Some quick notes about what I have studied so far about sequences (Note: this is an edit after I accepted the answer and realized my lack of understanding for this simple sequence):
Here I am working with sequences of Real numbers, therefore, for that sequence, I defined a function $a$, such that
$$a: \mathbb{N}\to \mathbb{R}$$
$$a:n \mapsto a(n) = a_n$$
The sequence is denoted as $\{a_n\}_{n\in\mathbb{N}}$, such that $\mathbb{N} = \{1, 2, 3, …\}$.
Given sequence $\left\{6(-\frac{5}{6})^n\right\} _{n=1}^\infty$
Once a sequence $\{a_n\}_{n\in\mathbb{N}}$ is convergent for $L\in\mathbb{R}$ when $n\to\infty$, therefore
$$\lim _{n\to\infty} a_n = L$$
this is for only if for a given $\epsilon > 0$, so we can find a $N_0 \in \mathbb{N}$ such that $\forall n > N_0$ we have $|a_n – L| < \epsilon $.
Basically, a sequence is convergent if it tends to a Real number. On the other hand, it is divergent if it does not converge to a Real number. This is the reason why I tried to find the limit but I ended up not being able to solve it.
I know that for the sequence $a_n$, such that $a_n = 6\left(-\dfrac{5}{6}\right)^n$ we have.
$$a_n = 6\left(-\dfrac{5}{6}\right)^n = 6 \cdot\dfrac{(-5)^n}{6^n} = 6^{1-n}\cdot(-5)^n$$
According to Wolfram Alpha, $\lim _{n\to \infty }(6^{1-n}(-5)^n) = 0$ and thus the sequence converges to $0$. But I couldn't solve the limit. Wolfram solves it using exponential but it wasn't a straightforward solution for me.
How can I solve $\lim _{n\to \infty }(6^{1-n}(-5)^n)$ in a simpler and good way?
This might be a consequence of the first question, but how can I show that the sequence is convergent? What would be another approach?
Best Answer
We recognize a geometric sequence. We take $r=-5/6$ and see that $|r| < 1$ so $6 (-5/6)^n \to 0$ (see here for a proof).