TITLE: Is there a connection between the sum of $\sum_{n=1}^\infty \frac{\ln(n+1)-\ln(n)}{n}$ and the Riemann zeta function?
QUESTION [6 upvotes]: It can be shown (see the bottom of my post) that for every integer $p\geq 0$, this beautiful identity holds:
\begin{align}
\int_1^\infty\frac{1}{x^{p+2}\lfloor x\rfloor}\text{ }dx &= -1+\frac{1}{p+1}\sum_{m=2}^{p+2}\zeta(m)\\
&= -1+\frac{\zeta(2)+\zeta(3)+\zeta(4)+\cdots+\zeta(p+2)}{p+1}
\end{align}
I find it pretty interesting that the quantity
$$\frac{\zeta(2)+\zeta(3)+\zeta(4)+\cdots+\zeta(p+2)}{p+1}$$
is precisely the arithmetic mean of the numbers $\zeta(2),\zeta(3),\zeta(4),\dots,\zeta(p+2)$. Anyway, notice that the integral
$$\int_1^\infty\frac{1}{x^{p+2}\lfloor x\rfloor}\text{ }dx$$
also converges for $p=-1$. This is intuitively plausible because $\lfloor x\rfloor$ grows roughly like $x$, so
$$\frac{1}{x^{-1+2}\lfloor x\rfloor}=\frac{1}{x\lfloor x\rfloor}$$
behaves like $1/x^2$ as $x\to\infty$, yielding a convergent integral. To clear up any doubts, here's a rigorous proof:
The sequence
$$\left\lbrace\int_1^{k+1}\frac{1}{x\lfloor x\rfloor}\text{ }dx\right\rbrace_{k=0}^\infty$$
is strictly increasing because the integrand is strictly positive over $[1,k+1]$. Now,
\begin{align}
\int_1^{k+1}\frac{1}{x\lfloor x\rfloor}\text{ }dx &= \int_1^2\frac{1}{x\lfloor x\rfloor}\text{ }dx+\int_2^3\frac{1}{x\lfloor x\rfloor}\text{ }dx+\cdots+\int_k^{k+1}\frac{1}{x\lfloor x\rfloor}\text{ }dx\\
&= \sum_{n=1}^k \int_n^{n+1}\frac{1}{x\lfloor x\rfloor}\text{ }dx\\
&= \sum_{n=1}^k \int_n^{n+1}\frac{1}{x\cdot n}\text{ }dx\\
&= \sum_{n=1}^k \frac{1}{n}\int_n^{n+1}\frac{1}{x}\text{ }dx\\
&= \color{green}{\sum_{n=1}^k \frac{\ln(n+1)-\ln(n)}{n}}\\
&= \sum_{n=1}^k \frac{1}{n}\ln\left(\frac{n+1}{n}\right)\\
&= \sum_{n=1}^k \frac{1}{n}\ln\left(1+\frac{1}{n}\right)\\
&< \sum_{n=1}^k \frac{1}{n}\cdot\frac{1}{n}\\
&< \sum_{n=1}^\infty \frac{1}{n^2}
\end{align}
so the terms of the sequence are bounded above. It follows by the monotonic sequence theorem that
$$\left\lbrace\int_1^{k+1}\frac{1}{x\lfloor x\rfloor}\text{ }dx\right\rbrace_{k=0}^\infty$$
is a convergent sequence. $\blacksquare$
Now, the equation
$$\int_1^\infty\frac{1}{x^{p+2}\lfloor x\rfloor}\text{ }dx=-1+\frac{1}{p+1}\sum_{m=2}^{p+2}\zeta(m)$$
doesn't make sense for $p=-1$ because substituting $-1$ for $p$ involves division by zero. Nevertheless, given the obvious connection between
$$\int_1^\infty\frac{1}{x^{p+2}\lfloor x\rfloor}\text{ }dx$$
and $\zeta(n)$ for $n\in\mathbb{N}$, I can't help but wonder if there's a link between
\begin{align}
\int_1^\infty\frac{1}{x\lfloor x\rfloor}\text{ }dx &= \lim_{k\to\infty}\left(\int_1^{k+1}\frac{1}{x\lfloor x\rfloor}\text{ }dx\right)\\
&= \lim_{k\to\infty}\left(\sum_{n=1}^k \frac{\ln(n+1)-\ln(n)}{n}\right)\\
&= \sum_{n=1}^\infty \frac{\ln(n+1)-\ln(n)}{n}
\end{align}
and the zeta function. Maybe it’s the exact value of zeta evaluated at some other point in the complex plane?
