f(x)=4x^2 +1 for all x ∈ R x_n =Σ_(k=1) ^n (1/(k^2 +3k+6)) for all n ∈N lim_(n→∞) f(x


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Question Number 56707 by gunawan last updated on 22/Mar/19

$f (x) = 4 x^{2} + 1 for all x \in R$ $x_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{k^{2} + 3 k + 6} for all n \in N$ $\lim_{n \to \infty} f (x_{n}) = . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Mar/19

	$j u s t t r y i n g t o j u s t i f y . . .$ $x_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{k^{2} + 3 k + 6}$ $x_{n} = (\frac{1}{1^{2} + 3 \times 1 + 6} + \frac{1}{2^{2} + 3 \times 2 + 6} + \frac{1}{3^{2} + 3 \times 3 + 6} + . . + \frac{1}{n^{2} + 3 n + 6})$ $= (\frac{1}{10} + \frac{1}{16} + \frac{1}{24} + \frac{1}{34} + \frac{1}{46} + . . . + \frac{1}{n^{2} + 3 n + 6})$ $n o w$ $(\frac{1}{10} + \frac{1}{16} + \frac{1}{24} + \frac{1}{34} + . . . + \frac{1}{n^{2} + 3 n + 6}) < n$ $n o w f (x_{n}) = 4 x_{n}^{2} + 1$ $4 x_{n}^{2} + 1 < 4 n^{2} + 1$ $b u t \lim_{n \to \infty} (4 x_{n}^{2} + 1) < \lim_{n \to \infty} (4 n^{2} + 1)$ $i t i s n o t s o l u t i o n b u t i h e i m a g e o f m y t h o u g h t$
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