Please help Calculate this limit Help me find: 
$$
\lim_{n\rightarrow\infty}\sqrt[n]{2^n\cdot3^0+2^{n-1}\cdot3^1+2^{n-2}\cdot3^2+\cdots+2^0\cdot3^n}
$$
 A: Hint: Multiply the thing inside the root sign by $3-2$. (It might look nicer if you first reverse the order of summation.) Then you will want to bring a $3$ "outside."
Alternately, but equivalently, note that the sum inside the root sign is a finite geometric series with common ratio $\dfrac{3}{2}$.  Find the sum, and again bring a $3$ outside the root sign. 
Or else bring a $3$ outside immediately. (Multiply on the outside by $3$, divide on the inside by $3^n$.)
A: Hint:
$$ \lim_{n\rightarrow\infty}\sqrt[n]{2^n\cdot3^0+2^{n-1}\cdot3^1+2^{n-2}\cdot3^2+\cdots+2^0\cdot3^n} = \lim_{n\rightarrow\infty}\sqrt[n]{3^n\sum_{k=0}^{n}\left( \frac{2}{3}\right)^k } $$
$$ = 3\lim_{n\rightarrow\infty}\sqrt[n]{\sum_{k=0}^{n}\left( \frac{2}{3}\right)^k }. $$
Now, there is a well known formula for the sum under the root you can use it and then evaluate the limit.
Added: Here is the formula for the sum
$$\sum_{k=0}^{n}\left( \frac{2}{3}\right)^k= 3-3\, \left( \frac{2}{3} \right) ^{n+1}. $$
A: Note that each summand $2^k3^{n-k}\leq 3^k3^{n-k}= 3^n$, so the expression inside the $n$-th root, say $f(n)$, is bounded by $(n+1)3^n$. But on the other hand $f(n)\geq 2^03^n=3^n$. So, we have 
$$3^n\leq f(n)\leq (n+1)3^n.$$ Hence, $$3\leq \sqrt[n]{f(n)}\leq (\sqrt[n]{n+1})3.$$
So, by the squeeze theorem you get $\lim_{n\to\infty}\sqrt[n]{f(n)}=3.$
