Convergence of $\sum\sum_{k, n=1}^\infty\frac{1}{(n+3)^{2k}}$ Convergence of $\sum\sum_{k, n=1}^\infty\frac{1}{(n+3)^{2k}}$.
What I tried:
For the iterated summation, $\sum_{n=1}^\infty(\sum_{k=1}^\infty\frac{1}{(n+3)^{2k}})=\sum_{n=1}^\infty\lim_{k\to\infty}\frac{1-(\frac{1}{n+3})^{2k}}{1-(\frac{1}{n+3})^2}=\sum_{n=1}^\infty\frac{1}{1-(\frac{1}{n+3})^2}$.
But when $n\to\infty$, $\frac{1}{1-(\frac{1}{n+3})^2}\to 1\neq 0$, so the double summation diverges.
Is this proof right? And for the a general double series to converge, is it necessary that the iterated summation also converges?
 A: The idea is completely right!—you just made a typo involving the first term of each geometric series:
$$
\sum_{n=1}^\infty \bigg( \sum_{k=1}^\infty\frac{1}{(n+3)^{2k}} \bigg) = \sum_{n=1}^\infty\lim_{K\to\infty}\frac{\frac{1}{(n+3)^2}-\frac{1}{(n+3)^{2K+2}}}{1-\frac{1}{(n+3)^2}}=\sum_{n=1}^\infty\frac{\frac{1}{(n+3)^2}}{1-\frac{1}{(n+3)^2}}=\sum_{n=1}^\infty\frac{1}{(n+3)^2-1}.
$$
I'm guessing you can determine whether this series converges or diverges.
For your last question: when the terms of a double series are nonnegative, its convergence is equivalent to the convergence of either iterated series (this is "Tonelli's theorem"). In general, however, one needs some assumptions to convert a double series to an iterated series. Look for "Fubini's theorem" (it's usually stated for double integrals, but it holds for double series as well).
A: The geometric series doesn't start at $k = 0$, so the numerator is not $1$.  (That is, the numerator is the first term in the series, which is not $1$.)
\begin{align*}
 \sum_{k=1}^\infty \frac{1}{(n+3)^{2k}} &= \sum_{k=1}^\infty \left( (n+3)^{-2} \right)^k  \\
    &= \frac{(n+3)^{-2}}{1 - (n+3)^{-2}}  \\
    &= \frac{1}{n^2+6n+8}  \text{,}  
\end{align*}
so your argument is not quite right.
Hopefully, you can use the comparison test to resolve the convergence of the resulting series in $n$.
For this particular series, since all the terms are positive, you are free to rearrange the series as you like, so turning it into an iterated series is fine.  (The collection of sums of finite subsets of the set of terms is indifferent to what order you imagine you are summing them.)  If the multiple series is not absolutely convergent, the question of rearranging it into an iterated series is a little more delicate.  (This should not be a surprise.  Contrast absolutely convergent and conditionally convergent and think about the Riemann rearrangement theorem.)
Fubini's theorem is the usual tool to justify such a rearrangement.  And when use the counting measure to switch back and forth between sums and integrals, the Fubini hypotheses are equivalent to requiring absolute convergence of the multiple series.
