# Question #2538f

Dec 27, 2017

My hunch is to use the Ratio Test to prove the $\Leftarrow$ direction and the Divergence Test to prove the contrapositive of the $\Rightarrow$ direction (which also proves the $\Rightarrow$ implication). Not sure if it will work, but here goes.

#### Explanation:

($\Leftarrow$) Suppose that $\left({u}_{n}\right)$ is a sequence that converges to a negative number $u$ and that $a > 1$. Let ${b}_{n} = {a}^{{u}_{1} + {u}_{2} + \cdots + {u}_{n}} = {a}^{{u}_{1}} \cdot {a}^{{u}_{2}} \cdots {a}^{{u}_{n}}$.

Then $| {b}_{n + 1} \frac{|}{|} {b}_{n} | = \frac{{a}^{{u}_{1}} \cdot {a}^{{u}_{2}} \cdots {a}^{{u}_{n}} \cdot {a}^{{u}_{n + 1}}}{{a}^{{u}_{1}} \cdot {a}^{{u}_{2}} \cdots {a}^{{u}_{n}}} = {a}^{{u}_{n + 1}} \to {a}^{u}$ as $n \to \infty$.

Since $a > 1$ and $u < 0$, $| {a}^{u} | = {a}^{u} < 1$.

Therefore, by the Ratio Test, the series ${\sum}_{n = 1}^{\infty} \left({a}^{{u}_{1} + {u}_{2} + \cdots + {u}_{n}}\right)$ converges.

Therefore, the $\Leftarrow$ direction is proven.

($\Rightarrow$) Suppose that $\left({u}_{n}\right)$ is a sequence that converges to a negative number $u$ and that $a \le 1$. Choose $N \in \mathbb{N}$ so that ${u}_{n} < 0$ for all $n > N$.

Let $A = {a}^{{u}_{1} + {u}_{2} + \cdots + {u}_{N}} = {a}^{{u}_{1}} \cdot {a}^{{u}_{2}} \cdots {a}^{{u}_{N}}$ and note that $A \ne 0$.

Next, suppose $n > N$. Then ${a}^{{u}_{1} + {u}_{2} + \cdots {u}_{n}} = {a}^{{u}_{1}} \cdot {a}^{{u}_{2}} \cdots {a}^{{u}_{N}} \cdot {a}^{{u}_{N + 1}} \cdots {a}^{{u}_{n}} = A \cdot {a}^{{u}_{N + 1}} \cdots {a}^{{u}_{n}}$. Since $a \le 1$ and ${u}_{N + 1} < 0$, ${u}_{N + 2} < 0$, ..., ${u}_{n} < 0$, it follows that ${a}^{{u}_{N + 1}} \cdots {a}^{{u}_{n}}$ is the product of a bunch of numbers that are all $\ge 1$.

Since $A \ne 0$, it follows that $A \cdot {a}^{{u}_{N + 1}} \cdots {a}^{{u}_{n}}$ does not converge to zero as $n \to \infty$. But this means that ${a}^{{u}_{1} + {u}_{2} + \cdots {u}_{n}}$ does not converge to zero as $n \to \infty$.

By the Divergence Test, the series ${\sum}_{n = 1}^{\infty} \left({a}^{{u}_{1} + {u}_{2} + \cdots + {u}_{n}}\right)$ diverges.

Thus, the contrapositive of the $\Rightarrow$ implication is proved, meaning the original implication itself is also proved.