The sequence $a_n = \sqrt[n]{(3^n+5^n)}$ is convergent?
Tried to resolve applying the the limit $n\to\infty$ but couldn't figure out how to finish it, any idea?
Thanks
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6$5^n<3^n+5^n<2\cdot 5^n$. Use the Squeeze Theorem. – David Mitra Jul 30 '14 at 14:06
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this is the square root of something to the power of $1/n$? are you sure you typed this correctly? – 5xum Jul 30 '14 at 14:07
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The (1/n) is the index of the n-root – arthurg Jul 30 '14 at 14:12
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So do you mean $a_n = \sqrt[n]{(3^n+5^n)}$? – zarathustra Jul 30 '14 at 14:15
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@FPE yes, exactly – arthurg Jul 30 '14 at 14:16
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@DavidMitra, I got the idea, but, there is any reason for 2*5^n or can be any constant? – arthurg Jul 30 '14 at 14:17
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It's simplest to use $3^n+5^n< 5^n+5^n=2\cdot 5^n$, $n>0$. – David Mitra Jul 30 '14 at 14:19
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The upper bound has to be something that converges to the same limit as the lower bound, for the squeeze theorem to be useful. It could have been $3\cdot 5^n$, for example. – zarathustra Jul 30 '14 at 14:19
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There is no problem using those in the squeeze theorem without the n-square? – arthurg Jul 30 '14 at 14:23
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The inequality carries over to the "n-square"; $(5^n)^{1/n}< a_n < 2^{1/n} (5^n)^{1/n}$, or $5< a_n<2^{1/n}\cdot 5$. Note $2^{1/n}\rightarrow1$ as $n\rightarrow\infty$. – David Mitra Jul 30 '14 at 14:37
2 Answers
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Get out the term with maximal absolute value
$(3^n + 5^n)^{\frac{1}{n}} = [5^n((\frac{3}{5})^n +1)]^{\frac{1}{n}} =5((\frac{3}{5})^n +1)^{\frac{1}{n}} \to 5(0+1)^0 = 5$
Petite Etincelle
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Here is an approach.
$$ a _n =e^{\frac{1}{n}\ln( 3^n+5^n )}= e^{\frac{1}{n}\ln(5^n(1+(3/5)^n) )}= e^{\frac{1}{n}\ln(5^n) +\frac{1}{n}\ln(1+(3/5)^n) ) } $$
$$ \sim {5}\,e^{\frac{1}{n} ( 3/5)^n } \longrightarrow_{n\to \infty} {5}. $$
Note: we used the Taylor series
$$ \ln(1+t) = t+O(t^2) $$
Mhenni Benghorbal
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I think you made a couple of mistakes or the post has been amended in between. There is no $\frac{1}{2}$ and the equivalent is wrong. – Matt B. Jul 30 '14 at 15:27
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@MattB.: It is not my fault. The post has been edited. Check the last edit on it and you will see that. Thanks for the comment – Mhenni Benghorbal Jul 30 '14 at 15:29
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