# Calculate $\lim_{k \rightarrow \infty } \sqrt[k]{k(k+1)}$

How do I calculate the following limit: $$\displaystyle\lim_{k \rightarrow \infty } \sqrt[k]{k(k+1)}$$

The only limit identity that I know which closely resembles this is $$\displaystyle\lim_{k \rightarrow \infty } \sqrt[k]{k}=1$$.

Edit: This question came in context of finding the radius of convergence of $$\displaystyle\sum_{k=1}^\infty \dfrac{2^k z^{2k}}{k^2+k}$$.

Attempt: Clearly, the ratio test here is inconclusive. So, I tried Cauchy-Hadamard Formula.

For the general form of a power series, this says that $$R=\displaystyle \dfrac{1}{\limsup_{k \rightarrow \infty} \sqrt[k]{|a_k|}}$$.

But now $$\displaystyle \lim_{k \rightarrow \infty} \sqrt[k]{\dfrac{2^k}{k^2+k}}=2 \lim_{k \rightarrow \infty} \sqrt[k]{\dfrac{1}{k^2+k}}=2$$ as per the calculations done by @gimusi, @Key Flex.

Doubt: The answer at the back of the book gives $$R=\frac{1}{\sqrt{2}}$$.

We have that

$$\sqrt[k]{k(k+1)}= \sqrt[k]{k} \,\sqrt[k]{k+1} \to 1 \cdot 1=1$$

indeed

$$\sqrt[k]{k+1}=e^{\frac{\ln k}{k+1}}\to e^0=1$$

and more in general for any polynomial $$p_n(k)$$ we have

$$\sqrt[k]{p_n(k)} \to 1$$

by the same proof.

• How does this help? What is $\displaystyle\lim_{k \rightarrow \infty} \sqrt[k]{k+1}$? Equivalently, what is $\displaystyle\lim_{k \rightarrow \infty} \sqrt[k]{1+\frac{1}{k}}$ Nov 4, 2018 at 22:59
• How do you evaluate $\lim_{k \rightarrow \infty } \sqrt[k]{k}=1$?
– user
Nov 4, 2018 at 23:00
• @user330477 We have $\sqrt[k]{k+1}=e^{\frac{\log k}{k+1}}$? Does the $+1$ change something?
– user
Nov 4, 2018 at 23:01
• @user330477 The key fact is that when k is large the "+1" is completely negligeble.
– user
Nov 4, 2018 at 23:04
• Note that $$(1+\frac{1}{k})^{\frac{1}{k}}$$ is in the form $1^0$ which is not an indeterminate form.
– user
Nov 4, 2018 at 23:11

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \begin{align} &\bbox[10px,#ffd]{\lim_{k \to \infty}\root[{\large k}]{k\pars{k + 1}}} = \exp\pars{\lim_{k \to \infty} {\ln\pars{k\bracks{k + 1}} \over k}} \\[5mm] = &\ \exp\pars{\lim_{k \to \infty}{\ln\pars{\bracks{k + 1}\bracks{k + 2}} - \ln\pars{k\bracks{k + 1}} \over \bracks{k + 1} - k}} \label{1}\tag{1} \\[5mm] = &\ \exp\pars{\lim_{k \to \infty}\ln\pars{1 + {2 \over k}}} = \bbx{\large 1} \end{align}

In expression \eqref{1}, I used the Stolz-Ces$$\mathrm{\grave{a}}$$ro Theorem.

HINT

$$\lim_{k\to\infty}\sqrt[k]{a_k}=\lim_{k\to\infty}\frac{a_{k+1}}{a_k}$$ if the second limit exists.

$$\lim_{k\rightarrow\infty}\sqrt[k]{k(k+1)}=\lim_{k\rightarrow\infty}((k^2+k)^{\frac1k})=\lim_{k\rightarrow\infty}\left(k^2\left(1+\dfrac1k\right)\right)^{\frac1k}=\lim_{k\rightarrow\infty}\left(1+\dfrac1k\right)^{\frac 1k}\cdot k^{\frac2k}=1\cdot1=1$$

• You forgot the $\frac{1}{k}$ over $1+\frac{1}{k}$. This limit is what is causing me a lot of trouble. Nov 4, 2018 at 23:05
• @user330477 I factored out $k^2$ from $k^2+k$, my computation is correct Nov 4, 2018 at 23:07
• No, your computation is not correct. The $\frac{1}{k}$ power should also be over $1+\frac{1}{k}$? Nov 4, 2018 at 23:08
• @user330477 see the double parentheses, $(())$. It means $1+\dfrac1k$ is also included Nov 4, 2018 at 23:13
• @KeyFlex You lost an exponent here $$\ldots=\lim_{k\rightarrow\infty}\left(1+\dfrac1k\right)^{\color{red}{\frac1k}}\cdot k^{\frac2k}=1\cdot1=1$$
– user
Nov 4, 2018 at 23:17

$$(k)^{1/k}(k+1)^{1/(k+1)} \lt$$

$$(k(k+1))^{1/k} \lt$$

$$(k)^{1/k}(2)^{1/k}(k^{1/k}).$$

Take the limit.