I was trying to find the convergence of 3 series, I don't know if the procedure I used is correct.
- $$\sum_{n=1}^\infty (-1)^n \cos(1/n) $$I had no idea how to proceed with this excercise
- $$\sum_{n=1}^\infty (2^n +4^n)/(2e)^n ∼ \sum_{n=1}^\infty = 4^n / e^n = (4/e)^n$$ I used comparision test, so $4/e > 1$ so the series diverges
- When $x\in\mathbb{R}$ $$\sum_{n=1}^\infty \log(1+(1/n^\frac{x}{2}) ∼\sum_{n=1}^\infty 1/n^\frac{x}{2} $$ The series converges if $x/2 > 1, x>2$
- When $x\in\mathbb{R}$ $$\sum_{n=1}^\infty e^{3nx}$$ $\sqrt[n]{e^{3nx}} = e^{3x} = (e^3)^x$ I used te root test so the series converges if $e^3 \leq t$ for $t\in (0,1)$, which is not true so the series diverges.
Some observations.
What is the following limit? $$\lim_{n \to \infty}\left|(-1)^n \cos \frac1n\right| $$
Note that $$ \frac{2^n +4^n}{(2e)^n}=\frac{1 +2^n}{e^n}=\frac{1}{e^n}+\left(\frac{2}{e}\right)^n. $$
For which values of $x$ do we have the following inequality? $$ |e^{3x}|<1.$$