I.S.I. B.Math.(Hons.) Admission test : 2010 Problem 5

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I.S.I. B.Math.(Hons.) Admission test : 2010 Problem 5

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#1 h Apr 10, 2012, 7:42 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $a_1>a_2>.....>a_r$ be positive real numbers .
Compute $\lim_{n\to \infty} (a_1^n+a_2^n+.....+a_r^n)^{\frac{1}{n}}$

This post has been edited 1 time. Last edited by mynamearzo, Apr 12, 2012, 4:25 PM

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#2 h Apr 10, 2012, 8:42 AM • 2 Y

Y by Adventure10 and 1 other user

$\lim_{n\to\infty}(a_{1}^{n}+a_{2}^{n}+.....+a_{r}^{n})^{\frac{1}{n}} = \lim _{n\to\infty} a_1 \left[ 1 + \left( \frac{a_2}{a_1} \right)^n + ... + \left( \frac{a_r}{a_1} \right)^n \right]^{\frac1n} = ...$

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#3 h Apr 10, 2012, 8:46 AM • 6 Y

Y by Dattier, integrated_JRC, ayan_mathematics_king, Adventure10, Rombo, and 1 other user

Try squeezing,

$(a^{n}_{1})^{\frac{1}{n}} \le (a_{1}^{n}+a_{2}^{n}+.....+a_{r}^{n})^{\frac{1}{n}} \le (r a^{n}_{1}) ^{\frac{1}{n}}$

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#4 h Apr 10, 2012, 11:29 AM • 1 Y

Y by Adventure10

Virgil Nicula wrote:

$\lim_{r\to 0} W_r =e^L$ , where $L=\lim_{r\to 0}\frac 1r\cdot \left(\frac{a^r_1+a^r_2+...+a^r_n}{n}-1\right)= \lim_{r\to 0}\frac 1n\cdot \sum_{k=1}^n\frac{a_k^r-1}{r}=$

$\frac 1n\cdot\sum_{k=1}^n\ln a_k=\ln\sqrt [n]{\prod_{k=1}^na_k}\implies$ $L=\ln\sqrt [n]{\prod_{k=1}^na_k}$ . Thus, $\lim_{r\to 0} W_r=e^L=\sqrt [n]{\prod_{k=1}^na_k}=\left(\prod_{k=1}^na_k\right)^{\frac 1n}$ .

I used the remarkable limit $\lim_{x\to 0}\frac {a^x-1}{x}=\ln a$ and the rule :

if $\lim_{x\to x_0}f(x)=1$ and $\lim_{x\to x_0}g(x)=\infty$ , then $\lim_{x\to x_0}f(x)^{g(x)}\stackrel{1^{\infty}}{\ =\ }e^L$ , where $L=\lim_{x\to x_0}g(x)\cdot [f(x)-1]$ .

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#5 h Apr 10, 2012, 12:13 PM • 1 Y

Y by Adventure10

Where did Virgil Nicula write this?

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#6 h Apr 10, 2012, 12:22 PM • 1 Y

Y by Adventure10

I have posted a month before here.
virgil sir proof is nice as usual

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#7 h Apr 10, 2012, 12:26 PM • 1 Y

Y by Adventure10

oops looks like those two are different

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#9 h Jan 15, 2018, 2:20 PM • 1 Y

Y by Adventure10

$\lim_{n\to\infty}(a_{1}^{n}+a_{2}^{n}+.....+a_{r}^{n})^{\frac{1}{n}} = \lim _{n\to\infty} a_1 \left[ 1 + \left( \frac{a_2}{a_1} \right)^n + ... + \left( \frac{a_r}{a_1} \right)^n \right]^{\frac1n}$ . Now denote $\frac{a_2}{a_1}=\frac{1}{x_1} , \frac{a_3}{a_1}=\frac{1}{x_2}......,\frac{a_r}{a_1}=\frac{1}{x_{r-1}}$ and then put the limit u will get the ans tobe $a_1$ .

Correct me if i am wrong.

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#10 h Jan 15, 2018, 3:46 PM • 2 Y

Y by Adventure10, Mango247

vickyricky wrote:

$\lim_{n\to\infty}(a_{1}^{n}+a_{2}^{n}+.....+a_{r}^{n})^{\frac{1}{n}} = \lim _{n\to\infty} a_1 \left[ 1 + \left( \frac{a_2}{a_1} \right)^n + ... + \left( \frac{a_r}{a_1} \right)^n \right]^{\frac1n}$ . Now denote $\frac{a_2}{a_1}=\frac{1}{x_1} , \frac{a_3}{a_1}=\frac{1}{x_2}......,\frac{a_r}{a_1}=\frac{1}{x_{r-1}}$ and then put the limit u will get the ans tobe $a_1$ .

Correct me if i am wrong.

Look at others' works before you post your solution. This is same as the post #2 in this thread.

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#11 h Jan 15, 2018, 3:50 PM • 2 Y

Y by Adventure10, Mango247

jrc1729 wrote:

vickyricky wrote:

$\lim_{n\to\infty}(a_{1}^{n}+a_{2}^{n}+.....+a_{r}^{n})^{\frac{1}{n}} = \lim _{n\to\infty} a_1 \left[ 1 + \left( \frac{a_2}{a_1} \right)^n + ... + \left( \frac{a_r}{a_1} \right)^n \right]^{\frac1n}$ . Now denote $\frac{a_2}{a_1}=\frac{1}{x_1} , \frac{a_3}{a_1}=\frac{1}{x_2}......,\frac{a_r}{a_1}=\frac{1}{x_{r-1}}$ and then put the limit u will get the ans tobe $a_1$ .

