Question

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

Answer 1

Answer:

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

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ple calculation,

e−nn∑k=0nkk!=e−nn!n∑k=0(nk)nk(n−k)!(1)⋯=e−nn!n∑k=0(nk)nk∫∞0tn−ke−tdt=e−nn!∫∞0(n+t)ne−tdt(2)⋯=1n!∫∞ntne−tdt=1−1n!∫n0tne−tdt(3)⋯=1−√n(n/e)nn!∫√n0(1−u√n)ne√nudu.

We remark that

In (1), we utilized the famous formula n!=∫∞0tne−tdt.In (2), the substitution t+n↦t is used.In (3), the substitution t=n−√nu is used.

Then in view of the Stirling's formula, it suffices to show that

∫√n0(1−u√n)n

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Answer 2

Answer:

∫√n0(1−u√n)n

Step-by-step explanation:

ple calculation,

e−nn∑k=0nkk!=e−nn!n∑k=0(nk)nk(n−k)!(1)⋯=e−nn!n∑k=0(nk)nk∫∞0tn−ke−tdt=e−nn!∫∞0(n+t)ne−tdt(2)⋯=1n!∫∞ntne−tdt=1−1n!∫n0tne−tdt(3)⋯=1−√n(n/e)nn!∫√n0(1−u√n)ne√nudu.

We remark that

In (1), we utilized the famous formula n!=∫∞0tne−tdt.In (2), the substitution t+n↦t is used.In (3), the substitution t=n−√nu is used.

Then in view of the Stirling's formula, it suffices to show that

∫√n0(1−u√n)n