Splitting up an infinite sum

Question

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?

Answer 1

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.

Answer 2

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 $n \ge 4$.

Answer 3

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 $(a_n) \notin \ell^2$.