vrotterigzl

Answered

2022-06-24

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

Answer & Explanation

Trey Ross

Expert

2022-06-25Added 30 answers

Let ${x}_{n}$ be the sequence given by

$\begin{array}{rl}{x}_{n}& =\frac{1}{n}(\sum _{k=1}^{n}\frac{n+1-k}{k}-\mathrm{log}(k))\\ \\ & =\frac{1}{n}(\sum _{k=1}^{n}\frac{n}{k}+\frac{1}{k}-1-\mathrm{log}(k))\\ \\ \text{(1)}& & =-1+\underset{\to 0\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{as}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\to \mathrm{\infty}}{\underset{\u23df}{\frac{1}{n}\sum _{k=1}^{n}\frac{1}{k}}}+\underset{\to \gamma \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{as}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\to \mathrm{\infty}}{\underset{\u23df}{\sum _{k=1}^{n}\frac{1}{k}-\mathrm{log}(n)}}-\underset{\to -1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{as}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\to \mathrm{\infty}}{\underset{\u23df}{\frac{1}{n}\sum _{k=1}^{n}\mathrm{log}(k/n)}}\end{array}$

Therefore, we find that

$\underset{n\to \mathrm{\infty}}{lim}{x}_{n}=\gamma $

$\begin{array}{rl}{x}_{n}& =\frac{1}{n}(\sum _{k=1}^{n}\frac{n+1-k}{k}-\mathrm{log}(k))\\ \\ & =\frac{1}{n}(\sum _{k=1}^{n}\frac{n}{k}+\frac{1}{k}-1-\mathrm{log}(k))\\ \\ \text{(1)}& & =-1+\underset{\to 0\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{as}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\to \mathrm{\infty}}{\underset{\u23df}{\frac{1}{n}\sum _{k=1}^{n}\frac{1}{k}}}+\underset{\to \gamma \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{as}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\to \mathrm{\infty}}{\underset{\u23df}{\sum _{k=1}^{n}\frac{1}{k}-\mathrm{log}(n)}}-\underset{\to -1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{as}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\to \mathrm{\infty}}{\underset{\u23df}{\frac{1}{n}\sum _{k=1}^{n}\mathrm{log}(k/n)}}\end{array}$

Therefore, we find that

$\underset{n\to \mathrm{\infty}}{lim}{x}_{n}=\gamma $

Emmy Knox

Expert

2022-06-26Added 10 answers

By the Stolz-Cesàro theorem

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

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

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