Question for the estimation of $\sum_{i=1}^x \frac{1}{w+i}$ as $x \to \infty$.

Question

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.

Answer 1

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 $x \to \infty$,

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

Answer 2

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 $\approx 1.54041$

Answer 3

$$ \sum_{i=1}^{x}\frac{1}{w+i}=\text{area of rectangles}\leq\int_{0}^{x}\frac{1}{w+t}dt=\ln\left(w+x\right)-\ln\left(w\right) $$