Calculus 2 - by Vlad Vasilescu (W)

Calculus Level 4

lim n 1 + 2 n + 2 2 n + 2 3 n + + 2 n n n \large \lim_{n\to\infty} \dfrac{1 + \sqrt[n]{2} + \sqrt[n]{2^2} + \sqrt[n]{2^3} + \cdots + \sqrt[n]{2^n} }{n}

Find the limit to 2 decimal places.


The answer is 1.44.

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3 solutions

Guilherme Niedu
Nov 22, 2016

L = lim n 1 + 2 n + 2 2 n + 2 3 n + + 2 n n n L = \large \displaystyle \lim_{n \to \infty} \dfrac{1 + \sqrt[n]{2} + \sqrt[n]{2^2} + \sqrt[n]{2^3} + \dots + \sqrt[n]{2^n}}{n}

By the geometric progression sum formula,

L = lim n 2 1 + 1 / n 1 n ( 2 1 / n 1 ) L = \lim_{n \to \infty} \dfrac{2^{1 + 1/n} - 1}{n(2^{1/n} - 1)}

Set m = 1 / n m = 1/n , we get

L = lim m 0 m ( 2 1 + m 1 ) 2 m 1 = ( lim m 0 2 1 + m 1 ) ( lim m 0 m 2 m 1 ) = lim m 0 m 2 m 1 \begin{array} {ll} L & = \large \displaystyle \lim_{m \to 0} \dfrac{m(2^{1+m}-1)}{2^m - 1} \\ &= \large \displaystyle \left (\lim_{m \to 0} 2^{1+m} - 1 \right ) \cdot \left (\lim_{m \to 0} \dfrac{m}{2^m - 1} \right ) \\ & = \large \displaystyle \lim_{m \to 0} \dfrac{m}{2^m - 1} \end{array}

Applying L'Hôpital, we get

L = lim m 0 1 2 m l n ( 2 ) = 1 l n ( 2 ) = 1.44... L = \lim_{m \to 0} \dfrac{1}{2^m \cdot ln(2)} = \dfrac{1}{ln(2)} = \boxed{1.44... }

Can you clarify what you mean by "AP formula"? I believe that you're using something differnt from arithmetic progression sum . I think what you want it:

( 1 x ) ( 1 + x + x 2 + + x n ) = ( 1 x n + 1 ) . (1-x)(1+x+x^2 + \ldots + x^n) = ( 1 - x^{n+1} ).

Calvin Lin Staff - 4 years, 6 months ago

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I meant GP sum formula, my bad

Guilherme Niedu - 4 years, 6 months ago

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Ah ic. I've edited that into the solution.

In future, you can edit your solution directly by hitting the "edit" button at the bottom of the solution.

Calvin Lin Staff - 4 years, 6 months ago

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@Calvin Lin I was trying to, and I did it before with other solutions of other problems, but my browser did not let me save the editions

Guilherme Niedu - 4 years, 6 months ago

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@Guilherme Niedu Hm, can you elaborate on "browser did not let me save the editions"? What happened then?

Calvin Lin Staff - 4 years, 6 months ago

@Calvin Lin Thanks anyway!

