what is the complexity of this algorithm , can i sum it to infinity?

Question

Func(n)
{
   i = n
   while(i>=1)
      g(i);
      i = i/3;
}

what is the complexity of this algorithm? (while the complexity of g(n) is theta(n²)) I assumed for bigger n's you say that the complexity is

n² + (n/3)² + (n/3²)² + (n/3³)²..... to infinity.

And the answer is theta(n²). Is that true?

Answer 1

Lets look at the series that we have got in our hands.

=> n² + (n²/3) + (n/3)² + (n/3²)² + (n/3³)²..... 
taking n² common
=> n² * [ 1 + (1/3) + (1/3)² + (1/3²)² + (1/3³)²..... ]

As [ 1 + (1/3) + (1/3)² + (1/3²)² + (1/3³)²..... ] is a decreasing series, it is equal to 1.

Thus the answer is O(n²)

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EDIT 1:

Prove for the sum of series [ 1 + (1/3) + (1/3)² + (1/3²)² + (1/3³)²..... ], is below.