# find the limit of ${(n + 1)^{{1 \over {\sqrt n }}}}$

I really tried to think of a way to find the limit of this sequence but couldn't think of something useful. I tried to use Bernoulli's inequality to squeeze the sequence, but it didn't come in handy.

$$\mathop {\lim }\limits_{n \to \infty } {(n + 1)^{{1 \over {\sqrt n }}}} = ?$$

• Does taking the logarithm clear it up? – Daniel Fischer Nov 25 '13 at 19:06
• As $n\to\infty$ or $n\to 0$? – Thomas Andrews Nov 25 '13 at 19:06
• I cannot use log, for we didn't reached that in our class. – DavidJack Nov 25 '13 at 19:08
• @DavidJack How did you define exponentiation then...? – Pedro Tamaroff Nov 25 '13 at 19:09
• In some cases we used Bernoulli's inequality to get away from the exponent, In others we used other principles and theorems which doesn't invlove $log$. – DavidJack Nov 25 '13 at 19:11

You can do this without logarithms. First' a lemma:

Lemma: For an arbitrary $\varepsilon > 0$ there exists an $n_0 \in \mathbb{N}$ such that for every $n \geq n_0$ we have $$n^\frac{1}{n} < 1 + \varepsilon.$$

To prove the lemma, note that the inequality is equivalent to $$n < (1 + \varepsilon)^n,$$ and this is true when $n$ is large enough by binomial expansion.

Now, all the terms in your sequence are greater than $1$. You can use this, the lemma above, and the definition of limits to prove that the sequence converges to $1$.

• Great! how did you think about it? – DavidJack Nov 25 '13 at 19:19
• @DavidJack This is what you inevitably come to if you try to prove using only definitions that $n^{1/n}$ converges to $1$. But I knew that it converges to $1$ in advance, by taking the logarithm. So I used logarithms implicitly to find the answer, but excluded them from the "official" reasoning. – Dan Shved Nov 25 '13 at 19:27

It converges to $1$

firstly you see

$(1+n)^{\frac{1}{\sqrt n}} \geq 1$

For the other bound

$$(1+n)^{\frac{1}{\sqrt n}} \leq (1+n)^{\frac{1}{\sqrt n}} \leq (2n)^{\frac{1}{\sqrt n}}$$

you know that $2^{\frac{1}{\sqrt n} } \rightarrow 1$

For the other part you can use $\rm{AM-GM}$ $$n^{\frac{1}{\sqrt n}}\leq ({\underbrace {1 \cdot 1 \ldots 1}_{[\sqrt n]-3}} \cdot n^{\frac{1}{3} }\cdot n^{\frac{1}{3}} \cdot n^{\frac{1}{3}})^{\frac{1}{[\sqrt n]}} \leq \frac{ [\sqrt n]-3 +3n^{\frac{1}{3}}}{ [\sqrt n]}\rightarrow 1$$ Where $[\cdot]$ is the integral part.