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