# Existence and value of $\lim_{n\to\infty} (\ln\frac{x}{n}+\sum_{k=1}^n \frac{1}{k+x})$ for $x>0$

Does the limit $$W(x)=\lim_{n\to\infty} \left(\ln\frac{x}{n}+\sum_{k=1}^n \frac{1}{k+x} \right)$$ exist for all $x>0$? If so, what is the limit $$\lim_{x\to\infty}W(x)?$$

• I assume the upper limit of the summation is $n$ and not infinity. – user17762 Apr 5 '13 at 17:57

We have $$\sum_{k=1}^n \dfrac1{k+x} = \int_{1^-}^{n^+} \dfrac{d \lfloor t \rfloor}{t+x} = \left. \dfrac{\lfloor t \rfloor}{t+x} \right\vert_{t=1^-}^{t=n^+} + \int_{1^-}^{n^+} \dfrac{\lfloor t \rfloor}{(t+x)^2} dt = \dfrac{n}{n+x} + \int_{1^-}^{n^+} \dfrac{\lfloor t \rfloor}{(t+x)^2} dt$$ Now $$\int_{1^-}^{n^+} \dfrac{\lfloor t \rfloor}{(t+x)^2} dt = \int_1^{n^+} \dfrac{t}{(t+x)^2} dt - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt$$ $$\int_1^{n^+} \dfrac{t}{(t+x)^2} dt = \int_1^{n} \dfrac{dt}{t+x} - x \int_1^n \dfrac{dt}{(t+x)^2} = \log(n+x) - \log(1+x) -x \left(\dfrac1{1+x} - \dfrac1{n+x}\right)$$ Hence, we get that \begin{align} \log(x/n) + \sum_{k=1}^n \dfrac1{k+x} & = \log\left(\dfrac{x}n \right) + \dfrac{n}{n+x} + \log \left(\dfrac{n+x}{1+x}\right) -\dfrac{x}{1+x} + \dfrac{x}{n+x} - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt\\ & = - \dfrac{x}{1+x} + 1 + \log\left(\dfrac{x}n \cdot \dfrac{n+x}{1+x}\right) - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt\\ & = \dfrac1{1+x} + \log\left(\dfrac{x}n \cdot \dfrac{n+x}{1+x}\right) - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt \end{align} Now letting $n \to \infty$, we get that $$W(x) = \dfrac1{1+x} + \log \left(\dfrac{x}{1+x}\right) - \underbrace{\int_1^{\infty} \dfrac{\{t\}}{(t+x)^2} dt}_{\text{Converges since }\{t\} \in [0,1)}$$ $$0 \leq \overbrace{\int_1^{\infty} \dfrac{\{t\}}{(t+x)^2} dt}^{g(x)} \leq \int_1^{\infty} \dfrac1{(t+x)^2} dt = \dfrac1{1+x}$$ There might be some name for $g(x)$ (Probably some of the number theorists on this website might be able to identity this). For instance, $g(0) = 1-\gamma$, where $\gamma \approx 0.57721$ is the Euler Mascheroni constant. Now $$\lim_{x \to \infty} W(x) = 0 + \log(1) + 0 = 0$$

Another method is as follows. From here, we have \begin{align} \sum_{k=1}^n \left(\dfrac1k - \dfrac1{x+k} \right) & = \sum_{k=1}^n \int_0^1 (y^{k-1} - y^{x+k-1})dy\\ & = \int_0^1 (1-y^x) \sum_{k=1}^n y^{k-1} dy\\ & = \int_0^1 (1-y^x) \dfrac{1-y^n}{1-y} dy \end{align} Hence, we have \begin{align} \log(x/n) + \sum_{k=1}^n \dfrac1{k+x} & = \log(x/n) + \sum_{k=1}^n \left(\dfrac1{k+x} - \dfrac1k \right) + \sum_{k=1}^n \dfrac1k\\ & = \log(x) + \sum_{k=1}^n \dfrac1k - \log(n) + \sum_{k=1}^n \left(\dfrac1{k+x} - \dfrac1k \right)\\ & = \log(x) + \sum_{k=1}^n \dfrac1k - \log(n) - \int_0^1 (1-y^x) \dfrac{1-y^n}{1-y} dy \end{align} Now letting $n \to \infty$, we get that $$W(x) = \log(x) + \gamma - \int_0^1 \dfrac{1-y^x}{1-y} dy$$ Now as $x \to \infty$, we have $$\int_0^1 \dfrac{1-y^x}{1-y} dy = \log(x) + \gamma + \mathcal{O}(1/x)$$ Hence, we get that $$\lim_{x \to \infty} W(x) = 0$$

