Is this limit obvious that it is tended to Euler's constant?
$$\lim_{n \to \infty}\sqrt{\sum_{k=1}^{n}H_{k}\left(\frac{1}{k}+\frac{1}{k+1}\right)}-\ln n=\gamma$$
Where $H_k$ is the Harmonic number and $\gamma$; Euler's constant
A limit tends to Euler’s constant
euler-mascheroni-constantlimitssequences-and-series
Related Solutions
We have \begin{align} \sum\limits_{k = 1}^m {\frac{1}{k}} - \log m &= \sum\limits_{k = 1}^m {\frac{1}{k}} - \log \prod\limits_{k = 2}^m {\frac{k}{{k - 1}}} \\ &= \sum\limits_{k = 1}^m {\frac{1}{k}} - \sum\limits_{k = 2}^m {\log \frac{k}{{k - 1}}} \\&= 1 + \sum\limits_{k = 2}^m {\left[ {\frac{1}{k} - \log \frac{k}{{k - 1}}} \right]} \\ &= 1 + \sum\limits_{k = 2}^\infty {\left[ {\frac{1}{k} - \log \frac{k}{{k - 1}}} \right]} - \sum\limits_{k = m + 1}^\infty {\left[ {\frac{1}{k} - \log \frac{k}{{k - 1}}} \right]} \\ &= 1 + \sum\limits_{k = 2}^\infty {\left[ {\frac{1}{k} + \log \left( {1 - \frac{1}{k}} \right)} \right]} - \sum\limits_{k = m + 1}^\infty {\left[ {\frac{1}{k} + \log \left( {1 - \frac{1}{k}} \right)} \right]} . \end{align} By Taylor's theorem $$ \frac{1}{k} + \log \left( {1 - \frac{1}{k}} \right) = - \frac{1}{{2k^2 }} + \mathcal{O}\!\left( {\frac{1}{{k^3 }}} \right), $$ whence the infinite series is convergent and we can write $$ \sum\limits_{k = 1}^m {\frac{1}{k}} - \log m = \gamma - \sum\limits_{k = m + 1}^\infty {\left[ {\frac{1}{k} + \log \left( {1 - \frac{1}{k}} \right)} \right]} , $$ with some constant $\gamma$. By Taylor's formula, $$ \frac{1}{k} + \log \left( {1 - \frac{1}{k}} \right) = - \sum\limits_{j = 2}^\infty {\frac{1}{{jk^j }}} , $$ hence \begin{align} \sum\limits_{k = 1}^m {\frac{1}{k}} - \log m - \gamma = \sum\limits_{k = m + 1}^\infty {\sum\limits_{j = 2}^\infty {\frac{1}{{jk^j }}} } = \sum\limits_{j = 2}^\infty {\frac{1}{j}\sum\limits_{k = m + 1}^\infty {\frac{1}{{k^j }}} } = \sum\limits_{j = 2}^\infty {\frac{1}{{j!}}\sum\limits_{k = m + 1}^\infty {\frac{{(j - 1)!}}{{k^j }}} } . \end{align} By the Euler integral $$ \frac{{(j - 1)!}}{{k^j }} = \int_0^{ + \infty } {e^{ - kt} t^{j - 1} dt} , $$ whence, using the geometric series and the Taylor series of the exponential function, \begin{align} \sum\limits_{k = 1}^m {\frac{1}{k}} - \log m - \gamma &= \sum\limits_{j = 2}^\infty {\frac{1}{{j!}}\sum\limits_{k = m + 1}^\infty {\int_0^{ + \infty } {e^{ - kt} t^{j - 1} dt} } } \\& = \sum\limits_{j = 2}^\infty {\frac{1}{{j!}}\int_0^{ + \infty } {\frac{{e^{ - (m + 1)t} }}{{1 - e^{ - t} }}t^{j - 1} dt} } \\ &= \int_0^{ + \infty } {\frac{{e^{ - (m + 1)t} }}{{1 - e^{ - t} }}\frac{1}{t}\sum\limits_{j = 2}^\infty {\frac{{t^j }}{{j!