calculus-II-find-convergence-of-n-2-1-n-ln-3-n-ln-n-

Question Number 140671 by mnjuly1970 last updated on 11/May/21

\dots \dots c a l c u l u s \dots . (I I) \dots . .

f i n d c o n v e r g e n c e o f ::

ϕ := \underset{n = 2}{\sum^{\infty}} \frac{1}{n ({l n}^{3} (n) + l n (n))}

Answered by mathmax by abdo last updated on 11/May/21

φ (x) = \frac{1}{x (\log^{3} x + logx)} (x > 2) \Rightarrow φ^{^{'}} (x) = - \frac{\log^{3} x + logx + x (\frac{{3 \log}^{2} x}{x} + \frac{1}{x})}{x^{2} {(\log^{3} x + logx)}^{2}}

= - \frac{\log^{3} x + logx + {3 \log}^{2} x + 1}{x^{2} {(\log^{3} x + logx)}^{2}} < 0 \Rightarrow φ is decreazing on [2, + \infty [\Rightarrow

so Σ (\dots) and \int_{e}^{\infty} \frac{dt}{t (\log^{3} t + logt)} are the same nature of convergence

Φ = \int_{e}^{\infty} \frac{dt}{t (\log^{3} t + logt)} =_{logt = x} \int_{1}^{\infty} \frac{e^{x} dx}{e^{x} (x^{3} + x)}

= \int_{1}^{\infty} \frac{dx}{x^{3} + x} = \int_{1}^{\infty} \frac{dx}{x (x^{2} + 1)} = \int_{1}^{\infty} (\frac{1}{x} - \frac{x}{1 + x^{2}}) dx

= {[\ln ∣ \frac{x}{\sqrt{1 + x^{2}}} ∣]}_{1}^{\infty} < + \infty \Rightarrow Φ converges \Rightarrow Σ u_{n} converge \dots

Commented by mnjuly1970 last updated on 12/May/21

g r e a t s i r m a x \dots