Summation with power
$begingroup$
Please help me with the below Summation
$sum_{r=0}^{infty}2^r times n^{frac{1}{2^r}}$
While solving problem based on finding time complexity, I came upto this expression. I am unable to move further. Please help.
summation
$endgroup$
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$begingroup$
Please help me with the below Summation
$sum_{r=0}^{infty}2^r times n^{frac{1}{2^r}}$
While solving problem based on finding time complexity, I came upto this expression. I am unable to move further. Please help.
summation
$endgroup$
add a comment |
$begingroup$
Please help me with the below Summation
$sum_{r=0}^{infty}2^r times n^{frac{1}{2^r}}$
While solving problem based on finding time complexity, I came upto this expression. I am unable to move further. Please help.
summation
$endgroup$
Please help me with the below Summation
$sum_{r=0}^{infty}2^r times n^{frac{1}{2^r}}$
While solving problem based on finding time complexity, I came upto this expression. I am unable to move further. Please help.
summation
summation
asked Jan 1 at 8:53
user3767495user3767495
4028
4028
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1 Answer
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$begingroup$
The infinite sum doesn’t converge:
$$lim_{r to infty} 2^rn^{frac{1}{2r}} = lim_{r to infty} 2^r(1) = +infty$$
$endgroup$
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The infinite sum doesn’t converge:
$$lim_{r to infty} 2^rn^{frac{1}{2r}} = lim_{r to infty} 2^r(1) = +infty$$
$endgroup$
add a comment |
$begingroup$
The infinite sum doesn’t converge:
$$lim_{r to infty} 2^rn^{frac{1}{2r}} = lim_{r to infty} 2^r(1) = +infty$$
$endgroup$
add a comment |
$begingroup$
The infinite sum doesn’t converge:
$$lim_{r to infty} 2^rn^{frac{1}{2r}} = lim_{r to infty} 2^r(1) = +infty$$
$endgroup$
The infinite sum doesn’t converge:
$$lim_{r to infty} 2^rn^{frac{1}{2r}} = lim_{r to infty} 2^r(1) = +infty$$
answered Jan 1 at 9:06
KM101KM101
6,0801525
6,0801525
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