrr, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The statement of the problem is incomplete; we don't know the starting value of k.

The formula for the infinite sum of a geometric sequence is
        first term
--------------------------
 1 minus the common ratio

The common ratio is clear: (-6/7).

But the first term depends on the starting value of k.

If the starting value is k=1, then the first term is %28-6%2F7%29%5E0=1, and the infinite sum is 1%2F%281-%28-6%2F7%29%29=1%2F%281%2B6%2F7%29=1%2F%2813%2F7%29=7%2F13

But if the starting value is k=0, then the first term is %28-6%2F7%29%5E%28-1%29=-7%2F6, and the first term is

And there is no reason the starting value couldn't be any other (positive or negative) integer.

ANSWER (maybe): 7/13

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
infinite sum of %28-6%2F7%29%5E%28k-1%29,
Geometric Series r = (-6/7), Assuming a%5B1%5D+=+1
sum%28+a%5Bi%5D%2C+i=1%2C+infinity+%29 = a%5B1%5D%2F%281-r%29+=+1%2F%2813%2F7%29+=+%28+7%2F13%29+

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.

When you request to compute the sum (finite or infinite), you MUST define the limits of summation, from and to.

In this problem, you defined the upper limit as infinity, but missed to define the lower limit.