Convergence of series $\sum_{n=1}^\infty\left[\left(1+\frac{1}{n}\right)^{n+1}-a\right]\sin{n}$ How to deduce if series is convergent (depending on parameter $a$):
$$
\sum_{n=1}^\infty\left[\left(1+\frac{1}{n}\right)^{n+1}-a\right]\sin{n}?
$$
If $a=e,$ we have that $\lim_{n\to\infty}a_n=0,$ but it is not sufficient to show that series converges.
I'm stuck with this today, not at my best. Any help is welcome. Thanks in advance.

If $a \neq e$, then $$\left[\left(1+\frac{1}{n}\right)^{n+1}-a\right]\sin{n}$$
does not tend to $0$ so the series cannot converge.
If $a=e$, you have
$$\left[\left(1+\frac{1}{n}\right)^{n+1}-e\right]\sin{n} = \left[e^{(n+1)\ln\left( 1 + \frac{1}{n}\right)}-e\right]\sin{n}=e\left[e^{(n+1)\ln\left( 1 + \frac{1}{n}\right)-1}-1\right]\sin{n}$$
Now
$$(n+1)\ln\left( 1 + \frac{1}{n}\right)-1 = (n+1)\left( \frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} + o\left( \frac{1}{n^3}\right)\right) - 1$$ $$=   \frac{1}{2n}  - \frac{1}{6n^2} + o\left( \frac{1}{n^2}\right)$$
So
$$\left[\left(1+\frac{1}{n}\right)^{n+1}-e\right]\sin{n} =e\left[ \frac{1}{2n}  - \frac{1}{24n^2} + o\left( \frac{1}{n^2}\right)\right]\sin{n}$$
Finally, the series
$$\sum \frac{\sin(n)}{n}$$
converges (by Dirichlet's test) ; and the series
$$\sum \frac{\sin(n)}{n^2}$$
is absolutely convergent. So your series converges.
