vrotterigzl

2022-06-24

Find $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\left(n+\frac{n-1}{2}+\frac{n-2}{3}+\cdots +\frac{2}{n-1}+\frac{1}{n}-\mathrm{log}\left(n!\right)\right)$

Trey Ross

Expert

Let ${x}_{n}$ be the sequence given by
$\begin{array}{rl}{x}_{n}& =\frac{1}{n}\left(\sum _{k=1}^{n}\frac{n+1-k}{k}-\mathrm{log}\left(k\right)\right)\\ \\ & =\frac{1}{n}\left(\sum _{k=1}^{n}\frac{n}{k}+\frac{1}{k}-1-\mathrm{log}\left(k\right)\right)\\ \\ \text{(1)}& & =-1+\underset{\to 0\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{as}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\to \mathrm{\infty }}{\underset{⏟}{\frac{1}{n}\sum _{k=1}^{n}\frac{1}{k}}}+\underset{\to \gamma \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{as}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\to \mathrm{\infty }}{\underset{⏟}{\sum _{k=1}^{n}\frac{1}{k}-\mathrm{log}\left(n\right)}}-\underset{\to -1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{as}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\to \mathrm{\infty }}{\underset{⏟}{\frac{1}{n}\sum _{k=1}^{n}\mathrm{log}\left(k/n\right)}}\end{array}$
Therefore, we find that
$\underset{n\to \mathrm{\infty }}{lim}{x}_{n}=\gamma$

Emmy Knox

Expert

By the Stolz-Cesàro theorem
$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}{x}_{n}& =& \underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{k=1}^{n}\left(\frac{n+1-k}{k}-\mathrm{log}\left(k\right)\right)\\ & =& \underset{n\to \mathrm{\infty }}{lim}\frac{\sum _{k=1}^{n}\left(\frac{n+1-k}{k}-\mathrm{log}\left(k\right)\right)}{n}\\ & =& \underset{n\to \mathrm{\infty }}{lim}\frac{\sum _{k=1}^{n+1}\left(\frac{n+2-k}{k}-\mathrm{log}\left(k\right)\right)-\sum _{k=1}^{n}\left(\frac{n+1-k}{k}-\mathrm{log}\left(k\right)\right)}{\left(n+1\right)-n}\\ & =& \underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n+1}\left(\frac{n+1-k}{k}+\frac{1}{k}-\mathrm{log}\left(k\right)\right)-\sum _{k=1}^{n}\left(\frac{n+1-k}{k}-\mathrm{log}\left(k\right)\right)\\ & =& \underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n+1}\frac{1}{k}+\sum _{k=1}^{n+1}\left(\frac{n+1-k}{k}-\mathrm{log}\left(k\right)\right)-\sum _{k=1}^{n}\left(\frac{n+1-k}{k}-\mathrm{log}\left(k\right)\right)\\ & =& \underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n+1}\frac{1}{k}-\mathrm{log}\left(n+1\right)\\ & =& \gamma \end{array}$

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