I have some difficulties solving this exercise:
\begin{align} a_n & = \sum_{k=1}^n \frac 1 {\sqrt {n^2+k}} \\[6pt] b_n & = n\\[6pt] \lim_{n→∞} (a_n)^{b_n} & = \text{?} \end{align}
I believe that I have to find the limit of $a_n$ first but then I have a limit of something at power infinity and I'm pretty stuck. I don't know how to find the an's limit and what to do next.
For $1 \le k \le n$, we have that $n < \sqrt{n^2+k} < \sqrt{n^2+n+\frac{1}{4}} = n+\frac{1}{2}$. Therefore, $$\frac{n}{n+1/2} = \sum_{k=1}^n \frac{1}{n+1/2} < \sum_{k=1}^n \frac{1}{\sqrt{n^2+k}} < \sum_{k=1}^n \frac{1}{n} = 1.$$ Thus, by the Squeeze Theorem, we have that $\lim_{n\to \infty} a_n = 1$. However, this only shows us that the limit is in the indeterminate form $1^\infty$.
To make more progress, we need more detailed information on the asymptotic behavior of $a_n$ as $n \to \infty$. To this end, let us write: $$1 - a_n = \sum_{k=1}^n \left( \frac{1}{n} - \frac{1}{\sqrt{n^2+k}} \right) = \sum_{k=1}^n \frac{\sqrt{n^2+k}-n}{n\sqrt{n^2+k}} = \sum_{k=1}^n \frac{k}{n\sqrt{n^2+k}(\sqrt{n^2+k}+n)}.$$ Now, in the final sum, we have $$\frac{k}{n (n+1/2) (2n+1/2)}< \frac{k}{n\sqrt{n^2+k}(\sqrt{n^2+k}+n)} < \frac{k}{n \cdot n (2n)}.$$ Taking the sum, we get $$\frac{n(n+1)/2}{n(n+1/2)(2n+1/2)} < 1 - a_n < \frac{n(n+1)/2}{n\cdot n(2n)}.$$ Looking at both ends, they are both asymptotically equivalent to $\frac{1}{4n}$. More formally, by the Squeeze Theorem we can conclude that $\lim_{n\to \infty} n(1 - a_n) = \frac{1}{4}$, or equivalently: $$a_n = 1 - \frac{1}{4n} + o\left(\frac{1}{n}\right).$$
It now follows that $\log(a_n) = -\frac{1}{4n} + o(1/n)$, so $\log(a_n^{b_n}) = n \log(a_n) = -\frac{1}{4} + o(1)$. In other words, $\log(a_n^{b_n}) \to -\frac{1}{4}$ as $n \to \infty$, from which you should hopefully be able to conclude what the limit of $a_n^{b_n}$ is.