Correct me if i am wrong.

Look at others' works before you post your solution. This is same as the post #2 in this thread.

But he did'nt give the full solution right.

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#12 h Jan 15, 2018, 4:05 PM • 1 Y

Y by Adventure10

Okay okay. The solution was incomplete but trivial. Don't post anymore in these sacred threads.

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#13 h Jan 15, 2018, 4:59 PM • 1 Y

Y by Adventure10

mathproof wrote:

Try squeezing,

$(a^{n}_{1})^{\frac{1}{n}} \le (a_{1}^{n}+a_{2}^{n}+.....+a_{r}^{n})^{\frac{1}{n}} \le (r a^{n}_{1}) ^{\frac{1}{n}}$

This is the usual (and probably best) proof. I believe this is a standard result in any analysis course.

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#14 h Dec 18, 2019, 3:47 AM • 2 Y

Y by Adventure10, Mango247

Answer
"proof*

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#15 h Dec 18, 2019, 7:52 AM • 1 Y

Y by Adventure10

Alternative solution:
If we apply the Cauchy-D'Alembert formula, we get: $\underset{n\to \infty }{\mathop{\lim }}\,{{(a_{1}^{n}+a_{2}^{n}+.....+a_{r}^{n})}^{\frac{1}{n}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{a_{1}^{n+1}+a_{2}^{n+1}+.....+a_{r}^{n+1}}{a_{1}^{n}+a_{2}^{n}+.....+a_{r}^{n}}=\underset{n\to \infty }{\mathop{\lim }}\,{{a}_{1}}\frac{1+{{\left( \frac{{{a}_{2}}}{{{a}_{1}}} \right)}^{n+1}}+...+{{\left( \frac{{{a}_{r}}}{{{a}_{1}}} \right)}^{n+1}}}{1+{{\left( \frac{{{a}_{2}}}{{{a}_{1}}} \right)}^{n}}+...+{{\left( \frac{{{a}_{r}}}{{{a}_{1}}} \right)}^{n}}}={{a}_{1}}$

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#16 h Dec 29, 2019, 11:19 AM • 1 Y

Y by Adventure10

Yeah...
..'

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#17 h Mar 5, 2024, 6:16 PM • 1 Y

Y by aesthetics

take common $a_1$ and (1+x)^n = 1+nx for small x

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#18 h Jun 2, 2024, 5:24 PM

Y by

$\ln{\left (\lim\limits_{n\to\infty} (a_1^n + a_2^n + \cdots a_r^n)^{\frac{1}{n}} \right) } = \lim\limits_{n\to\infty} \ln{ (a_1^n + a_2^n + \cdots a_r^n)^{\frac{1}{n}} } = \lim\limits_{n\to\infty} \frac{\ln{(a_1^n + a_2^n + \cdots a_r^n)}}{n}$
$=\lim\limits_{n\to\infty} \frac{a_1^n \ln{a_1} + a_2^n \ln{a_2} + \cdots a_r^n \ln{a_r}}{a_1^n + a_2^n + \cdots a_r^n}$ (by L'Hopital's rule)
$=\ln{a_1} \lim\limits_{n\to\infty} \frac{1}{1+(\frac{a_2}{a_1})^n + \cdots (\frac{a_r}{a_1})^n} + \ln{a_2} \lim\limits_{n\to\infty} \frac{1}{(\frac{a_1}{a_2})^n + 1 + \cdots (\frac{a_r}{a_2})^n} + \cdots + \ln{a_r} \lim\limits_{n\to\infty} \frac{1}{(\frac{a_1}{a_r})^n + (\frac{a_2}{a_r})^n + \cdots + 1}$
$= \ln{a_1} \cdot 1 + \ln{a_2} \cdot 0 + \cdots + \ln{a_r} \cdot 0 = \ln{a_1}$ .

Hence $\lim\limits_{n\to\infty} (a_1^n + a_2^n + \cdots a_r^n)^{\frac{1}{n}} = a_1$ .

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#19 h Apr 24, 2025, 4:46 AM

Y by

Claim
$\lim_{n \to \infty} (a_{1}^n + a_{2}^n + a_{3}^n + .... + a_{r}^n)^{1/n} =$ max $(a_{1},a_{2},a_{3},....,a_{r})$

Proof $\lim_{n \to \infty} (a_{1}^n + a_{2}^n + a_{3}^n + .... + a_{r}^1/n)^{1/n} = \lim _{n\to\infty} a_1 \left[ 1 + \left( \frac{a_2}{a_1} \right)^n + ... + \left( \frac{a_r}{a_1} \right)^n \right]^{\frac1n} = a_{1}$

$\boxed{\therefore \lim_{n \to \infty} (a_{1}^n + a_{2}^n + a_{3}^n + .... + a_{r}^n)^{1/n} = a_{1}}$

This post has been edited 5 times. Last edited by P162008, Apr 24, 2025, 5:10 AM
Reason: Typo

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