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Evaluate $\lim_{n \to \infty} ((15)^n +([(1+0.0001)^{10000}])^n)^{\frac{1}{n}}$ Here [.] denotes the greatest integer function.

My Try : I know how to solve this kind of problem :$\lim_{n \to \infty} ((a)^n +(b)^n)^{\frac{1}{n}}$ where $a, b \geq 0$. But here I can not find $([(1+0.0001)^{10000}])$?

Can anyone please help me out?

Thank You.

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4 Answers 4

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Hint: since $$ \lim_n \left(1 + \frac 1n\right)^n = \mathrm e \in [2,3], $$ so the integral part of your expression should be…?

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Since for all $n\in \mathbb{N}$

$$2\le\left(1+\frac{1}{n}\right)^n \le e < 3$$

we have that

$$\left((15)^n +\left[\left(1+\frac1{10000}\right)^{10000}\right]^n\right)^{\frac{1}{n}}= \left((15)^n +2^n\right)^{\frac{1}{n}}=15 \left(1 +(2/15)^n\right)^{\frac{1}{n}}\to 15$$

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  • $\begingroup$ Don't you have to mention that $\left(1+\frac{1}{n}\right)^n$ is an increasing sequence?@gimusi $\endgroup$
    – cmi
    Commented Aug 13, 2018 at 17:56
  • $\begingroup$ @cmi no it suffices to know that it’s bounded in between 2 and 3 $\endgroup$
    – user
    Commented Aug 13, 2018 at 17:59
  • $\begingroup$ Yea...I can think that way...@gimusi $\endgroup$
    – cmi
    Commented Aug 14, 2018 at 5:03
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Hint: $$\left(1+\frac{1}{n}\right)^n \le e < 3$$

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Let $a = 15, b = [(1+10^{-4})^{10000}]=3$. We have that a > b. So: $$ \begin{align} (a^n+b^n)^{\frac{1}{n}} &= \exp\left(\frac{1}{n}\ln(a^n+b^n)\right) \\ &= \exp\left(\frac{1}{n}\left(\ln(a^n)+\ln\left(1+\left(\frac{b}{a}\right)^n\right)\right)\right) \\ &= a\cdot\exp\left(\frac{1}{n}\left(\frac{b}{a}\right)^n+o\left(\frac{1}{n}\left(\frac{b}{a}\right)^n\right)\right) \\ & \underset{n\infty}{\longrightarrow}a \end{align} $$.

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  • $\begingroup$ @gimusi The question uses the floor function. So, while your answer is correct, you can skip the steps with the squeeze theorem :) $\endgroup$
    – Jam
    Commented Aug 13, 2018 at 17:34
  • $\begingroup$ @Jam ops sorry I lost that part, I fix. Thanks a lot $\endgroup$
    – user
    Commented Aug 13, 2018 at 17:42
  • $\begingroup$ Wrong integer part. $\endgroup$
    – Did
    Commented Aug 13, 2018 at 17:53

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