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[tex]\Huge\boxed{\tt\green{✠Answer}}[/tex]
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[tex] \pmb{\sf{\gray{According \: to \: the \: Question }}}[/tex]
[tex] \bold{ \gamma = \displaystyle \lim_{x \to \infty} \Bigg(\sum\limits_{k=1}^{n} \frac{1}{k} - \displaystyle \ln \: n \Bigg)}[/tex]
[tex]\longrightarrow\bold{ \gamma = \displaystyle \lim_{x \to \infty} \bigg(1 + \frac{1}{2} + \frac{1}{3}. \: . \: . \: . \frac{1}{n} - \displaystyle \ln \: n \bigg)}[/tex]
[tex] \pmb{\sf{\gray{ Multiply \: and \: divide \: 'n' \:,}}} \\ \pmb{\sf{\gray{denominator \: and \: numerator }}}[/tex]
[tex]\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)}[/tex]
[tex]\pmb{\sf{\gray{By \: taking \: in \: common }}}[/tex]
[tex]\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)}[/tex]
[tex]\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)}[/tex]
[tex]\pmb{\sf{\gray{Now \: given \: limit \: x \longrightarrow \infty \: by \: putting,}}}[/tex]
[tex] \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)}
[/tex]
[tex] \sf{ \longrightarrow \infty = \frac{1}{ \infty } \infty = \frac{ \infty }{ \infty } }[/tex]
[tex] \sf{ \longrightarrow \infty = \frac{1}{ \infty } \infty = \frac{ \bcancel\infty }{ \bcancel{\infty } }}[/tex]
[tex] \huge \boxed{ \tt{ \gamma = 0}}[/tex]
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