Question

$\huge \bold{ \gamma = \displaystyle \lim_{x \to \infty} (\sum\limits_{k=1}^{n} \frac{1}{k} - \displaystyle \ln \: n)}$
$\bold{Gøogle's \: recommendation:)}$
$\bold \red{answer \: it}$
​

Answer 1

$\large\color{lime}\boxed{\colorbox{black}{Answer : - }}$

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

$\gamma (k + 1) - \gamma (k) = \frac{1}{k} = > \sum\limits_{k=1}^{n} \frac{1}{k} </p><p>$

$= \gamma \: (n + 1) + \gamma$

$= \gamma (1) = - \gamma$

$2) \: \gamma (x) = log(x) - \frac{1}{2x} - 0( \frac{1}{ {x}^{2} } )$

$now :$

$lim_{x \to \infty}(\sum\limits_{k=1}^{n} \frac{1}{k} - \displaystyle log(n) = lim_{x \to \infty} \: \gamma (n + 1) + \gamma - 10y(n)</p><p></p><p>[tex] \gamma = 0.5772$