Guilherme Niedu - 4 years, 6 months ago
Chew-Seong Cheong
Nov 22, 2016

Relevant wiki: Riemann Sums

L = lim n 1 + 2 n + 2 2 n + 2 3 n + . . . + 2 n n n = lim n 1 n k = 0 n 2 k n By Riemann sums = a b 2 x d x = 0 1 2 x d x a = lim n 0 n = 0 , b = lim n n n = 1 = 2 x ln 2 0 1 = 2 1 ln 2 = 1 ln 2 1.44 \begin{aligned} L & = \lim_{n \to \infty} \frac {1+\sqrt[n]{2}+\sqrt[n]{2^2}+\sqrt[n]{2^3}+...+\sqrt[n]{2^n}}n \\ & = \lim_{n \to \infty} \frac 1n \sum_{k=0}^n 2^\frac kn & \small \color{#3D99F6}\text{By Riemann sums} \\ & = \int_{\color{#3D99F6}a}^{\color{#3D99F6}b} 2^x \ dx = \int_{\color{#3D99F6}0}^{\color{#3D99F6}1} 2^x \ dx & \small \color{#3D99F6} a = \lim_{n \to \infty} \frac 0n = 0, \ b = \lim_{n \to \infty} \frac nn = 1 \\ & = \frac {2^x}{\ln 2} \bigg|_0^1 = \frac {2-1}{\ln 2} = \frac 1{\ln 2} \approx \boxed{1.44} \end{aligned}

The limit of summation should be from k=1 to k=n or k=0 to k-n-1.

I think first of all 1/n should be separated which will become zero.then rest of the things will be same as above

Kushal Bose - 4 years, 6 months ago

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L = lim n 1 + 2 n + 2 2 n + 2 3 n + . . . + 2 n n n = lim n 2 0 n + 2 1 n + 2 2 n + 2 3 n + . . . + 2 n n n = lim n 1 n k = 0 n 2 k n \begin{aligned} L & = \lim_{n \to \infty} \frac {1+\sqrt[n]{2}+\sqrt[n]{2^2}+\sqrt[n]{2^3}+...+\sqrt[n]{2^n}}n \\ & = \lim_{n \to \infty} \frac {2^\frac {\color{#D61F06}0}n +2^\frac 1n+2^\frac 2n+2^\frac 3n+...+2^\frac {\color{#3D99F6}n}n}n \\ & = \lim_{n \to \infty} \frac 1n \sum_{k=\color{#D61F06}0}^{\color{#3D99F6}n} 2^\frac kn \end{aligned}

Chew-Seong Cheong - 4 years, 6 months ago

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But the limit of summation should be k=1 to k=n

Kushal Bose - 4 years, 6 months ago

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@Kushal Bose What do you mean the limit of summation? I am quite sure my solution is correct. Refer to Riemann sums.

Chew-Seong Cheong - 4 years, 6 months ago

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@Chew-Seong Cheong Your solution is correct but I am saying about lower limit and upper limit of summation that is it should be k=1 to k=n

Kushal Bose - 4 years, 6 months ago

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@Kushal Bose No, I have shown that 1 = 2 0 n 1 = 2^\frac {\color{#D61F06}0}n , therefore the limit s are k = 0 k=0 to k = n k=n .

Chew-Seong Cheong - 4 years, 6 months ago

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@Chew-Seong Cheong But riemann sum is valid for k=1 to k=n or k=0 to k=n-1

Kushal Bose - 4 years, 6 months ago

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@Kushal Bose No, I have shown that a b 2 x d x \int_a^b 2^x \ dx . a = lim n 0 n a = \lim_{n \to \infty} \frac 0n and b = lim n n n b = \lim_{n \to \infty} \frac nn . The limits of integration follow those of summation. Refer to the wiki I have attached. I learned from it.

Even if k = 1 k=1 to k = n 1 k=n-1 , still a = 0 a=0 and b = 1 b=1 because n n \to \infty .

Chew-Seong Cheong - 4 years, 6 months ago

You are everywhere 😂

Vlad Vasilescu - 4 years, 6 months ago

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Is it good or bad that I am everywhere?

Chew-Seong Cheong - 4 years, 6 months ago

You come up with different but very good solutions , so I guess it's good .👍

Vlad Vasilescu - 4 years, 6 months ago

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Glad that you like it.

Chew-Seong Cheong - 4 years, 6 months ago
Vlad Vasilescu
Nov 21, 2016

And one more thing your handwriting is really beautiful!

Prakhar Bindal - 4 years, 6 months ago

Did the same! . the infamous Riemann's sum technique

Prakhar Bindal - 4 years, 6 months ago

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