$$a_n = \frac{e^{n!\tan (n\pi)}-1+n^2\sin n}{\log (1+3^{n\log n})\log (n+5^n)}, n \in \mathbb{N}$$
This sequence:
- Has minimum but no maximum
- Has neither maximum nor minimum
- Has both maximum and minimum
- Has maximum but no minimum
I have tried by looking at the limit of this sequence, and since $\tan(n\pi) = 0$ $\forall n \in \mathbb{N}$, the $e^{\tan \pi}$ seems to be constantly equal to 1. The two logarithms at the denominator approach $+\infty$, but the numerator doesn't seem to approach a number due to the sine.
$a_1 = \frac{\sin 1}{\log 2 \log 6}$, however that's not enough for me to draw any conclusions on maxima and/or minima without being able to evaluate the limit.
This kind of exercise is the one I tend to have the most problems with. When I have to determine whether there is an upper/lower bond, or maxima/minima I tend to struggle the most with it.
Any hints?
$e^{n!\tan (n\pi)}$ will be exactly equal to $1$ for all integral values since $\tan n\pi$ is exactly equal to $0$. Hence, the fraction becomes $\frac{n^2\sin n}{\ln (1+3^{n\ln n})\ln (n+5^n)}$.
$\implies L=\lim\limits_{n\to\infty}\frac{\sin n}{\frac{\ln(1+3^{n\ln n})}{n}\frac{\ln(n+5^n)}{n}}$
For any integral value of $n, \sin n$ does not vanish and oscillates between $(-1,1)$. For the denominator, both separate factors can be resolved using L'Hospital's Rule.
$\lim\limits_{n\to\infty}\frac{\ln(1+3^{n\ln n})}{n}=\lim\limits_{n\to\infty}\frac{\frac{1}{1+3^{n\ln n}}\cdot3^{n\ln n}\cdot(1+\ln n)\ln3}{1}=\lim\limits_{n\to\infty}\ln3(1+\ln n)\to\infty$ Similarly, the other factor tends to a finite value. Hence, the denominator tends to $\infty$ and $L\to0$.
Therefore, the sequence converges to $0$.