<s>[INST] Trying to compute a Riemann-sum as an integral where the $\lim_{n\to\infty} a\cdot n +\sum_{n^2}^{4n^2} \dfrac{1}{k+n}$
I have the following problem: I want to figure out , for what value of a does this limit converge and what value does it then converge to.
I'm assuming that I'm supposed to rewrite the sum as an integral where I can evaluate the integral in the form of "n's". Given the integral $\int_{1}^{4}\dfrac{1}{\sqrt{x}}dx = [2\sqrt{x}]_{1}^{4}$, where $x= k+n$. I think I can finish the problem. 
However, I am unsure of how to get the integral $\int_{1}^{4}\dfrac{1}{\sqrt{x}}dx$
Any help will be grateful.
Thanks for your time.
 [/INST] Unless there is some miraculous cancellation -- and there will not be -- the term $a \cdot n$ with $a \neq 0$ will cause divergence. So let's take $a = 0$ and focus on the behavior of the sum.  This would be a typical Riemann sum for $f(x) = (1 + x)^{-1}$ if not for the limits that are given.
Nevertheless, we can find upper and lower bounds and apply the squeeze theorem to find the limit.
For $k \leqslant x \leqslant k+1,$ we have
$$\frac{1}{k+1+n} \leqslant \int_k^{k+1} \frac{1}{n+x} \, dx = \int_{k/n}^{(k+1)/n} \frac{1}{1+u} \, du  \leqslant \frac{1}{k+n}.$$
Summing from $p = n^2$ to $q = 4n^2$ we have
$$ \int_{p/n}^{(q+1)/n} \frac{1}{1+u} \, du  = \sum_{k=p}^q\int_{k/n}^{(k+1)/n} \frac{1}{1+u} \, du \leqslant   \sum_{k=p}^q \frac{1}{k+n}.$$
Summing from $p-1$ to $q-1$ we have
$$ \sum_{k=p}^q \frac{1}{k+n} = \sum_{k=p-1}^{q-1} \frac{1}{k+1+n} \leqslant \sum_{k=p-1}^{q-1}\int_{k/n}^{(k+1)/n} \frac{1}{1+u} \, du =  \int_{(p-1)/n}^{q/n} \frac{1}{1+u} \, du  .$$
Hence,
$$\int_{p/n}^{(q+1)/n} \frac{1}{1+u} \, du   \leqslant \sum_{k=p}^q \frac{1}{k+n} \leqslant \int_{(p-1)/n}^{q/n} \frac{1}{1+u} \, du.  $$
Integrating we find, 
$$\log\left( \frac{n + q +1 }{n +p} \right)   \leqslant \sum_{k=p}^q \frac{1}{k+n} \leqslant \log\left( \frac{n + q}{n + p -1}\right).  $$
Substituting $p = n^2$ and $q = 4n^2$ we get,
$$\log\left( \frac{n + 4n^2 +1 }{n + n^2} \right)   \leqslant \sum_{k=n^2}^{4n^2} \frac{1}{k+n} \leqslant \log\left( \frac{n + 4n^2}{n + n^2 -1}\right).  $$
Applying the squeeze theorem, we find the limit
$$\lim_{n \to \infty} \sum_{k=n^2}^{4n^2} \frac{1}{k+n} = \log(4).$$
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