How to find the limit of the sequence $$ \left(\frac{2}{2p+1}\right)^{1/p} $$ as $p \to \infty$?

Is it just $1$?

  • $\begingroup$ @Mark: no, the limit is for $p\to\infty$. $\endgroup$ – Yves Daoust Nov 16 at 21:45

Using L'Hospital,

$$\lim_{p\to\infty}\frac{\log(p+\frac12)}p=\lim_{p\to\infty}\frac1{p+\frac12}=0$$ and the initial limit is $1$.

  • $\begingroup$ Excuse me very much for this comment: the question is for a sequence. Therefore we can not apply the de l'Hôpital's rule. (+1) $\endgroup$ – Sebastiano Nov 16 at 21:23
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    $\begingroup$ @Sebastiano: the claim is true for the real function, hence also for the subsequence of naturals. $\endgroup$ – Yves Daoust Nov 16 at 21:24
  • $\begingroup$ Recall that a function $f:[0,\infty)\to\mathbb R$ is satisfies $\lim_{x\to\infty}f(x) = L$ if and only if for each increasing sequence $x_n$ with $x_n\to\infty$, $\lim_{n\to\infty} f(x_n)=L$. $\endgroup$ – Math1000 Nov 16 at 21:32
  • $\begingroup$ @Math1000: you needn't invoke such a "strong" result. The convergence of $f(x)$ in the reals is granted. Then so is the convergence for any unbounded sequence. $\endgroup$ – Yves Daoust Nov 16 at 21:35
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    $\begingroup$ Fair enough, I was just being thorough. Also I notice a typo and it is too late to edit my comment :( $\endgroup$ – Math1000 Nov 16 at 21:38

Hint: you can write your sequence (I write $n\in \mathbb N$ instead of $p$) as $$\lim _{n\to \infty} (2^{\frac{1}{n}} (2n+1)^{-\frac{1}{n}})=\lim _{n\to \infty}(2^{\frac{1}{n}})\cdot \lim _{n\to \infty} (2n+1)^{-\frac{1}{n}}$$

$$=\lim _{n\to \infty} (2^{\frac{1}{n}})\cdot \lim _{n\to \infty}\left(e^{-\frac{1}{n}\ln \left(2n+1\right)}\right)=1\cdot 1=1$$

Remember that: $$\lim _{n\to \infty}-\frac{\ln (2n+1)}{n}=0\tag{1}$$ because $-\frac{\ln (2n+1)}{n}=-\frac{2\ln \sqrt{2n+1}}{n}<2\sqrt{\frac{2n+1}{n^2}}$ and being $\lim _{n\to \infty} 2\sqrt{\frac{2n+1}{n^2}}=0 \implies (1)$

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    $\begingroup$ Note that we can extract the limit from the quotient precisely because the limits are finite (and the limit of the denominator is nonzero). $\endgroup$ – Math1000 Nov 16 at 21:18
  • $\begingroup$ @Math1000 I totally agree with you. I hope I haven't made any mistakes. Could you add your comment in my answer please considering that it is important? $\endgroup$ – Sebastiano Nov 16 at 21:21
  • $\begingroup$ Remains to justify the limit of $(2n+1)^{1/n}$, and indeterminate form $\infty^0$. By the way, it is simpler to consider $(n+1/2)^{1/n}$ directly. $\endgroup$ – Yves Daoust Nov 16 at 21:28
  • $\begingroup$ @YvesDaoust Done! I hope that now is correct and more complete answer. $\endgroup$ – Sebastiano Nov 16 at 21:34
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    $\begingroup$ @Sebastiano Bound $2n+1$ from above by a multiple of $n$ to get rid of the constant. Use this to bound $\frac{\ln(2n+1)}{n}$ from above, then show that the bound goes to $0$. There's relevant proofs on the site if you get stuck. Hence, $\frac{\ln(2n+1)}{n}$ vanishes by the squeeze theorem, since it's positive. $\endgroup$ – Jam Nov 16 at 22:54

$\dfrac{1}{(3p)^{1/p}}< (\dfrac{2}{2p+1})^{1/p}<$


Take the limit.

Recall : $\lim_{p \rightarrow \infty} p^{1/p}=1.$


We have that


and since $2^{1/p}\to 2^0=1$ we need to consider

$$\lim_{p\to \infty}(2p+1)^{1/p} =1$$


$$\left[(2p+1)^{\frac1{2p+1}}\right]^{\frac{2p+1}p}\to 1^2 =1$$

since we need to recall the foundamental result as $x\to \infty$

$$x^\frac1x=e^{\frac{\log x}x} \to e^0=1$$

indeed by $x=e^y$ with $y\to \infty$

$$\frac{\log x}x=\frac y{e^y}\to 0$$


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