Consider for large integers $n$ the expression $\sin \left(\pi \sqrt{4 n^2+n}\right)$.

Since 

$\sqrt{4 n^2+n}=2 n \sqrt{1 + \frac{1}{4 n}}$

we can use the standard series for the square root and next the standard series for $sin$ to find a series expansion of this expression around $\infty$. That is a simple first year exercise.

I was unable to do this computation in an elegant way with *Mathematica*. I expected the following to work:

```
Series[Sin[Sqrt[n+4 n^2] π], {n,∞,3}, Assumptions->{n ∈ Integers}]

```

$\sin \left(2 \pi n+\frac{\pi }{4}-\frac{\pi }{64 n}+\frac{\pi
 }{512 n^2}-\frac{5 \pi }{16384
 n^3}+O\left(\left(\frac{1}{n}\right)^4\right)\right)$

To my surprise, the expansion is only inside `Sin` and the fact that `n` is an integer is not used.

I tried to use `Normal` and `TrigExpand` without success. Finally I found a rather dirty trick to obtain the result I was looking for:

```
Series[Sin[Sqrt[n + 4 n^2] π], {n, ∞, 3}]  /.  Sin[x_] :> Sin[x - 2 π n]

```

$\frac{1}{\sqrt{2}}-\frac{\pi }{64 \sqrt{2} n}+\frac{\frac{\pi
 }{512 \sqrt{2}}-\frac{\pi ^2}{8192
 \sqrt{2}}}{n^2}+\frac{-\frac{5 \pi }{16384 \sqrt{2}}+\frac{\pi
 ^2}{32768 \sqrt{2}}+\frac{\pi ^3}{1572864
 \sqrt{2}}}{n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$

Can this result be found in a more straightforward way?

**Edit**

Let us reconsider

```
Series[Sin[Sqrt[n + 4 n^2] π], {n, ∞, 3}, Assumptions -> {n ∈ Integers}]

```

$\sin \left(2 \pi n+\frac{\pi }{4}-\frac{\pi }{64 n}+\frac{\pi
 }{512 n^2}-\frac{5 \pi }{16384
 n^3}+O\left(\left(\frac{1}{n}\right)^4\right)\right)$

Henric Schumacher is quite right in his assumption that we do not find a series expansion, but only a series expansion wrapped in `Sin`, because `Sin` has an essential singularity at `Infinity`. The documentation of `Series`, under `Possible Issues`, shows something similar. So this should not have surprised me.

The assumption that `n`is an integer has not prevented the appearance of the term `2πn` in the argument of `Sin`. Without that term, the result would immediately further evaluate to the result we are looking for. So we have to get rid of this term.

The argument of `Sin` is a `SeriesData` expression, which is atomic. Therefore, the substitution `2πn->0` does not work. But assignment is possible:

```
sd=Series[Sin[Sqrt[n+4 n^2] π],{n,∞,3}];
sd[[1,3,1]]=0;
sd

```

$\frac{1}{\sqrt{2}}-\frac{\pi }{64 \sqrt{2} n}+\frac{\frac{\pi
 }{512 \sqrt{2}}-\frac{\pi ^2}{8192
 \sqrt{2}}}{n^2}+\frac{-\frac{5 \pi }{16384 \sqrt{2}}+\frac{\pi
 ^2}{32768 \sqrt{2}}+\frac{\pi ^3}{1572864
 \sqrt{2}}}{n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$

Another way of getting rid of the term `2πn` is to use `Normal` and `FullSimplify` with the assumption that `n` is an integer, as done in the answer of Mikado. After that, we need another call of `Series` to arrive at the desired result.

To summarize: at the moment, it seems that there is no straightforward way for finding this series expansion by *Mathematica*, while it is easily found with pen and paper.