Evaluation of limit to infinity

Question

I am supposed to calculate $$\lim_{n\rightarrow \infty } \sqrt[n]{2^{n}\cdot3^{0}+2^{n-1}\cdot3+...+2^{0}\cdot3^{n}}$$ I tried to calculate a few terms separately, but it was undefined and I have no idea what to do.

Can anyone help me?

Answer 1

The dominant term is $3^n$. So if you factor $3^n$, you get $$ \sum_{k=0}^n2^{n-k}3^{k}=3^n\,\sum_{k=0}^n \left(\frac23\right)^{n-k}=3^n\,\sum_{k=0}^n \left(\frac23\right)^{k}=3^n\,\frac{1-\left(\tfrac23\right)^{n+1}}{1-\tfrac23}=3^{n+1}\,\left(1-\left(\tfrac23\right)^{n+1}\right). $$ Then $$ \left(\sum_{k=0}^n2^{n-k}3^{k}\right)^{1/n}=3^{1+\frac1n}\,\left(1-\left(\tfrac23\right)^{n+1}\right)^{1/n}=3^{1+\frac1n}\,e^{\frac1n\,\log\left(1-\left(\tfrac23\right)^{n+1}\right)}\to3. $$

Answer 2

$$\sqrt[n]{\sum_{k=0}^{n}{2^{n-k}\cdot 3^k}}=\sqrt[n]{3^{n+1}-2^{n+1}} \overset{n\to \infty}\longrightarrow 3$$

Answer 3

Hint: Use that $$\sum_{i=0}^n2^{n-i}\times33^i=\frac{1}{31}(33^{n+1}-2^{n+1})$$

Answer 4

Hint: For each natural $n$,$$3^n<\sum_{k=0}^n2^k3^{n-k}<(n+1)3^n.$$