Seeing the pattern in the identity, my only "reasonable" idea was to try and find a link between the sum the series (I'll call it $S$) and $\zeta(1)$. This expression is undefined ($\zeta$ has a singularity at $1$), but its principal value exists and equals the Euler-Mascheroni constant:
$$\lim_{h\to 0}\frac{\zeta(1-h)+\zeta(1+h)}{2}=\gamma$$
We can't have an exact equality; $\gamma\approx 0.57$, but [WolframAlpha][2] says that $S\approx 1.25$. It's unlikely, but could these numbers be related in some other way? If not, are there any other possible links between $S$ and the zeta function? Any answer is greatly appreciated.
Proof of the top identity:
Start with this chain of equalities:
\begin{align}
\int_1^{k+1}\frac{1}{x^{p+2}\lfloor x\rfloor}\text{ }dx &= \int_1^2\frac{1}{x^{p+2}\lfloor x\rfloor}\text{ }dx+\int_2^3\frac{1}{x^{p+2}\lfloor x\rfloor}\text{ }dx+\cdots+\int_k^{k+1}\frac{1}{x^{p+2}\lfloor x\rfloor}\text{ }dx\\
&= \sum_{n=1}^k \int_n^{n+1}\frac{1}{x^{p+2}\lfloor x\rfloor}\text{ }dx\\
&= \sum_{n=1}^k \int_n^{n+1}\frac{1}{x^{p+2}\cdot n}\text{ }dx\\
&= \sum_{n=1}^k \frac{1}{n}\int_n^{n+1} x^{-p-2}\text{ }dx\\
&= \sum_{n=1}^k \frac{1}{n}\left(\frac{(n+1)^{-p-1}}{-p-1}-\frac{n^{-p-1}}{-p-1}\right)\\
&= \sum_{n=1}^k \frac{1}{n}\cdot -\frac{1}{p+1}\left(\frac{1}{(n+1)^{p+1}}-\frac{1}{n^{p+1}}\right)\\
&= \frac{1}{p+1}\sum_{n=1}^k \left(\frac{1}{n^{p+2}}-\frac{1}{n(n+1)^{p+1}}\right)\\
&= \frac{1}{p+1}\sum_{n=1}^k \frac{1}{n^{p+2}}-\frac{1}{p+1}\sum_{n=1}^k \frac{1}{n(n+1)^{p+1}}
\end{align}
To evaluate the second sum, we use the partial fraction decomposition
$$\frac{1}{n(n+1)^{p+1}}=\frac{1}{n}-\sum_{m=1}^{p+1}\frac{1}{(n+1)^m}$$
This gives
\begin{align}
\frac{1}{p+1}\sum_{n=1}^k \frac{1}{n^{p+2}}-\frac{1}{p+1}\sum_{n=1}^k \frac{1}{n(n+1)^{p+1}} &= \frac{1}{p+1}\sum_{n=1}^k \frac{1}{n^{p+2}}-\frac{1}{p+1}\sum_{n=1}^k \left(\frac{1}{n}-\sum_{m=1}^{p+1}\frac{1}{(n+1)^m}\right)\\
&\text{(applied partial fraction decomposition)}\\
&= \frac{1}{p+1}\sum_{n=1}^k \frac{1}{n^{p+2}}-\frac{1}{p+1}\sum_{n=1}^k \left(\frac{1}{n}-\frac{1}{n+1}-\sum_{m=2}^{p+1}\frac{1}{(n+1)^m}\right)\\
&\text{(separated }m=1\text{ term from the sum)}\\
&= \frac{1}{p+1}\sum_{n=1}^k \frac{1}{n^{p+2}}-\frac{1}{p+1}\sum_{n=1}^k\left(\frac{1}{n}-\frac{1}{n+1}\right)+\frac{1}{p+1}\sum_{n=1}^k\sum_{m=2}^{p+1}\frac{1}{(n+1)^m}\\
