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?
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1What do you mean by “it was undefined”? – José Carlos Santos Sep 25 '19 at 13:39
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@JoséCarlosSantos that function is not defined for n approaching to infinity, so limit is not defined too – Sep 25 '19 at 13:41
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1There is no $x$ in your expression. – José Carlos Santos Sep 25 '19 at 13:42
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@JoséCarlosSantos sorry, I edited my answer – Sep 25 '19 at 13:43
4 Answers
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$$\sqrt[n]{\sum_{k=0}^{n}{2^{n-k}\cdot 3^k}}=\sqrt[n]{3^{n+1}-2^{n+1}} \overset{n\to \infty}\longrightarrow 3$$
Analysis
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Hint: Use that $$\sum_{i=0}^n2^{n-i}\times33^i=\frac{1}{31}(33^{n+1}-2^{n+1})$$
Dr. Sonnhard Graubner
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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 $$ \left(\sum_{k=0}^n2^{n-k}3^{k}\right)^{1/n}=3^{1+\frac1n}\,\left(1-\left(\tfrac23\right)^{n+1}\right)^{1/n}=3^{1+\frac1n}\,e^{\frac1n\,\log\left(1-\left(\tfrac23\right)^{n+1}\right)}\to3. $$
Martin Argerami
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I factored $3^n$. Not sure if you are asking about the first equal sign. – Martin Argerami Sep 25 '19 at 13:58
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Oh, in the second sum the indices are $n,n-1,\ldots,0$. In the next one they are $0,1,\ldots,n$. Same indices. – Martin Argerami Sep 25 '19 at 14:01
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