Show that Ratio Test gives no information for $3^{-1} + 5^{-1} + 3^{-2} + 5^{-2} + 3^{-3} + 5^{-3} + \cdots$ 
Q: Examine the series:
\begin{align*}
  \frac{1}{3} + \frac{1}{5} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots \\
\end{align*}
Prove that the Root Test shows that the series converges while the Ratio test gives no information.

I can use the Root Test to show convergence as the question requests. I will omit that. My issue is that when I apply the Ratio Test, it also seems to show convergence, when the problem asks me to show that the Ratio Test gives no information.
Applying the Ratio Test:
\begin{align*}
  \lim\limits_{j \to \infty} \left| \frac{a_{j+1}}{a_j} \right| &= \lim\limits_{j \to \infty} \frac{\frac{1}{3^{j+1}} + \frac{1}{5^{j+1}}}{\frac{1}{3^j} + \frac{1}{5^j}} \\
  &= \lim\limits_{j \to \infty} \frac{\frac{1}{3} + \frac{1}{5} \cdot \left( \frac{3}{5} \right)^j}{1 + \left( \frac{3}{5} \right)^j} \\
  &= \frac{1}{3} < 1 \\
\end{align*}
And therefore the ratio test says that this series will converge which is not what the problem was asking for.
 A: Given the series is $$3^{-1} + 5^{-1} + 3^{-2} + 5^{-2} + 3^{-3} + 5^{-3} + \cdots=\dfrac13+\dfrac15+\dfrac{1}{3^2}+\dfrac1{5^2}+\dfrac1{3^3}+\dfrac1{5^3}+.......$$
Notice that from the above series we get
$$a_{2n-1}=\dfrac{1}{3^n}\mbox{ and }a_{2n}=\dfrac{1}{5^n}$$
Now consider taking $\dfrac{a_{2n+1}}{a_{2n}}$ and $\dfrac{a_{2n}}{a_{2n-1}}$
$$\dfrac{a_{2n+1}}{a_{2n}}=\dfrac{1}{3}\left(\dfrac{5}{3}\right)^n\mbox{ which is greater than }1$$
and now $$\dfrac{a_{2n}}{a_{2n-1}}=\left(\dfrac{3}{5}\right)^n\mbox{ which is less than }1$$
Therefore, from $\dfrac{a_{2n+1}}{a_{2n}}$ and $\dfrac{a_{2n}}{a_{2n-1}}$ we cannot conclude.
A: Any sum of the form
$\sum_{n=1}^{\infty} (a_n+b_n)
$
where
$\dfrac{b_n}{a_n} < 1$
and
$\dfrac{a_{n+1}}{b_n} > 1$
will behave this way.
If
$a_n = u^{2n-1}$
and
$b_n = v^{2n}$
then
$\dfrac{b_n}{a_n}
=\dfrac{v^{2n}}{u^{2n-1}}
=\frac1{u}(v/u)^{2n}
$
and
$\dfrac{a_{n+1}}{b_n}
=\dfrac{u^{2n+1}}{v^{2n}}
=u(u/v)^{2n}
$
so,
if $u > v$,
the ratio test will not work.
However,
if $0 < u, v < 1$,
the root test shows that
the sum converges.
