Convert to Riemann Sum $\lim_{N→∞}\sum^{N}_{n=-N}\left(\frac{1}{(N+in)}+\frac{1}{(N-in)}\right)$ The following limit has to be converted to Riemann Sum.
$$\lim_{N→∞}\sum^{N}_{n=-N}\left(\frac{1}{(N+in)}+\frac{1}{(N-in)}\right)$$
My attempt:
$$\lim_{N→∞}\sum^{N}_{n=-N}\left(\frac{2N}{N^{2}+n^{2}}\right)$$=$\lim_{N→∞}\sum^{N}_{n=-N}\left(\frac{2}{1+n^{2}/N^{2}}\right).1/N$.
Now, replacing $1/N$ by $dx$, $n^{2}/N^{2}$ by $x^{2}$ and summation by integral, we have
$$\lim_{N→∞}1/N\sum^{N}_{n=-N}\left(\frac{2}{1+n^{2}/N^{2}}\right)= \lim_{N→∞}[1-(-1)]/N\sum^{N}_{n=-N}\left(\frac{1}{1+n^{2}/N^{2}}\right)$$
$$=2\lim_{N→∞}[1-(-1)]/N\sum^{N}_{n=0}\left(\frac{1}{1+n^{2}/N^{2}}\right)=?$$
I feel that I am very close to the final answer which is $2\int_{-1}^{1} dt/(1+t^{2})$. But I am stuck after this step. please complete the rest, feel free to edit the question directly to complete the remaining steps.
With what I have done I am able to compare $a=-1$ and $b=1$, but I still need to get the general form $$\lim_{n\to \infty}\sum_{i=1}^{n}\frac{b-a}{n}\cdot f\left(a+i\frac{b-a}{n} \right)$$
 A: I suggest that you do not do the last steps, and thus not turn this into what you call the standard form. Look more carefully at
$$
\lim_{N\to+\infty}\sum_{n=-N}^N\frac{2}{1+(n/N)^2}\frac{1}{N}.
$$
Here, $n/N$ that should be replaced by $t$ varies from $-1$ to $1$. $1/N$ can be replaced by $dt$, so it is indeed the Riemann sum of
$$
\int_{-1}^1\frac{2}{1+t^2}\,dt.
$$
Update
Since you seem to know that you want to replace $n/N$ by your variable and you ask for the limits of integration here comes a clarification of why the limits are $-1$ and $1$ (also read the comments by @mgn): 
Since the sum runs from $n=-N$ to $n=N$ the ratio $n/N$ will run from $-1$ to $1$.
Yet another update
Let us put the sum into the so-called standard form (although, I strongly suggest one should not do this all the time):
With $k=n+N$ and then $\tilde N=2N$, we have
$$
\begin{aligned}
\sum_{n=-N}^N\frac{2}{1+(n/N)^2}\frac{1}{N}&= \sum_{k=0}^{2N}\frac{2}{1+\bigl(\frac{k-N}{N}\bigr)^2}\frac{1}{N}\\
&=\sum_{k=0}^{\tilde N} \frac{2}{1+\bigl(-1+k\frac{1-(-1)}{\tilde N}\bigr)^2}\frac{1-(-1)}{\tilde N}\\
&=\frac{2}{\tilde N}+\sum_{k=1}^{\tilde N} \frac{2}{1+\bigl(-1+k\frac{1-(-1)}{\tilde N}\bigr)^2}\frac{1-(-1)}{\tilde N}
\end{aligned}
$$
The term $2/\tilde N$ tends to zero as $\tilde N$ (and $N$) tends to infinity, and the second term is exacly in the form called standard.
