Splitting up an infinite sum I am playing with the following example trying to determine if $\frac{1}{n}-\frac{1}{\sqrt{n}}$ is in $\ell^2$.
Here is what I have,
\begin{align}
\sum\limits_{n=1}^\infty \left( \frac{1}{n}-\frac{1}{\sqrt{n}}\right)^2 &= \sum\limits_{n=1}^\infty \left(\frac{\sqrt{n}-n}{n\sqrt{n}}\right)^2\\
&=\sum\limits_{n=1}^\infty\frac{n-2n\sqrt{n}+n^2}{n^3}\\
&=\sum\limits_{n=1}^\infty\frac{1}{n^2}-2\sum\limits_{n=1}^\infty\frac{1}{n\sqrt{n}}+\sum\limits_{n=1}^\infty\frac{1}{n}
\end{align}
The first summand is in $\ell^2$, the second summand is in $\ell^2$, but the third summand is not. Thus the sequence $\frac{1}{n}-\frac{1}{\sqrt{n}}$ is not in $\ell^2$. I think I may be making an error when I split the sums. Allowing such splitting would result in things like,
\begin{align}
\sum\limits_{n=1}^\infty\frac{1}{n^2} &= \sum\limits_{n=1}^\infty\frac{1-n^2}{n^2}+\sum\limits_{n=1}^\infty\frac{n^2}{n^2}
\end{align}
What am I missing here?
 A: You could note that$$\sum_{n=1}^\infty\frac{n-2n\sqrt n+n^2}{n^3}=\sum_{n=1}^\infty\frac{1-2\sqrt n+n}{n^2}$$and that$$\lim_{n\to\infty}\frac{\frac{1-2\sqrt n+n}{n^2}}{\frac1n}=\lim_{n\to\infty}\frac{1-2\sqrt n+n}n=1.$$Therefore, your series diveres, by the comparison test.
A: Put $a_n=\frac{1}{n}-\frac{1}{\sqrt{n}}, b_n =\frac{1}{n}$ and $c_n=\frac{1}{\sqrt{n}}$.
Suppose that $(a_n) \in \ell^2$. Since $(b_n) \in \ell^2$ and since $\ell^2$ is a vector space, we get that $(c_n)=(b_n)-(a_n) \in \ell^2$, a contradiction. Hence $(a_n) \notin \ell^2$.
A: It is not in $\ell^2$:
\begin{align}
\sum_{n=1}^\infty\frac{n-2n\sqrt n+n^2}{n^3}&=\sum_{n=1}^\infty\frac{1-2\sqrt n+n}{n^2} \\
&= \sum_{n=1}^\infty\frac{(\sqrt{n}-1)^2}{n^2} \\
&\ge \sum_{n=1}^3\frac{(\sqrt{n}-1)^2}{n^2} +  \sum_{n=4}^\infty\frac{\left(\frac12\sqrt{n}\right)^2}{n^2} \\
&= \sum_{n=1}^3\frac{(\sqrt{n}-1)^2}{n^2} + \frac14 \sum_{n=4}^\infty\frac1n \\
&= +\infty
\end{align}
because $\sqrt{n}-1 \ge \frac12 \sqrt{n}$ for all $n \ge 4$.
