Sequence of Partial Sums is Convergent How do I show that the series
$\sum_{n=1}^{\infty} \frac1{(2n-1)^{n}} + \frac1{(2n)^{3 }} $  is convergent? I'm trying to use Comparison Test but I'm having a hard time looking for a convergent series that is "larger" than the given series.
 A: Note that since $\sum_{n=1}^{\infty} \frac{1}{n^2} $ is convergent(see here), this implies that $\sum_{n=1}^{\infty}  \frac1{(2n)^{3 }} $ is convergent. 
Also, for$\sum_{n=2}^{\infty} \frac1{(2n-1)^{n}} \le  \sum_{n=2}^{\infty} \frac{1}{n^2} $
Therefore $\sum_{n=1}^{\infty}  \frac1{(2n)^{3 }} +\frac1{(2n-1)^{n}} \le 2(\sum_{n=2}^{\infty} \frac{1}{n^2}) +1$ 
This implies the series is convergent. 
A: Well you can break apart the summation into two. The second summation can be rewritten as $\frac{1}{8}\sum_{n=1}^\infty \frac{1}{n^3}$, and any series with a common term $\frac{1}{n^k}$ is convergent for $k > 1$ (see this). As for the first part, you can compare it to a geometric series.
\begin{equation}
1 + \frac{1}{2^2} + \frac{1}{5^3} + \cdots < 1 + \frac{1}{2^2} + \frac{1}{2^3} + \cdots
\end{equation}
The right hand side(excluding the one) is a geometric series, and each term is greater than that of its corresponding right side term, since $\frac{1}{(2n - 1)^n} \leq \frac{1}{2^n}$ if $n > 1$. Therefore the second sum converges(as it is a geometric series with a common ratio less than 1). Since both sums converge, then by the linearity of the summation the whole sum must converge.
