Calculate $\lim_{n\to \infty}(n+1)^{\frac{1}{\sqrt{n}}}$

Question

Calculate the following limit: $$\lim_{n\to \infty}(n+1)^{\frac{1}{\sqrt{n}}}$$

I have tried to use the squeeze theorem and other convergence tests but all failed.

Please, any help?

score 5 · Accepted Answer · answered Jan 27 '18 at 12:26

5

Note that

$$1\le(n+1)^{1/\sqrt n}\le(2n)^{1/\sqrt n}=2^{1/\sqrt n}((\sqrt n)^{1/\sqrt n})^2$$

If we take $2^{1/x}$ and $x^{1/x}\to1$ as $x\to\infty$ for granted, then

$$2^{1/\sqrt n}((\sqrt n)^{1/\sqrt n})^2\to1\cdot1^2=1$$

and the Squeeze Theorem does the rest.

answered Jan 27 '18 at 12:26

Barry Cipra

79,832

MSE reference for the fact that $\lim n^{\frac{1}{n}}=1$: https://math.stackexchange.com/questions/76330/prove-sequence-a-n-n1-n-is-convergent – Salech Alhasov Jan 27 '18 at 12:34
Absolutely love this answer. Thank you so much! – Mister Bister Jan 27 '18 at 12:35

Atmos · Answer 2 · 2018-01-27T12:15:52.933

4

Use that $$ \left(n+1\right)^{1/\sqrt{n}}=e^{\left(1/\sqrt{n}\right)\ln\left(1+n\right)} $$ And $$ \ln\left(1+n\right)=\ln\left(n\right)+\ln\left(1+\frac{1}{n}\right) $$ So it becomes $$ \left(n+1\right)^{1/\sqrt{n}}=e^{\frac{\ln(n)}{\sqrt{n}}+\frac{\ln\left(1+\frac{1}{n}\right)}{\sqrt{n}}} $$ Using exponential properties $$ \left(n+1\right)^{1/\sqrt{n}}=e^{\frac{\ln(n)}{\sqrt{n}}}\times e^{\frac{\ln\left(1+\frac{1}{n}\right)}{\sqrt{n}}} $$ First term tends to $e^{0}=1$ because the power $\sqrt{n}$ is " stronger " than logarithm. And for the second use that $$ \ln\left(1+\frac{1}{n}\right)=\frac{1}{n}+o\left(\frac{1}{n}\right) $$ Hence $$ \frac{\displaystyle \ln\left(1+\frac{1}{n}\right)}{\sqrt{n}}=\frac{1}{n^{3/2}}+o\left(\frac{1}{n^{3/2}}\right) $$ Hence

$$ \left(n+1\right)^{1/\sqrt{n}}\underset{(+\infty)}{\sim}e^{1/n^{3/2}}\underset{n \rightarrow +\infty}{\rightarrow}1 $$

edited Jan 27 '18 at 12:15

answered Jan 27 '18 at 12:04

Atmos

7,369

the solution is true? – W. mu Jan 27 '18 at 12:08
Why it would not ? Explain me – Atmos Jan 27 '18 at 12:09
Can you explain the last step? – Mister Bister Jan 27 '18 at 12:11
1

I've edited it, tell me if it's all clear now. ^^ – Atmos Jan 27 '18 at 12:16
The final solutions is right. But it's wrong that you use the symbol ~, which represent asymptotic expansion of some function, or the infinitesimal of the same order. The first term is $e^{2/n^{1/2}}$, the final result can be written as $(n+1)^{1/\sqrt{n}}$~$e^{2/n^{1/2}}$. I think it is more easier to use the L' hospital rule. – W. mu Jan 27 '18 at 12:30
That's how we write it in France, where l'Hospiral rule is not even taught ! – Atmos Jan 27 '18 at 13:20

score 1 · Answer 3 · answered Jan 27 '18 at 12:03

1

write your Limit in the form $$e^{\lim_{n\to \infty}\frac{\ln(n+1)}{\sqrt{n}}}$$

answered Jan 27 '18 at 12:03

Dr. Sonnhard Graubner

95,283

What an answer, Brilliant!!! – Jan 27 '18 at 12:06

Calculate $\lim_{n\to \infty}(n+1)^{\frac{1}{\sqrt{n}}}$

3 Answers3