}}} dt} \\ &= \int_0^{ + \infty } {\frac{{e^{ - mt} }}{{e^t - 1}}\frac{{e^t - t - 1}}{t}dt} \\ & = \int_0^{ + \infty } {e^{ - mt} \left( {1 - \frac{t}{{e^t - 1}}} \right)\frac{1}{t}dt} . \end{align} Now for $0<t<2\pi$, $$ \left( {1 - \frac{t}{{e^t - 1}}} \right)\frac{1}{t} = \frac{1}{2} - \sum\limits_{n = 1}^\infty {\frac{{B_{2n} }}{{(2n)!}}t^{2n - 1} } , $$ with $B_n$ being the Bernoulli numbers. Noting that our function tends to zero at infinity and employing Taylor's theorem, we have that $$ \left| {\left( {1 - \frac{t}{{e^t - 1}}} \right)\frac{1}{t} - \left( {\frac{1}{2} - \sum\limits_{n = 1}^{N - 1} {\frac{{B_{2n} }}{{(2n)!}}t^{2n - 1} } } \right)} \right| \le C_N t^{2N - 1} $$ for $t>0$ and each positive $N$ with a suitable positive constant $C_N$. Therefore, using the Euler integral, \begin{align} \sum\limits_{k = 1}^m {\frac{1}{k}} - \log m - \gamma &= \int_0^{ + \infty } {e^{ - mt} \left( {\frac{1}{2} - \sum\limits_{n = 1}^{N - 1} {\frac{{B_{2n} }}{{(2n)!}}t^{2n - 1} } } \right)dt} + \mathcal{O}(1)\int_0^{ + \infty } {e^{ - mt} t^{2N - 1} dt} \\ &= \frac{1}{2}\int_0^{ + \infty } {e^{ - mt} dt} - \sum\limits_{n = 1}^{N - 1} {\frac{{B_{2n} }}{{(2n)!}}\int_0^{ + \infty } {e^{ - mt} t^{2n - 1} dt} } \\ &\quad \, + \mathcal{O}(1)\int_0^{ + \infty } {e^{ - mt} t^{2N - 1} dt} \\ &= \frac{1}{{2m}} - \sum\limits_{n = 1}^{N - 1} {\frac{{B_{2n} }}{{2n}}\frac{1}{{m^{2n} }}} + \mathcal{O}\! \left( {\frac{1}{{m^{2N} }}} \right). \end{align} Re-arranging and subtracting $1/m$ from both sides gives $$ \sum\limits_{k = 1}^{m - 1} {\frac{1}{k}} = \log m + \gamma - \frac{1}{{2m}} - \sum\limits_{n = 1}^{N - 1} {\frac{{B_{2n} }}{{2n}}\frac{1}{{m^{2n} }}} + \mathcal{O}\!\left( {\frac{1}{{m^{2N} }}} \right). $$ Taking $N=2$ yields your approximation.
With the help of @ParesseuxNguyen, I try to solve it with Stolz-Cesaro's lemma:
$\quad\lim\limits_{x\to\infty}\frac{\sum_{i=1}^x(\sum_{j=1}^i\frac1j-\ln i-\gamma)}{\sum_{i=1}^x\frac1i}\\ =\liminf\limits_{i\to\infty}\frac{\sum_{j=1}^i\frac1j-\ln i-\gamma}{\frac1i}\\ =\limsup\limits_{i\to\infty}\frac{\sum_{j=1}^i\frac1j-\ln i-\gamma}{\frac1i}\\ =\lim \limits_{i\to\infty}\frac{\sum_{j=1}^i\frac1j-\ln i-\gamma}{\frac1i}\\ =\lim \limits_{i\to\infty}\frac i{2i}=\frac12$.
(Since $H_i-\ln i-\gamma\sim\frac1{2i}$, so the last equality is right.)
Best Answer
Note that using Telescoping Series $$ \begin{align} \sum_{k=1}^nH_k\left(\frac1k+\frac1{k+1}\right) &=\sum_{k=1}^nH_k\left(H_{k+1}-H_{k-1}\right)\tag1\\ &=H_{n+1}H_n-H_1H_0\tag2\\[9pt] &=H_{n+1}H_n\tag3 \end{align} $$ Furthermore, since Harmonic Numbers are monotonically increasing, $$ H_n^2\le H_{n+1}H_n\le H_{n+1}^2\tag4 $$ and by the definition of the Euler-Mascheroni Constant, $$ \lim_{n\to\infty}(H_n-\log(n))=\lim_{n\to\infty}(H_{n+1}-\log(n))=\gamma\tag5 $$ The Squeeze Theorem completes the answer.