A calculus problem by Jose Sacramento

Let us consider the summant $\dfrac {\sin n + 1}{(2n)!} \le \dfrac 2{(2n)!} < \dfrac 1{n!}$ . Since $\displaystyle \sum_{n=1}^\infty \frac 1{n!} = e-1$ converges, therefore by direct comparison, $\displaystyle \sum_{n=1}^\infty \frac {\sin n + 1}{(2n)!}$ also converges. Therefore the answer is $\boxed{\text{True}}$ .

No mathematical formula, but just logical reasoning (I may be wrong, please correct me if so :) ): We have two parts: The numerator and the denominator.

>> The sinus-function is a periodic function and just repeats forever (adding one just moves the whole function to more positive values -> think of adding one to any function in a coordinate system, the form of the function stays the same because +1 is a constant and does not change). >> The (2n)! part however will get bigger and bigger if n gets larger.

Therefore a somewhat "constant" function is devided by more and more bigger values, the result will get smaller and smaller till it reaches some kind of "balance"/"limit". So it converges.