Is $\lim_{x\to \infty}{\sum_{i,h=1}^x \frac{1}{i^h} - x-\ln{x}}$ equal to $\gamma$, the Euler-Mascheroni constant? If so, why? I was looking at the sum $\sum_{i,h=1}^x \frac{1}{i^h}$ on Desmos, and I realized it seemed to converge to the line $y=x$. When I subtracted x from it and increased the bounds, it seemed to be converging close to the Euler-Mascheroni constant. Unfortunately, it quickly gets difficult to calculate for large x, so the best estimate I could get with Desmos was for $x=10000$, for which $\sum_{i,h=1}^x \frac{1}{i^h} - x-\ln{x}$ is approximately 0.577165, which is very close to the Euler-Mascheroni constant. Is this actually converging to the constant, or just to something close to it? I would guess that it does, due to this series's clear similarity to the harmonic series, however it is interesting that the constant still appears even with the added exponentiation in the denominator.
For clarity, an alternate notation for the sum above would be $${\sum_{i=1}^x}{\sum_{h=1}^x {1\over i^h}}$$
 A: Result: $$\lim_{N\to\infty}\sum_{j,k=1}^N\frac{1}{j^k}-N-\ln(N)=\gamma.$$
Proof:
Splitting apart the sum, we have:
$$
\underbrace{\left(\sum_{j=1}^N\frac{1}{j}-\ln(N)\right)}_{\to\gamma\text{ as }N\to\infty}+\left(\sum_{j=1}^N\sum_{k=2}^N\frac{1}{j^k}-N\right).
$$
Since we know the first term converges to $\gamma$, we can focus on the second term. Notice that we have $N-1$ terms with $j=1$, so we can further reduce the double sum to
$$
\left(\sum_{j=2}^N\sum_{k=2}^N\frac{1}{j^k}\right)-1.
$$
The inner sum is just a finite geometric series, so we can write it in closed form as
$$
\sum_{j=2}^N\frac{(1/j)^2-(1/j)^{N+1}}{1-(1/j)}=\sum_{j=2}^N\frac{1}{j(j-1)}-\frac{1}{j^N(j-1)}.
$$
Here, we have
$$
\sum_{j=2}^N\frac{1}{j(j-1)}=1-\frac{1}{N}\quad\text{and}\quad\sum_{j=2}^N\frac{1}{j^N(j-1)}\ll\frac{\ln(N)}{2^N}
$$
Thus, as $N\to\infty$, the left-hand sum converges to $1$ while the right-hand sum converges to $0$. In particular,
$$
\lim_{N\to\infty}\left(\sum_{j=2}^N\sum_{k=2}^N\frac{1}{j^k}\right)-1=0.
$$
Therefore, the original double sum above also converges to $0$, giving the conclusion to the proof.
A: $$\lim_{x\to \infty} \sum_{i = 1}^{x} 1/i - \ln(x) = \gamma$$
$$\sum_{i = 1}^{\infty} 1/i^2 = \pi^2/6$$
$\sum_{i = 1}^{\infty} 1/i^k = $ convergent and tends to $1$ for $k>1$ and as $k$ goes to $\infty$. Everyone of these $1$s will add up to offset $x$. 
@will is correct. 
Write down the terms with power $1$ in a row and power $2$ in a row below that and so on. 
$1 + \frac{1}{2} + \frac{1}{3} + ... \ - 1 - \ln(x)$
$1 + \frac{1}{2^2} + \frac{1}{3^2} + ... - 1 $
$1 + \frac{1}{2^3} + \frac{1}{3^3} + ... - 1 $
...
The first column of $1$s will cancel out the penultimate column of $-1$s. 
Second column is $1/2 + 1/2^2 + ... = 1$
Third column is $1/2$. Fourth column is $1/3$. And so on. 
So we are left with $1+1/2+1/3+...1/n - log(n) = \gamma$ as $n \to \infty$
