Is it alright to substitute $0$ for $1/n$ in this limit problem? 
Evaluate: $\lim\limits_{n \to \infty}\displaystyle\sum_{r=1}^{n} \dfrac{1}{2n + 2r-1}$

To solve this I used the following approach: 
\begin{align} S &= \lim\limits_{n \to \infty}\sum_{r=1}^{n} \dfrac{\frac 1n}{2 + 2\frac rn-\frac1n} \\
&= \lim\limits_{n \to \infty}\displaystyle\sum_{r=1}^{n} \dfrac{\frac 1n}{2 + 2\frac rn-\color{red}{0}} = \dfrac 12\int _0^1 \dfrac{1}{1+x}\,\mathrm{d}x \\ &= \ln(\sqrt 2)\end{align}
Though the answer is correct, I am unsure about my second step in which I have replaced $1/n$ by $0$. Is that allowed? I have tried it in some other problems too and it works. 
 A: By definition of Riemann integral we have $$\int_{0}^{1}f(x)\,dx=\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n}f(t_r)$$ where $t_r$ lies between $(r-1)/n$ and $r/n$. For the current question the given sum matches $f(x) =\dfrac{1}{2(1+x)}$ and $t_r=(2r-1)/2n$. Thus the desired limit is indeed equal to $(1/2)\log 2$.

The notion of a Riemann sum is far more general than usual calculus texts will lead you to believe (most calculus texts either keep $t_r=(r-1)/n$ or $t_r=r/n$). The sum in question is a Riemann sum as explained above.

Also in mathematics one can't replace $A$ by $B$ unless $A=B$. If you see some replacements which work out fine but look mysterious then you just need to understand that some steps are missing.

The technique can be used to handle sums which are not exactly Riemann sums but differ slightly from it. See this answer for an example. 
A: No, this is not allowed, because for all you know that $\frac 1 n$ could have some subtle interaction with the rest of the formula which cancels out its going to zero. For example in the limit
$$\lim_{n\to \infty}n\frac1n$$
we can't just replace $\frac1n$ with $0$ (yielding a limit of $0$) because the presence of the factor $n$ cancels that out. How do you know there isn't something you didn't spot in your formula which produces similar behavior? Maybe the $\frac rn$ or the $\frac 1n$ in the numerator has a similar "cancelling" effect on the $\frac 1n$ which you're trying to replace with $0$.
A: I think it is better and safer to bound the limit from both sides as following:$$\alpha n+2r<2n+2r-1<2n+2r$$for every $\alpha<2$ and $n$ sufficiently large. Therefore$$\sum_{r=1}^{n}{1\over 2 n+2r}<\sum_{r=1}^{n}{1\over 2n+2r-1}<\sum_{r=1}^{n}{1\over \alpha n+2r}$$Now fix $\alpha<2$. Therefore by integrating we obtain$${1\over 2}\ln 2\le\text{the desired limit}\le {1\over 2}\ln{2+\alpha\over\alpha}$$Since the latter holds for every $\alpha<2$, then we can write $$\text{the desired limit}={1\over 2}\ln 2$$
A: "That" is not allowed in general.
It works here because you can squeeze the sum by two Riemann sums for the same function $\frac{1}{2(1+x)}$ on the same interval $[0,1]$:
$$\small\frac{1}{2}\sum_{r=1}^{n} \dfrac{1}{1 + \frac{r}{n}}\cdot \frac{1}{n} = \sum_{r=1}^{n} \dfrac{1}{2n + 2r} \leq \sum_{r=1}^{n} \dfrac{1}{2n + 2r-1} \leq \sum_{r=1}^{n} \dfrac{1}{2n + 2r-2} = \frac{1}{2}\sum_{r=1}^{n} \dfrac{1}{1 + \frac{r-1}{n}}\cdot \frac{1}{n}$$
