I am trying to solve $T(n)=T(\frac{n}{2}) + T(\frac{n}{3}) +n $ where n is integer and we want to find the aproximate running time Θ of the tree $T(n)$ where $T(1)=Θ(1)$, At the first level the sum is $n$ at the second it is $\frac{5}{6}n$ at the third $(\frac{5}{6})^2n$ and etc.The tree of the problem can be bounded by 2 trees $S_{1}$ and $S_{2}$. or in other words $S_{1} \le T(n) \le S_{2} $ The sum for first is $ \sum \limits_{n = 1}^{ log_2n} (\frac{5}{6})^in $
and The sum for second is $ \sum \limits_{n = 1}^{ log_3n} (\frac{5}{6})^in $.
But how can I find the Θ of $T(n)$.I think the answer should be $Θ( \sum \limits_{n = 1}^{ log_3n} (\frac{5}{6})^in ) $. but how should I simplify this sum? I was thinking since $Θ( \sum \limits_{n = 1}^{ log_3n} (\frac{5}{6})^in ) $ $\le $ $Θ( \sum \limits_{n = 1}^{ infinity} (\frac{5}{6})^in ) $ =5n then the answer is $ Θ(5n) =Θ(n)$