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\noindent {\bf Question}

\noindent Let $\sum_{n=1}^\infty a_n$ be a convergent series of
positive terms.  Show that for each $s\ge 1$, the series
$\sum_{n=1}^\infty a_n^s$ is also convergent.

\medskip

\noindent {\bf Answer}

\noindent By the contrapositive to the $n^{th}$ term test for
divergence, since the series $\sum_{n=1}^\infty a_n$ converges, we
have that $\lim_{n\rightarrow\infty} a_n =0$.  In particular,
taking $\varepsilon = 1$ and remembering that each $a_n
&gt;0$, there exists $M$ so that $0 &lt; a_n &lt;1$ for all $n &gt;M$.   Since $0
&lt; a_n &lt; 1$ for $n &gt;M$ and since $s\ge 1$, we have that $a_n^s &lt;
a_n$ for $n &gt;M$, and so by the second comparison test, we have
that $\sum_{n=M+1}^\infty a_n^s$ converges by comparison to
$\sum_{n=M+1}^\infty a_n$.  Since $\sum_{n=M+1}^\infty a_n^s$
converges, we see that $\sum_{n=0}^\infty a_n^s$ converges, as
desired.

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