Help me prove $\lim_{n→∞}\left(\sum_{k=1}^{n}\frac{1}{k} -\log n \right) >1/2$

Question

I am a high school student. I was able to prove that the limit of $\sum_{k=1}^{n}\frac{1}{k}-\log n$ converges. However, I can’t prove the convergence value is greater than $\frac{1}{2}$. I tried a lot and was able to prove it is smaller than $1$. However, I can't greater than $\frac{1}{2}$.

Answer 1

Copied from this answer:

Note that $$ \begin{align} \frac1n-\log\left(\frac{n+1}n\right) &=\int_0^{1/n}\frac{t\,\mathrm{d}t}{1+t}\\ &\ge\int_0^{1/n}\frac{t}{1+\frac1n}\,\mathrm{d}t\\[3pt] &=\frac1{2n(n+1)} \end{align} $$ Therefore, $$ \begin{align} \gamma &=\sum_{n=1}^\infty\left(\frac1n-\log\left(\frac{n+1}n\right)\right)\\ &\ge\sum_{n=1}^\infty\frac1{2n(n+1)}\\[3pt] &=\sum_{n=1}^\infty\frac12\left(\frac1n-\frac1{n+1}\right)\\[6pt] &=\frac12 \end{align} $$

Answer 2

We can apply the Euler-MacLaurin Formula, which says that $$ \sum_{n=a}^b f(n) = \int_a^b f(x) dx + \frac{f(a)+f(b)}{2} + \sum_{k=1}^\infty \frac{B_{2k}}{(2k)!}(f^{(2k-1)}(b)-f^{(2k-1)}(a))+R $$

In this case, $f(n) = 1/n$ and the above summation reduces to

$$ \sum_{k=1}^n \frac{1}{n} = \log n + \frac{1}{2} + \frac{1}{2n} + Other_{terms} $$

Show that the $Other_{terms}$ is positive. For this you need the definition of Bernouli numbers, the $(2k-1)$-th derivative of $1/x$ and the definition of $R$

Answer 3

$$\gamma\stackrel{\text{def}}{=}\lim_{n\to +\infty}\left(H_n-\log n\right)=\sum_{n\geq 1}\left(\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right)\tag{A}$$ Now we may exploit the following facts to derive an integral representation for $\gamma$: $$ \frac{1}{n}=\int_{0}^{+\infty}e^{-nx}\,dx,\qquad \log(n)=\int_{0}^{+\infty}\frac{e^{-x}-e^{-nx}}{x}\,dx\quad(\text{Frullani})\tag{B}$$

$$ \gamma = \int_{0}^{+\infty}\left(\frac{1}{e^x-1}-\frac{1}{x e^x}\right)\,dx =\int_{0}^{1}\left(\frac{1}{\log x}+\frac{1}{1-x}\right)\,dx$$ By the (generalized) Shafer-Fink inequality we may derive arbitrarily accurate uniform approximation for the $\arctan$ and $\text{arctanh}$ functions, hence for the logarithm. We have, for instance, $\log(x)\geq \frac{3(x^2-1)}{x^2+4x+1}$ for any $x\in(0,1)$. By plugging in such inequality in the previous integral representation we get $$ \gamma \leq \frac{1+\log 2}{3} = 0.564382\ldots \tag{C1}$$ while by plugging in the improved inequality $$\forall x\in(0,1),\qquad \log(x)\leq \frac{45(x^2-1)}{7+32(x+1)\sqrt{x}+x(12+7x)}$$ we get: $$ \gamma \geq \frac{71-65\log 2}{45} = 0.576565\ldots\tag{C2} $$ This method can be used for producing even sharper approximations, see here.

Answer 4

$$ \left(\sum_{k=1}^{n}\frac{1}{k} -\log (n+1) \right) $$ has the same limit but is strictly increasing in $n.$ As soon as $n = 6$ the expression above is larger than $\frac{1}{2}.$

An alternative, using a little less, is if you can prove $$ e^3 > 20. $$ YES. $$ e > 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} = \frac{326}{120} = \frac{163}{60}. $$ Now, $$ \left( \frac{163}{60} \right)^3 = \frac{4330747}{216000} > 20, $$ as $20 \cdot 216000 = 4320000 < 4330747.$ $$ e^3 > 20. $$ And so $$ \log 20 < 3. $$ We have, which can be estimated by hand, $H_{19} > 3.54,$ after which $ \log 20 < 3 $ tells us that $\gamma > 0.54$

Fri Nov 10 11:54:52 PST 2017
1  harmonic  1  above  1  below  0.306853
2  harmonic  1.5  above  0.806853  below  0.401388
3  harmonic  1.83333  above  0.734721  below  0.447039
4  harmonic  2.08333  above  0.697039  below  0.473895
5  harmonic  2.28333  above  0.673895  below  0.491574
6  harmonic  2.45  above  0.658241  below  0.50409
7  harmonic  2.59286  above  0.646947  below  0.513416
8  harmonic  2.71786  above  0.638416  below  0.520633
9  harmonic  2.82897  above  0.631744  below  0.526383
10  harmonic  2.92897  above  0.626383  below  0.531073
11  harmonic  3.01988  above  0.621982  below  0.534971
12  harmonic  3.10321  above  0.618304  below  0.538261
13  harmonic  3.18013  above  0.615184  below  0.541076
14  harmonic  3.25156  above  0.612505  below  0.543512
15  harmonic  3.31823  above  0.610179  below  0.54564
16  harmonic  3.38073  above  0.60814  below  0.547516
17  harmonic  3.43955  above  0.606339  below  0.549181
18  harmonic  3.49511  above  0.604736  below  0.550669
19  harmonic  3.54774  above  0.603301  below  0.552007
20  harmonic  3.59774  above  0.602007  below  0.553217
21  harmonic  3.64536  above  0.600836  below  0.554316
22  harmonic  3.69081  above  0.599771  below  0.555319
23  harmonic  3.73429  above  0.598797  below  0.556238
24  harmonic  3.77596  above  0.597904  below  0.557082
25  harmonic  3.81596  above  0.597082  below  0.557862
26  harmonic  3.85442  above  0.596323  below  0.558583
27  harmonic  3.89146  above  0.59562  below  0.559252
28  harmonic  3.92717  above  0.594967  below  0.559875
29  harmonic  3.96165  above  0.594358  below  0.560456
30  harmonic  3.99499  above  0.59379  below  0.561
31  harmonic  4.02725  above  0.593258  below  0.561509
32  harmonic  4.0585  above  0.592759  below  0.561988
33  harmonic  4.0888  above  0.592291  below  0.562438
Fri Nov 10 11:54:52 PST 2017

Answer 5

Maybe you can show that $\sum_{k=1}^{\color{red}{n-1}}\frac{1}{k}-\log n$ converges to the same limit, but is increasing in $n.$ Then you can get arbitrarily good lower bounds by calculating this sequence for large enough $n.$