Question for the estimation of $\sum_{i=1}^x \frac{1}{w+i}$ as $x \to \infty$. I have a question of the estimation of this summation:
$$ \frac{1}{w+1}+\frac{1}{w+2}+\cdots+\frac{1}{w+x}$$
Which is:
$$\sum_{i=1}^x \frac{1}{w+i}$$
What I have tried: applying limit to the summation:
$$\lim_{x \rightarrow \infty }\sum_{i=1}^x \frac{1}{w+i}$$
Which becomes:
$$\int_1^x \frac{1}{w+y}\,dy$$
Which equals:
$$\ln{(w+x)}-\ln{(w+1)}$$
However, I found that my answer is wrong and other people's answer is:
$$\ln(w+x)-\ln(w)$$
If I try this in MS Excel, the error for my answer is bigger than the others.  
Can anyone explain at which part I did it wrong?
And what is the name for this approximation? Is it Riemann sum approximation?
Thank you! I am sorry if this is a duplicate. I tried to search but I didn't find any duplicate for this specific question.
 A: Assume $w>-1$ and $x \in \mathbb{N}$. 
One may recall the classic identity
$$
\Gamma(y+1)=y \:\Gamma(y),\quad y>0. \tag1
$$ from which, setting $\psi (y):= \left(\log \Gamma (y) \right)'$, one deduces 
$$
\psi(y+1)-\psi(y)=\frac1y,\quad y>0. \tag2
$$ Next, putting  $y=w+k$ and summing $(2)$ from $k=1$ to $k=x$, terms telescope, one gets
$$
\sum_{k=1}^x\frac1{w+k}=\psi(w+x+1)-\psi(w+1). \tag3
$$ Then, as $x \to \infty$, one may use the digamma asymptotic expansion
${\bf 6.3.18}$:
$$
\psi(x+1)=\ln x+O\left( \frac1x\right)
$$ yielding, as $x \to \infty$,

$$
\sum_{k=1}^x\frac1{w+k}=\ln x-\psi(w+1) +O\left( \frac1x\right).
$$

A: You can also use harmonic numbers since  $$S_x=\sum_{i=1}^x \frac{1}{w+i}=H_{w+x}-H_w$$ Now, using the asymptotics you would get $$S_x=\left(\log \left(x\right)+\gamma-H_w
   \right)+\frac{w+\frac{1}{2}}{x}-\frac{6 w^2+6 w+1}{12
   x^2}+O\left(\frac{1}{x^3}\right)$$
For illustration purposes, using $w=123$ and $x=456$, the exact result would be $\approx 1.54592$ while the abovea pproximation gives $\approx 1.54041$
A: 
$$
\sum_{i=1}^{x}\frac{1}{w+i}=\text{area of rectangles}\leq\int_{0}^{x}\frac{1}{w+t}dt=\ln\left(w+x\right)-\ln\left(w\right)
$$