Let us prove why, as $x \to \infty$, we have $$\int_0^1 \dfrac{1-y^x}{1-y} dy = \log(x) + \gamma + \mathcal{O}(1/x)$$ The proof is the same as before. We have $$\sum_{k=1}^n \dfrac1k = \int_0^1 \sum_{k=1}^n y^{k-1} dy = \int_0^1 \dfrac{1-y^n}{1-y} dy$$ But we know that $\displaystyle \sum_{k=1}^n \dfrac1k = \log(n) + \gamma + \mathcal{O}(1/n)$. Hence, we get that $$\int_0^1 \dfrac{1-y^n}{1-y} dy = \log(n) + \gamma + \mathcal{O}(1/n)$$ Replacing $n$ by $x$ and because the integral is a smooth function of $x$, we can conclude that $$\int_0^1 \dfrac{1-y^x}{1-y} dy = \log(x) + \gamma + \mathcal{O}(1/x)$$

• Or you can use Euler McLaurin summation. – Aryabhata Apr 5 '13 at 19:06
• @Aryabhata In some sense, isn't what I have used in the first proof? – user17762 Apr 5 '13 at 19:07
• Yes, I guess it is easier to read, though :). In any case, I am not sure if it gives the limit of $W(x)$ to be 0. – Aryabhata Apr 5 '13 at 19:10
• @Aryabhata Do you mean that Euler Maclaurin doesn't give the limit as $0$ or is something wrong in what I have written? – user17762 Apr 5 '13 at 19:13
• I meant Euler MacLaurin. There is some undetermined constant which creeps in usually. – Aryabhata Apr 5 '13 at 19:15

\begin{align} W(x) &=\lim_{n\to\infty}\left(\log(x/n)+\sum_{k=1}^n\frac1{k+x}\right)\\ &=\lim_{n\to\infty}\left(\log(x)+\left(\sum_{k=1}^n\frac1k-\log(n)\right)-\sum_{k=1}^n\left(\frac1k-\frac1{k+x}\right)\right)\\ &=\log(x)+\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1k-\log(n)\right)-\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)\tag{1}\\[6pt] &=\log(x)+\gamma-(\gamma+\psi(x+1))\\[12pt] &=\log(x)-\psi(x+1) \end{align} where $\psi(x)$ is the digamma function and $\gamma$ is the Euler-Mascheroni Constant.

Note that for $n\in\mathbb{Z}$, $(1)$ gives \begin{align} \lim_{\substack{n\to\infty\\n\in\mathbb{Z}}}W(n) &=\lim_{\substack{n\to\infty\\n\in\mathbb{Z}}}\left(\log(n)+\gamma-\sum_{k=1}^n\frac1k\right)\\ &=\gamma-\gamma\\[12pt] &=0\tag{2} \end{align} Taking derivatives, we get that $$W'(x)=\frac1x-\sum_{k=1}^\infty\frac1{(k+x)^2}\tag{3}$$ Comparing the sum in $(3)$ with the integral of $\frac1{x^2}$, we get $W'(x)$ is between $\frac1x{-}\frac1{x+1}$ and $\frac1x{-}\frac1{x+2}$.

Thus, $$\lim_{x\to\infty}W'(x)=0\tag{4}$$ Combining $(2)$ and $(4)$ yields $$\lim_{x\to\infty}W(x)=0$$

• @TCL: thanks. I often get that wrong. – robjohn Apr 5 '13 at 19:01

I think it does not always converge: in fact for x->0, W(x) tends to -infinity, it seems

First of all, look at the definition of the Euler-Mascheroni constant and observe that it is very similar to your problem

http://en.wikipedia.org/wiki/Euler%27s_constant

You must develop the part of with the logarith ln(x/n)= ln x - ln(n)

Then, in the limit you may drop the x from the sumatory and if you substract ln(n) it is exactly the Euler constant=0.577 aprox. But ln(x) tends to -infinity. Thus, it does not always converge.

Hope this helps,

David, Barcelona, Catalonia

• The first limit is considered for fixed $x$. Letting $x \to 0$ is not saying anything about that. What you say provides more clues for establishing convergence than for concluding divergence. – Lord_Farin Apr 5 '13 at 17:16