= Solution
A real <sequence> $(x_n)$ has <limit of a sequence> $x$ when
$$
\forall\varepsilon>0\ \exists N\in\mathbb N\ \forall n\geq N:\quad |x_n-x|<\varepsilon.
$$
A <series> $\sum_{n\geq1}x_n$ is a <convergent series> with sum $s$ when its <partial sums> $S_N=\sum_{n=1}^Nx_n$ have <limit of a sequence> $s$.
For the first <comparison test for series>, either of the two allowed bounds implies $0<x_n\leq a_n+b_n$. Hence
$$
0<\sum_{n=1}^Nx_n\leq\sum_{n=1}^{\infty}a_n+\sum_{n=1}^{\infty}b_n<\infty.
$$
The <partial sums> form a <monotone bounded sequence>, so they converge. \b[The <series> $\sum x_n$ is convergent.]
For the exponent $2$, use $1/n^2\leq1/[n(n-1)]=1/(n-1)-1/n$ for $n\geq2$. The resulting <telescoping series> bounds every <partial sum> by $2$, proving that the <series> converges by the <monotone bounded sequence> property. To prove divergence of the <harmonic series>, group the terms with $2^j<n\leq2^{j+1}$: their sum is at least $2^j/2^{j+1}=1/2$. Thus its <partial sums> are unbounded. If $\alpha\leq1$, then $n^{-\alpha}\geq n^{-1}$, so the <comparison test for series> proves
$$
\boxed{\sum_{n\geq1}n^{-2}<\infty,\qquad\sum_{n\geq1}n^{-\alpha}=\infty\quad(\alpha\leq1).}
$$
For the weighted square assumption, the finite-dimensional <Cauchy-Schwarz inequality> gives, for every $N$,
$$
\sum_{n=1}^Nx_n=\sum_{n=1}^N(nx_n)\frac1n
\leq\left(\sum_{n=1}^Nn^2x_n^2\right)^{1/2}\left(\sum_{n=1}^N\frac1{n^2}\right)^{1/2}
\leq\left(\sum_{n\geq1}n^2x_n^2\right)^{1/2}\left(\sum_{n\geq1}n^{-2}\right)^{1/2}<\infty.
$$
Again the positive <partial sums> form a <monotone bounded sequence>. This is the <weighted square roots of a summable sequence> argument, with $a_n=n^2x_n^2$. \b[The weighted square condition implies $\sum x_n<\infty$.]
\b[The converse is false.] Take $x_n=n^{-3/2}$. Each block $2^j\leq n<2^{j+1}$ contributes at most $2^j2^{-3j/2}=2^{-j/2}$, and the resulting <geometric series> converges. But $n^2x_n^2=1/n$, whose <harmonic series> diverges. This proves the counterexample without assuming the full <P-series> criterion.
Back to article page
