Evaluate $$\lim_{n \rightarrow\infty} \sqrt[n]{3^{n} +5^{n}}$$
Attempt:
The only sort of manipulation that has come to mind is: $$e^{\frac{1}{n}ln(e^{n\ln(3)} + e^{n\ln(5)})}$$
So what is the trick to successfully evaluate this?
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https://math.stackexchange.com/questions/326172/the-l-infty-norm-is-equal-to-the-limit-of-the-lp-norms – Guy Fsone Jan 28 '18 at 07:23
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By searching in Approach0 you can find a few copies of this question and some very similar question. For example: Find $\lim_{n\to \infty}\sqrt[n]{3^n+5^n}$ or Evaluate $\lim_{n \to \infty} \sqrt[n]{3^n+4^n}$ – Martin Sleziak Jan 28 '18 at 09:09
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And probably looking at various generalizations of this might be useful, too. Such as: How to show $\lim_{n \to \infty} \sqrt[n]{a^n+b^n}=\max {a,b}$? (and other posts linked there) or Convergence of $\sqrt[n]{x^n+y^n}$ (for $x, y > 0$) (and other posts linked there.) – Martin Sleziak Jan 28 '18 at 09:12
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hint: There is a standard trick....: $5^n < 3^n+5^n < 2\cdot 5^n$, and in general if $0 < a < b$, then $ \displaystyle \lim_{n \to \infty} \sqrt[n]{a^n+b^n} = b$
DeepSea
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more general https://math.stackexchange.com/questions/326172/the-l-infty-norm-is-equal-to-the-limit-of-the-lp-norms – Guy Fsone Jan 28 '18 at 07:24
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With the well known limit of the exponential $\lim\limits_{n \rightarrow\infty} \frac 1 {e^{n}}=0 $
$$\lim_{n \rightarrow\infty} \sqrt[n]{3^{n} +5^{n}}=5\lim_{n \rightarrow\infty} \sqrt[n]{ \left(\frac 35\right)^{n} +1}=5\lim_{n \rightarrow\infty} \sqrt[n]{ \frac 1 {e^{n(\ln 5 -\ln 3)}} +1}=5(0+1)^0=5$$
user577215664
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By Bernoulli's inequality, if $x > 0$ and $n \ge 1$, $(1+x)^n \ge 1+nx$.
Therefore $(1+x/n)^n \ge 1+x$ so that $(1+x)^{1/n} \le 1+x/n $.
Therefore, if $0 < a < b$ then $\sqrt[n]{a^n+b^n} =b\sqrt[n]{1+(a/b)^n} \le b(1+\frac{(a/b)^n}{n}) = b+\frac{b(a/b)^n}{n} \lt b+\frac{b}{n} $ since $a/b < 1$ and $\sqrt[n]{a^n+b^n} =b\sqrt[n]{1+(a/b)^n} \gt b $.
Thererfore $b \lt \sqrt[n]{a^n+b^n} \lt b + \frac{b}{n} $ so that $\lim_{n \to \infty} \sqrt[n]{a^n+b^n} = b $.
marty cohen
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1Very nice +1 for the mention of Bernouilli's famous inequality ... – user577215664 Jan 28 '18 at 05:11
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1I find it impressive how often that elementary inequality is enough. – marty cohen Jan 28 '18 at 05:51
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\begin{align*} \dfrac{1}{n}\cdot\log\left(\left(\dfrac{3}{5}\right)^{n}+1\right)\rightarrow 0, \end{align*} so \begin{align*} \left(\left(\dfrac{3}{5}\right)^{n}+1\right)^{1/n}\rightarrow e^{0}=1, \end{align*} and hence \begin{align*} \sqrt[n]{3^{n}+5^{n}}=5\left(\left(\dfrac{3}{5}\right)^{n}+1\right)^{1/n}\rightarrow 5. \end{align*}
user284331
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I've never seen this techinque used before. Could you explain it a bit more? I'm interested in the different approach. – D.C. the III Jan 28 '18 at 04:40
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1Take $\log$ first, and compute what's the limit, after that, take $e$ to recover back to the original one. – user284331 Jan 28 '18 at 04:40
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I did @user284331, but I am trying to see where that fraction form came from – D.C. the III Jan 28 '18 at 04:43
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1$(3^{n}+5^{n})^{1/n}=(5^{n}((3/5)^{n}+1))^{1/n}=5((3/5)^{n}+1)^{1/n}$. – user284331 Jan 28 '18 at 04:45
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The correct maniplulation by exponential is
$$\sqrt[n]{3^{n} +5^{n}}=(3^{n} +5^{n})^\frac1n=e^{\frac{\log (3^n+5^n)}{n}}\to5$$
indeed
$$\frac{\log (3^n+5^n)}{n}=\frac{\log 5^n+\log \left(1+\left(\frac{3}{5}\right)^n\right)}{n}=\frac{n\log 5+\log \left(1+\left(\frac{3}{5}\right)^n\right)}{n}=$$ $$=\log 5+\frac{\log \left(1+\left(\frac{3}{5}\right)^n\right)}{n}\to \log5+0=\log 5$$
user
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