How can i prove if the series $\sum_{n=1}^\infty x_n^2$ converges, then so does $\sum_{n=1}^\infty x_n^3$?

Question

if the series $\sum_{n=1}^\infty x_n^2$ converges, then so does $\sum_{n=1}^\infty x_n^3$

is there a simple proof for this question?

Answer 1

Having a sum converge requires that the terms converge to zero. That means that after some $N$ they are all less than $1$ in absolute value. The sum of $x_n^3$ up to $N$ is some constant. After that $|x_n^3| \lt |x_n^2|$. The sum of $x_n^2$ converges absolutely so the sum of $x_n^3$ does as well by the comparison test.

Answer 2

Use the comparison test for the tail of the sequence. By the $n$th term test, we know that $\sum_{k=1}^\infty x_n$ converges, $x_n \to 0$ as $ n\to\infty$. By the definition of convergenve, we may choose $\varepsilon =1$ and guarantee that there is some $K\in\mathbb{N}$ so that $n>K \implies |x_n|<\varepsilon=1$. Since $|x_n^3|<|x_n^2|$ we can use the comparison test for $\sum_{k=K+1}^\infty x_n^3$. For the first $K$ terms, we have a finite sum, and the tail converges by our comparison test, so the sum should converge.