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f-x-4x-2-1-for-all-x-R-x-n-k-1-n-1-k-2-3k-6-for-all-n-N-lim-n-f-x-n-




Question Number 56707 by gunawan last updated on 22/Mar/19
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_n )=...
$${f}\left({x}\right)=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\:\:\mathrm{for}\:\mathrm{all}\:{x}\:\in\:\mathbb{R} \\ $$$${x}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{6}}\:\mathrm{for}\:\mathrm{all}\:{n}\:\in\mathbb{N} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{f}\left({x}_{{n}} \right)=… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Mar/19
just trying to justify...  x_n =Σ_(k=1) ^n (1/(k^2 +3k+6))  x_n =((1/(1^2 +3×1+6))+(1/(2^2 +3×2+6))+(1/(3^2 +3×3+6))+..+(1/(n^2 +3n+6)))  =((1/(10))+(1/(16))+(1/(24))+(1/(34))+(1/(46))+...+(1/(n^2 +3n+6)))  now  ((1/(10))+(1/(16))+(1/(24))+(1/(34))+...+(1/(n^2 +3n+6)))<n    now f(x_n )=4x_n ^2 +1  4x_n ^2 +1<4n^2 +1  but lim_(n→∞)  (4x_n ^2 +1)<lim_(n→∞) (4n^2 +1)  it is not solution but ihe image of my thought
$${just}\:{trying}\:{to}\:{justify}… \\ $$$${x}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{6}} \\ $$$${x}_{{n}} =\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} +\mathrm{3}×\mathrm{1}+\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} +\mathrm{3}×\mathrm{2}+\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} +\mathrm{3}×\mathrm{3}+\mathrm{6}}+..+\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{6}}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{16}}+\frac{\mathrm{1}}{\mathrm{24}}+\frac{\mathrm{1}}{\mathrm{34}}+\frac{\mathrm{1}}{\mathrm{46}}+…+\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{6}}\right) \\ $$$${now} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{16}}+\frac{\mathrm{1}}{\mathrm{24}}+\frac{\mathrm{1}}{\mathrm{34}}+…+\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{6}}\right)<{n} \\ $$$$ \\ $$$${now}\:{f}\left({x}_{{n}} \right)=\mathrm{4}{x}_{{n}} ^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{4}{x}_{{n}} ^{\mathrm{2}} +\mathrm{1}<\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1} \\ $$$${but}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{4}{x}_{{n}} ^{\mathrm{2}} +\mathrm{1}\right)<\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${it}\:{is}\:{not}\:{solution}\:{but}\:{ihe}\:{image}\:{of}\:{my}\:{thought} \\ $$$$ \\ $$

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