Consider the sequence 1+1/2+...+1/n - log n. I have shown that it is strictly decreasing and bounded below by zero. I need to show that the limit of this sequence ( the Euler Constant) lies in (1/2,3/5).
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2If you did some calculus maybe you can condider some integral approximazions of 1/x=f(x) – mathreadler Jul 20 '18 at 16:33
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$ 1+1/2+...+1/n - \log (n+1)$ is strictly increasing and always below $ 1+1/2+...+1/n - \log n$ – Will Jagy Jul 20 '18 at 18:22
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Start with $\gamma =\lim_{n \to \infty} (\sum_{k=1}^n \frac1{k}-\ln(n)) $.
Let
$s_n =\sum_{k=1}^{n-1} \frac1{k}-\ln(n) $.
I will show by elementary means that $\frac{5}{16} \lt s_n \lt \frac{49}{64} $.
More refined bounds can be gotten by taking more terms in the approximations below.
In the limit, we get this:
$s_n \to \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n+1)}\zeta(n+1) $ as $n \to \infty$.
Obviously, nothing here is original, but it was fun deriving it.
$\begin{array}\\ s_n &=\sum_{k=1}^{n-1} \frac1{k}-\ln(n)\\ &=\sum_{k=1}^{n-1} \frac1{k}-\int_1^n \frac{dt}{t}\\ &=\sum_{k=1}^{n-1} \frac1{k}-\sum_{k=1}^{n-1}\int_k^{k+1} \frac{dt}{t}\\ &=\sum_{k=1}^{n-1} (\frac1{k}-\int_k^{k+1} \frac{dt}{t})\\ &=\sum_{k=1}^{n-1} (\int_k^{k+1} (\frac1{k}-\frac1{t})dt\\ &=\sum_{k=1}^{n-1} (\int_k^{k+1} (\frac{t-k}{tk})dt\\ &=\sum_{k=1}^{n-1} (\int_0^{1} (\frac{t}{(t+k)k})dt\\ &=\sum_{k=1}^{n-1} \frac1{k}\int_0^{1} \frac{t}{t+k}dt\\ &=\sum_{k=1}^{n-1} \frac1{k^2}\int_0^{1} \frac{t}{1+t/k}dt\\ &\lt\sum_{k=1}^{n-1} \frac1{k^2}\int_0^{1} t(1-\frac{t}{k}+\frac{t^2}{k^2})dt \qquad\text{since }\frac1{1+x} < 1-x+x^2\\ &=\sum_{k=1}^{n-1} \frac1{k^2}\int_0^{1} (t-\frac{t^2}{k}+\frac{t^3}{k^2})dt\\ &=\sum_{k=1}^{n-1} \frac1{k^2}(\frac12-\frac{1}{3k}+\frac1{4k^2})\\ &=\sum_{k=1}^{n-1} \frac{6k^2-4k+3}{12k^4}\\ &=\frac{5}{12}+\frac{19}{192}+\sum_{k=3}^{n-1} \frac{6k^2-4k+3}{12k^4}\\ &<\frac{33}{64}+\sum_{k=3}^{n-1} \frac{1}{2k(k-1)} \qquad\text{since } \frac{6k^2-4k+3}{12k^4}<\frac{1}{2k(k-1)}\\ &=\frac{33}{64}+\frac12\sum_{k=3}^{n-1} (\frac1{k-1}-\frac1{k})\\ &=\frac{33}{64}+\frac14\\ &=\frac{49}{64}\\ \text{and}\\ s_n &=\sum_{k=1}^{n-1} \frac1{k^2}\int_0^{1} \frac{t}{1+t/k}dt\\ &\gt\sum_{k=1}^{n-1} \frac1{k^2}\int_0^{1} t(1-\frac{t}{k})dt \qquad\text{since }\frac1{1+x} > 1-x\\ &=\sum_{k=1}^{n-1} \frac1{k^2}\int_0^{1} (t-\frac{t^2}{k})dt\\ &=\sum_{k=1}^{n-1} \frac1{k^2}(\frac12-\frac1{3k})\\ &=\sum_{k=1}^{n-1} \frac{3k-2}{6k^3}\\ &=\frac16+\frac{1}{12}+\sum_{k=3}^{n-1} \frac{3k-2}{6k^3}\\ &>\frac{7}{48}+\sum_{k=3}^{n-1} \frac{3k-3}{6(k-1)k(k+1)}\\ &=\frac{7}{48}+\sum_{k=3}^{n-1} \frac{1}{2k(k+1)}\\ &=\frac{7}{48}+\frac12\sum_{k=3}^{n-1} (\frac1{k}-\frac1{k+1})\\ &=\frac{7}{48}+\frac16\\ &=\frac{15}{48}\\ &=\frac{5}{16}\\ \end{array} $
