If $\sum x^6_n$ converges, then $\sum x^7_n$ converges too If $\displaystyle\sum x^6_n$ converges, then  $\displaystyle\sum x^7_n$ converges too.
 A: Hint: We deduce that $x_n^6 \to 0$, and furthermore that $\sum x_n^6$ converges absolutely.
Choose $N \in \Bbb N$ such that for all $n > N$, $|x_n| < 1$.  For all such $n$, we also have $|x_n|^7 \leq |x_n|^6$.
A: Note that this is false
if 6 and 7
are replaced by 5 and 6.
Here is an example where
$\sum x^5_n$
converges but
$\sum x^6_n$
does not.
Let
$x_n
=\dfrac{(-1)^n}{n^{1/6}}
$.
Then
$\sum x^5_n
=\sum \dfrac{(-1)^n}{n^{5/6}}
$
converges because 
it is an
alternating series with
decreasing terms
and
$\sum x^6_n
=\sum \dfrac{1}{n}
$
diverges because harmonic.
A: The statement is false if the terms are allowed to be complex. Take for example
$$
x_{6k+j} = \exp(2\pi i j/36) \cdot \frac1{(k+1)^{1/7}}
$$
for all $k$ and $0 \le j \le 5$.
Then partial sums $s_n = \sum_{k=0}^n x_k^6$ with $n=6k$ terms are equal to $0$, and the other partial sums are small: for example $|s_{6n+j}| \le 6/(n+1)^{1/7}$.
On the other hand, with a little effort we can show that
$$
\sum_{n=0}^{6k} x_n^7 = c \sum_{n=0}^k \frac{1}{n+1}
$$
where $c \approx (0.868-0.076i)$. This shows that the series $\sum x_n^7$ diverges.
