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We are given a collection of real numbers where a real number $a_{i}\neq 0$ occurs $n_{i}$ times. Let the collection be enumerated as $\left\{x_{1}, x_{2},...x_{n}\right\}$ so that $x_{1}=x_{2}=...=x_{n_{1}}=a_{1}$ and so on, and $n=\sum _{i}n_{i}$ is finite. What is 

$\displaystyle \lim_{k\rightarrow \infty}\left(\sum\limits_{i=1}^{n}\frac{1}{|x_{i}|^{k}}\right)^{-1/k} ?$

  1. $\max_{i}\left(n_{i}|a_{i}|\right)$
  2. $\min_{i} |a_{i}|$
  3. $\min_{i} \left(n_{i}|a_{i}|\right)$
  4. $\max_{i} |a_{i}|$
  5. None of the above
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$\displaystyle\lim_{k\rightarrow \infty}\left(\sum_{i=1}^{n}\frac{1}{|x_{i}|^{k}}\right)^{-1/k} $

$ =\displaystyle\lim_{k \to \infty} \left(\sum_{i} n_i\frac{1}{|a_i|^k}\right)^{-1/k}$ (The $i$ here is different from the previous expression and denote the $i$ unique real values repeated $n_i$ times each.

$=\displaystyle\lim_{k \to \infty} \left(\sum_{i} n_i |a_i|^{-k}\right)^{-1/k}$

Since, $k$ is very large $(\to \infty),$ the summation above will be dominated by the largest value for $n_i |a_i|^{-k}$ which will come for $\min |a_i |$ due to $-k$ in exponent. We can also say that the summation series converges to $n_i {(\min |a_i|)}^{-k}.$ So, our limit becomes

$=\displaystyle \lim_{k \to \infty} \left({n_i {(\min |a_i|)}^{-k}}\right)^{-1/k} = n_i^{-1/k} ((\min |a_i|)^{-k})^{-1/k}$
$\quad = n_i^0 \min |a_i| = \min |a_i|$

Correct Option: B.
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We get $2$ cases:

  1. $k$ is even and
  2. $k$ is odd.

$k$ is even implies

$\sum {\frac{1}{( x(i))^{k}}}$ is always positive and has maximum value.

when $k$ is odd, there might be negative values  $x(i)<0$;

now the total sum for all $k$,the sum results in its minimum value since $1 \div |x(i)|$ values will be subtracted when they are negative.

To find the value, it is given that all $x(n_i) = a(i)$.

therefor the sum results as $pow(sum(n(i) \div a(i))^k, -1 \div k)= pow(min (n(i) \div a(i)),-1)=max(n(i) \div a(i))$.

Hence $A$.

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