Problem Let the following sequence $f_n: \mathbb{R}^+ \rightarrow \mathbb{R}$ given by $$ f_n(x) = \sum\limits^n_{k=1} \frac{1}{\sqrt{n^2 +k^x}}$$ Show that $f_n$ converges to a function $f$ such that:
a) $\;f(2)=\ln(1+\sqrt2)$
b)$\;f(x)=1 \quad x\in[0,2)$
c)$\;f(x)=0 \quad x>2$
I'm out of ideas with this problem. What sort of trick I can use to deal of this sequence ?
For the first one, by Riemann sum approximation $$ f_n(2) = \sum_{k=1}^n \frac{1}{\sqrt{n^2+k^2}} = \frac{1}{n} \sum_{k=1}^n \frac{1}{\sqrt{1 + \big(\frac{k}{n}\big)^2}} \xrightarrow[n \to \infty]{} \int_0^1 \frac{1}{\sqrt{1+x^2}}dx $$ and you need to evaluate that integral (For example by trigonometric/hiperbolic substitution)
Now, if $x \in [0,2)$ then $x-2 <0$ so that $n^{x-2} \xrightarrow[n \to \infty]{} 0$ $$1 = \sum_{k=1}^n \frac{1}{\sqrt{n^2}} \le f_n(x) \le \sum_{k=1}^n\frac{1}{\sqrt{n^2+n^x}} = \frac{1}{\sqrt{1+ n^{x-2}}} \xrightarrow[n \to \infty]{}1$$ Similarly with the last one, for $x > 2$ you have $x-2>0$ so that $n^{x-2} \xrightarrow[n \to \infty]{} \infty$ and $$ 0 \le f_n(x) \le \sum_{k=1}^n \frac{1}{\sqrt{n^2+n^x}} = \frac{1}{\sqrt{1+n^{x-2}}} \xrightarrow[n \to \infty]{} 0$$