Note: From
$\begin{array}\\ s_n &=\sum_{k=1}^{n-1} \frac1{k^2}\int_0^{1} \frac{t}{1+t/k}dt\\ &=\sum_{k=1}^{n-1} \frac1{k^2}\int_0^{1} \sum_{n=0}^{\infty} \frac{(-1)^{n}t^{n}}{k^n}dt\\ &=\sum_{k=1}^{n-1} \frac1{k^2}\int_0^{1} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}t^n}{k^{n-1}}dt\\ &=\sum_{k=1}^{n-1} \frac1{k^2} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(n+1)k^{n-1}}\\ &= \sum_{n=1}^{\infty}\sum_{k=1}^{n-1} \frac1{k^2} \frac{(-1)^{n-1}}{(n+1)k^{n-1}}\\ &= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n+1)}\sum_{k=1}^{n-1} \frac1{k^2} \frac{1}{k^{n-1}}\\ &= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n+1)}\sum_{k=1}^{n-1} \frac{1}{k^{n+1}}\\ &\to \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n+1)}\sum_{k=1}^{\infty} \frac{1}{k^{n+1}} \qquad\text{as } n \to \infty\\ &= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n+1)}\zeta(n+1)\\ \end{array} $
marty cohen
- 107,799
0
Um... $H_n - \log n $ decreases. At the same time, $H_n - \log (n+1)$ increases, but is always a bit smaller than the first one. The difference between these two sequences is $\log \left( \frac{n+1}{n} \right)$ which approaches zero. Therefore there is a limit, which always lies between them. By the time $n\geq 6$ we know the limit is larger than $0.5,$ while $n \geq 22$ shows the limit is below $0.6 \; \; .$
Fri Jul 20 11:34:58 PDT 2018
n Hn Hn - log(n+1) Hn- log(n)
1 1 0.3068528194400547 1
2 1.5 0.4013877113318902 0.8068528194400547
3 1.833333333333333 0.4470389722134427 0.7347210446652235
4 2.083333333333333 0.4738954208992328 0.6970389722134425
5 2.283333333333333 0.4915738641052783 0.6738954208992329
6 2.45 0.5040898509446865 0.6582405307719448
7 2.592857142857143 0.5134156011773068 0.6469469938018293
8 2.717857142857143 0.520632565520923 0.6384156011773068
9 2.828968253968254 0.5263831609742078 0.6317436766320341
10 2.928968253968254 0.5310729811698831 0.6263831609742079
11 3.019877344877345 0.5349706950893443 0.6219820720789739
12 3.103210678210678 0.5382613207491413 0.6183040284226777
13 3.180133755133755 0.5410764255184968 0.6151843976722184
14 3.251562326562327 0.5435121254601167 0.6125049969470684
15 3.318228993228994 0.5456402709892125 0.6101787921267836
16 3.380728993228994 0.5475156491727775 0.6081402709892125
17 3.439552522640758 0.5491807647445937 0.606339178584542
18 3.495108078196314 0.5506690990298733 0.604736320300149
19 3.547739657143682 0.5520073835896913 0.6033006779772419
20 3.597739657143682 0.5532172194202589 0.6020073835896911
21 3.645358704762729 0.5543162514044133 0.6008362670393064
22 3.690813250217275 0.5553190342881251 0.5997707968589587
23 3.73429151108684 0.5562376807388945 0.5987972951576905
24 3.775958177753507 0.5570823528853062 0.597904347405561
25 3.815958177753507 0.5578616397320246 0.5970823528853062
n Hn Hn - log(n+1) Hn- log(n)
Fri Jul 20 11:34:58 PDT 2018
Will Jagy
- 139,541