what is least upper bound of the sequence {(1+(1/n))ln(1+(1/n))}


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Question Number 174389 by Best1 last updated on 31/Jul/22
$what is least upper bound of the sequence {(1+(1/n))ln(1+(1/n))}_(n=1) ^∞ ? A.0 B.ln2 C.ln5 D.2ln2$
$w h a t i s l e a s t u p p e r b o u n d o f t h e$ $s e q u e n c e {(1 + \frac{1}{n}) l n (1 + \frac{1}{n})}_{n = 1}^{\infty} ?$ $A . 0 B . l n 2 C . l n 5 D . 2 l n 2$
	Answered by Gazella thomsonii last updated on 31/Jul/22
	$Let′s think of a Sequence A_k Function as f(z) and Find Root f^((1)) (z)=0 for find inf [f(z),z_0 ] so , f(z)=(1+(1/z))ln(1+(1/z)) ((df(z))/dz)=−((1/z))^2 ln(1+(1/z))−((1+(1/z))/(1+(1/z)))∙((d )/dz)(1+(1/z))=0 ∴z_0 =−(e/(e−1)) inf[f(z),z=−(e/(e−1))]=lim_(x→z_0 ) f(z)=−(1/e) But, z∈Z^+ /{0} we don′t need inf[f(z)] when z is less than 0 inf[f(z),z=∞]=0 (becauselim_(z→∞) f(z)=0) sup is not Exist. (because lim_(z→0) f(z)=∞, f^((1)) (a)<0 ,a∈Z^+ /{0})$
	${Let}^{'} s think of a Sequence A_{k} Function as f (z)$ $and Find Root f^{(1)} (z) = 0 for find \inf [f (z), z_{0}]$ $so, f (z) = (1 + \frac{1}{z}) \ln (1 + \frac{1}{z})$ $\frac{d f (z)}{d z} = - {(\frac{1}{z})}^{2} \ln (1 + \frac{1}{z}) - \frac{1 + \frac{1}{z}}{1 + \frac{1}{z}} \cdot \frac{d}{d z} (1 + \frac{1}{z}) = 0$ $∴ z_{0} = - \frac{e}{e - 1}$ $\inf [f (z), z = - \frac{e}{e - 1}] = \lim_{x \to z_{0}} f (z) = - \frac{1}{e}$ $But, z \in Z^{+} / {0} we {don}^{'} t need \inf [f (z)] when z is less than 0$ $\inf [f (z), z = \infty] = 0 (because \lim_{z \to \infty} f (z) = 0)$ $\sup is not Exist . (because \lim_{z \to 0} f (z) = \infty, f^{(1)} (a) < 0, a \in Z^{+} / {0})$
	Commented by Best1 last updated on 31/Jul/22

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