Evaluate $\lim\limits_{n \to \infty}\sum\limits_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2+n-nk}}$

Question

$$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2+n-nk}}$$

How to consider it?

Answer 1

Here's my solution. Please correct me if I'm wrong.

First, let's introduce and prove a result, which will be applied soon. That is

$$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n^2-n+nk}}=2.$$

Consider using the squeeze theorem. Notice that \begin{align*} \frac{1}{\sqrt{n+1-k}+\sqrt{n-k}}\geq\frac{1}{2\sqrt{n+1-k}}&\geq\frac{1}{\sqrt{n+1-k}+\sqrt{n+2-k}}, \end{align*} which implies \begin{align*} \sqrt{n+1-k}-\sqrt{n-k}\geq\frac{1}{2\sqrt{n+1-k}}&\geq\sqrt{n+2-k}-\sqrt{n+1-k}. \end{align*} Therefore $$2\geq\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+n-nk}}\geq\frac{2\left(\sqrt{n+1}-1\right)}{\sqrt{n}}\to 2(n \to \infty)$$ As per the squeeze theorem, the conclusion is followed.

Now, let's manage to solve the present problem. Since $\sqrt[k]{k}$ is decreasing for $k\geq 2$, and converges to $1$ as $k \to \infty$. Thus $$\forall \varepsilon>0,\exists N>0,\forall n>N,\sqrt[k]{k}\leq1+\varepsilon.$$ Hence \begin{align*} \sum_{k=1}^{n}\frac{1}{\sqrt{n^2-n+nk}}&\leq\sum_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}\\&=\sum_{k=1}^{N}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}+\sum_{k=N+1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}\\ &\leq\sum_{k=1}^{N}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}+(1+\varepsilon)\sum_{k=N+1}^{n}\frac{1}{\sqrt{n^2-n+nk}}\\ &\leq\sum_{k=1}^{N}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}+(1+\varepsilon)\sum_{k=1}^{n}\frac{1}{\sqrt{n^2-n+nk}}. \end{align*} Take the limits as $n \to \infty$ of both sides. We obtain $$2\leq \lim_{n \to \infty}\sum_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}\leq 2(1+\varepsilon).$$ Since $\varepsilon$ is arbitary, we can conclude $$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}=2,$$ which is what we want.

Answer 2

We have that

$$\frac{1}{\sqrt{n^2+n-nk}}\le \frac{\sqrt[k]{k}}{\sqrt{n^2+n-nk}}\le \frac{(e-1)\frac{\log k}k+1}{\sqrt{n^2+n-nk}}$$

and since

$$\frac{1}{\sqrt{n^2+n-nk}}-\frac{1}{\sqrt{n^2-nk}}=$$

$$=\frac{n}{\sqrt{n^2+n-nk}\sqrt{n^2-nk}\left(\sqrt{n^2+n-nk}+\sqrt{n^2-nk}\right)}=$$

$$=\frac{\frac1{n^2}}{\sqrt{1+\frac1n-\frac kn}\sqrt{1-\frac k n}\left(\sqrt{1+\frac1n-\frac kn}+\sqrt{1-\frac kn}\right)}$$

$$\le\frac1{n^2} \frac{1}{\sqrt{1-\frac kn}\sqrt{1-\frac k n}\left(\sqrt{1-\frac kn}+\sqrt{1-\frac kn}\right)}$$

therefore

$$ \lim\limits_{n \to \infty}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{n^2+n-nk}}=\lim\limits_{n \to \infty} \frac1n\sum\limits_{k=1}^{n}\frac{1}{\sqrt{1-\frac kn}}=\int_0^1 \frac{1}{\sqrt{1-x}}dx=2$$

then we need to show that

$$ \lim\limits_{n \to \infty}\sum\limits_{k=1}^{n}\frac{\frac{\log k}k}{\sqrt{n^2+n-nk}}=0$$

Answer 3

Given any $\varepsilon>0$, ,there exists a number $N(\varepsilon)$ such that whenever $k>N(\varepsilon)$, $1-\epsilon<k^{1/k}<1+\epsilon$. So, for the given $\varepsilon$ and corresponding $N(\varepsilon)$ we write

$$\sum_{k=1}^n \frac{k^{1/k}}{\sqrt{n^2+n-nk}}=\sum_{k=1}^{N(\varepsilon)} \frac{k^{1/k}}{\sqrt{n^2+n-nk}}+\sum_{k=N(\varepsilon)+1}^n \frac{k^{1/k}}{\sqrt{n^2+n-nk}}\tag1$$

With $N(\varepsilon)$ fixed, we see that the first sum on the right-hand side of $(1)$ approaches $0$ as $n\to \infty$.

Next, we analyze the limit of the second sum on the right-hand side of $(2)$. We begin by writing

$$\sum_{k=N(\varepsilon)}^{n-1} \frac{(1-\epsilon)\frac1n}{\sqrt{1-k/n}}<\sum_{k=N(\varepsilon)+1}^n \frac{k^{1/k}}{\sqrt{n^2+n-nk}}< \sum_{k=N(\varepsilon)}^{n-1} \frac{(1+\epsilon)\frac1n}{\sqrt{1-k/n}}\tag2$$

Observing that $\sum_{k=N(\varepsilon)}^{n-1}\frac{\frac1n}{\sqrt{1-k/n}}$ is simply the Riemann sum of $\int_{0}^{1}\frac{1}{\sqrt{1-x}}\,dx=2$, we see that given any $\varepsilon>0$, there exists a number $N(\varepsilon)$ such that

$$2(1-\varepsilon)<\lim_{n\to\infty}\sum_{k=N(\varepsilon)+1}^n \frac{k^{1/k}}{\sqrt{n^2+n-nk}}<2(1+\varepsilon)\tag3$$

Putting it all together, we conclude that

$$\lim_{n\to\infty}\sum_{k=1}^n \frac{k^{1/k}}{\sqrt{n^2+n-nk}}=2$$

And we are done!