Question

$\huge \bold{ \gamma = \displaystyle \lim_{x \to \infty} \Bigg(\sum\limits_{k=1}^{n} \frac{1}{k} - \displaystyle \ln \: n \Bigg)}$



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$\sf{No \: pic \: answers}$
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Answer 1

$\begin{gathered}{ \orange{\underbrace{{\Huge{\textsf{\textbf{\red{Answer}}}}}}}}\end{gathered}$

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$\pmb{\sf{\gray{According \: to \: the \: Question }}}$

$\bold{ \gamma = \displaystyle \lim_{x \to \infty} \Bigg(\sum\limits_{k=1}^{n} \frac{1}{k} - \displaystyle \ln \: n \Bigg)}$

$\longrightarrow\bold{ \gamma = \displaystyle \lim_{x \to \infty} \bigg(1 + \frac{1}{2} + \frac{1}{3}. \: . \: . \: . \frac{1}{n} - \displaystyle \ln \: n \bigg)}$

$\pmb{\sf{\gray{ Multiply \: and \: divide \: 'n' \:,}}} \\ \pmb{\sf{\gray{denominator \: and \: numerator }}}$

$\longrightarrow\bold{ \gamma = \displaystyle \lim_{x \to \infty} \bigg( \frac{n}{1n} + \frac{n}{2n} + \frac{n}{3n} ... \: ... + \frac{n}{n^{2} } - \displaystyle \frac{n \: \ln \: n}{n} \bigg)}$

$\pmb{\sf{\gray{By \: taking \: in \: common }}}$

$\longrightarrow\bold{ \gamma = \displaystyle \lim_{x \to \infty} \bigg( \frac{1}{n} \bigg( \frac{n}{1} + \frac{n}{2} + \frac{n}{3} ... \: ... + \frac{n}{n } - \displaystyle n \: \ln \: n \bigg)}$

$\tt{So,}\longrightarrow\bold{ \gamma = \displaystyle \lim_{x \to \infty} \bigg( \frac{1}{x} \bigg( \frac{x}{1} + \frac{x}{2} + \frac{x}{3} ... \: ... + 1 - \displaystyle x\: \ln \:x\bigg)}$

$\pmb{\sf{\gray{Now \: given \: limit \: x \longrightarrow \infty \: by \: putting,}}}$

$\tt{So,} \longrightarrow\bold{ \gamma = \displaystyle \lim_{x \to \infty} \bigg( \frac{1}{ \infty } \bigg( \frac{ \infty }{1} + \frac{ \infty }{2} + \frac{ \infty }{3} ... \: ... + 1 - \displaystyle \infty \: \ln \: \infty \bigg)}$

$\sf{ \longrightarrow \infty = \frac{1}{ \infty } \infty = \frac{ \infty }{ \infty } }$

$\sf{ \longrightarrow \infty = \frac{1}{ \infty } \infty = \frac{ \bcancel\infty }{ \bcancel{\infty } }}$

$\huge \boxed{ \tt{ \gamma = 0}}$

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Answer 2

$\huge \mathcal \colorbox{lavenderblush}{{ \color{b} \huge \purple{➳ SoLuTiOn \:➳ }}}$

. . . See the attachment . . .