# Thread: Sequences, Series and Convergence.

1. ## Sequences, Series and Convergence.

For each of the following sequences, determine the convergence or divergence. If the sequence converges, find it's limit:

1) {$\displaystyle {a_n}$} $\displaystyle =$ {$\displaystyle {1+\frac{(-1)^n}{n}}$}

2) {$\displaystyle {a_n}$} $\displaystyle =$ {$\displaystyle {\frac{3n^{-1} - 5n^{2}}{2n(7n-9n^{-2})}}$}

Thanks!

2. Originally Posted by qzno
For each of the following sequences, determine the convergence or divergence. If the sequence converges, find it's limit:

1) {$\displaystyle {a_n}$} $\displaystyle =$ {$\displaystyle {1+\frac{(-1)^n}{n}}$}

2) {$\displaystyle {a_n}$} $\displaystyle =$ {$\displaystyle {\frac{3n^{-1} - 5n^{2}}{2n(7n-9n^{-2})}}$}

Thanks!
Do you want to know if the sequences converge or if the series converge?

1. The sequence $\displaystyle a_n = 1 + \frac{(-1)^n}{n}$ is clearly convergent.

As $\displaystyle n \to \infty, \frac{(-1)^n}{n} \to 0$.

Therefore $\displaystyle a_n \to 1$

2. The sequence $\displaystyle a_n = \frac{3n^{-1} - 5n^{2}}{2n(7n-9n^{-2})}$ can be rewritten as

$\displaystyle a_n = -\frac{5}{14} - \frac{1}{7n^2 - 9}$.

As $\displaystyle n \to \infty, \frac{1}{7n^2 - 9} \to 0$.

Therefore

$\displaystyle a_n \to -\frac{5}{14}$.