&\text{(distributive property)}\\
&= \frac{1}{p+1}\sum_{n=1}^k \frac{1}{n^{p+2}}+\frac{1}{p+1}\sum_{n=1}^k\left(\frac{1}{n+1}-\frac{1}{n}\right)+\frac{1}{p+1}\sum_{m=2}^{p+1}\sum_{n=1}^k\frac{1}{(n+1)^m}\\
&\text{(rewrote the second sum for telescoping and interchanged the double sum on the right)}\\
&= \frac{1}{p+1}\sum_{n=1}^k \frac{1}{n^{p+2}}+\frac{1}{p+1}\left(\frac{1}{k+1}-1\right)+\frac{1}{p+1}\sum_{m=2}^{p+1}\sum_{n=2}^{k+1}\frac{1}{n^m}\\
&\text{(evaluated telescoping sum and performed an index-shift in the double sum)}\\
&= \frac{1}{p+1}\sum_{n=1}^k \frac{1}{n^{p+2}}+\frac{1}{(p+1)(k+1)}-\frac{1}{p+1}+\frac{1}{p+1}\sum_{m=2}^{p+1}\left(-1+\sum_{n=1}^{k+1}\frac{1}{n^m}\right)\\
&\text{(rewrote middle term and completed zeta sum in double sum)}\\
&= \frac{1}{p+1}\sum_{n=1}^k \frac{1}{n^{p+2}}+\frac{1}{(p+1)(k+1)}-\frac{1}{p+1}+\frac{1}{p+1}\sum_{m=2}^{p+1}(-1)+\frac{1}{p+1}\sum_{m=2}^{p+1}\sum_{n=1}^{k+1}\frac{1}{n^m}\\
&\text{(distributive property)}\\
&= \frac{1}{p+1}\sum_{n=1}^k \frac{1}{n^{p+2}}+\frac{1}{(p+1)(k+1)}-\frac{1}{p+1}-\frac{p+1-2+1}{p+1}+\frac{1}{p+1}\sum_{m=2}^{p+1}\sum_{n=1}^{k+1}\frac{1}{n^m}\\
&\text{(evaluated sum over the constant)}\\
&= \frac{1}{(p+1)(k+1)}-1+\frac{1}{p+1}\sum_{n=1}^k \frac{1}{n^{p+2}}+\frac{1}{p+1}\sum_{m=2}^{p+1}\sum_{n=1}^{k+1}\frac{1}{n^m}
\end{align}
Letting $k\to\infty$ yields the desired result.
\begin{align}
\int_1^\infty\frac{1}{x^{p+2}\lfloor x\rfloor}\text{ }dx &= \lim_{k\to\infty}\left(\frac{1}{(p+1)(k+1)}-1+\frac{1}{p+1}\sum_{n=1}^k \frac{1}{n^{p+2}}+\frac{1}{p+1}\sum_{m=2}^{p+1}\sum_{n=1}^{k+1}\frac{1}{n^m}\right)\\
&= -1+\frac{1}{p+1}\zeta(p+2)+\frac{1}{p+1}\sum_{m=2}^{p+1}\zeta(m)\\
&= -1+\frac{1}{p+1}\sum_{m=2}^{p+2}\zeta(m)
\end{align}
Q.E.D.

REPLY [7 votes]: If we expand the logarithm into its Maclaurin series and change the order of summations, we obtain
$$
\sum\limits_{k = 1}^\infty  {\frac{1}{n}\log \left( {1 + \frac{1}{n}} \right)}  = \sum\limits_{k = 1}^\infty  {\frac{{( - 1)^{k + 1} }}{k}\zeta (k + 1)}  = \int_0^1\frac{\psi (z + 1) + \gamma}{z}dz
$$
where $\psi$ is the digamma function. The integral shares some similarities with http://dlmf.nist.gov/25.5.E17. A faster converging version is
$$
\log 2 + \sum\limits_{k = 1}^\infty  {\frac{{( - 1)^{k + 1} }}{k}(\zeta (k + 1) - 1)}
$$
giving the value $1.257746887\ldots$. See also Integral of digamma function.