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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 $R=\frac{1}{\sqrt{2}}$.

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5 Answers 5

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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.

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  • $\begingroup$ How does this help? What is $\displaystyle\lim_{k \rightarrow \infty} \sqrt[k]{k+1}$? Equivalently, what is $\displaystyle\lim_{k \rightarrow \infty} \sqrt[k]{1+\frac{1}{k}}$ $\endgroup$
    – user330477
    Nov 4, 2018 at 22:59
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    $\begingroup$ How do you evaluate $\lim_{k \rightarrow \infty } \sqrt[k]{k}=1$? $\endgroup$
    – user
    Nov 4, 2018 at 23:00
  • $\begingroup$ @user330477 We have $\sqrt[k]{k+1}=e^{\frac{\log k}{k+1}}$? Does the $+1$ change something? $\endgroup$
    – user
    Nov 4, 2018 at 23:01
  • $\begingroup$ @user330477 The key fact is that when k is large the "+1" is completely negligeble. $\endgroup$
    – user
    Nov 4, 2018 at 23:04
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    $\begingroup$ Note that $$(1+\frac{1}{k})^{\frac{1}{k}}$$ is in the form $1^0$ which is not an indeterminate form. $\endgroup$
    – user
    Nov 4, 2018 at 23:11
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.

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HINT

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

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

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

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