The correct option is A 12 ∑_k=1^Nk1+k^2+k^4 =12∑_k=1^N2k(1+k+k^2)(1 k+k^2) =12∑_k=1^N11 k+k^2 11+k+k^2 =12[11 13+13 17+17 113+…+11 N+ ...

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Factorization Method Form to Remove Indeterminate Form

limN→∞∑k=1Nk1...

Question

$lim N \to \infty N \sum k = 1 \frac{k}{1 + k^{2} + k^{4}}$ = ?

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\frac{1}{2}

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Solution

The correct option is A $\frac{1}{2}$
$N \sum k = 1 \frac{k}{1 + k^{2} + k^{4}}$
$= \frac{1}{2} N \sum k = 1 \frac{2 k}{(1 + k + k^{2}) (1 - k + k^{2})}$
$= \frac{1}{2} N \sum k = 1 \frac{1}{1 - k + k^{2}} - \frac{1}{1 + k + k^{2}}$
$= \frac{1}{2} [\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{7} + \frac{1}{7} - \frac{1}{13} + \dots + \frac{1}{1 - N + N^{2}} - \frac{1}{1 + N + N^{2}}]$
$= \frac{1}{2} [1 - \frac{1}{1 + N + N^{2}}]$
$∴ lim N \to \infty N \sum k = 1 \frac{k}{1 + k^{2} + k^{4}} = lim N \to \infty \frac{1}{2} [1 - \frac{1}{1 + N + N^{2}}]$
$= \frac{1}{2}$

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Similar questions

L = lim n \to \infty \frac{1}{n^{4}} [1. (n \sum k = 1 k) + 2. (n \sum k = 1 k) + 3. (n \sum k = 1 k) + . . . . . . n .1]

n \sum k = 1 \frac{k}{k^{4} + \frac{1}{4}}

L = lim n \to \infty n \frac{1}{4} [1. (n \sum k = 1 k) + 2. (n - 1 \sum k = 1 k) + 3. (n - 2 \sum k = 1 k) + . . . + n .1]

S_{n} = n \sum k = 1 \frac{k}{(k - 1)^{4 / 3} + (k^{2} - 1)^{2 / 3} + (k + 1)^{4 / 3}}

lim n \to \infty \frac{S_{n}}{n^{2 / 3}} = \frac{1}{p} .

p

S_{n} = n \sum k = 1 a_{n}

lim n \to \infty a_{n} = a

lim n \to \infty \frac{S_{n + 1} - S_{n}}{\sqrt{\sum_{1}^{n} k}} =

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