I'm hoping for some help on the following question. I haven't gotten very far:
Let $\{h_n\}_{n\geq 1}$ be a sequence of vectors in a Hilbert space $H$ with the property that $(h_n-h_m)\perp h_m$ for $n\geq m$. Show that
$$ \sum_{n\geq1}\frac{h_n}{||h_n||^2} \ \text{ converges in $H$ }\iff \sum_{n\geq1}\frac{n}{||h_n||^2}<\infty. $$
Playing with the given property it's easy to show $(h_n,h_m)=||h_m||^2$ for $n\geq m$, and hence via Cauchy-Schwarz we can get $||h_n||\geq ||h_m||$ for $n\geq m$. Unfortunately, beyond this, I am stuck. Any sort of help would be greatly appreciated. Thanks!
These are the ideas that should be used.
Let us do some massage to what we have.
$$||h_{n+1}-h_n||^2=||h_{n+1}||^2+||h_n||^2-2(h_{n+1},h_n)=||h_{n+1}||^2-||h_n||^2$$
We also have, for $n\geq m$,
$$\begin{align}(h_{n+1}-h_n,h_{m+1}-h_m)&=(h_{n+1},h_{m+1})+(h_n,h_m)-(h_{n+1},h_m)-(h_n,h_{m+1})\\&=||h_{m+1}||^2+||h_m||^2-||h_m||^2-||h_{m+1}||^2\\&=0\end{align}$$
for $n\geq m$
$$(h_{n+1}-h_n,h_{m})=(h_{n+1},h_m)-(h_n,h_m)=||h_m||^2-||h_m||^2=0$$.
and for $n>m$
$$(h_{m+1}-h_m,h_{n})=(h_{m+1},h_n)-(h_m,h_n)=||h_{m+1}||^2-||h_m||^2.$$
To simplify notation let us put $||h_n||^2=r_n$
We use the following transformation of the sum we have ($\sum \frac{h_n}{r_n}$)
I will only do it with a small partial sum.
$$\frac{h_1}{r_1}+\frac{h_2}{r_2}+\frac{h_3}{r_3}=h_1\left(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\right)+(h_2-h_1)\left(\frac{1}{r_2}+\frac{1}{r_3}\right)+(h_3-h_2)\left(\frac{1}{r_3}\right)$$
You know, summation by parts (Abel's transformation) taking the factor $h_n$ for the finite difference and $\frac{1}{r_n}$ for the summation.
Now we compute norm squared and bound it with (use for this the computation of $||h_{n+1}-h_n||^2$ and the orthogonality of the differences)
$$\begin{align}r_1\left(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\right)^2&+(r_2-r_1)\left(\frac{1}{r_2}+\frac{1}{r_3}\right)^2+(r_3-r_2)\left(\frac{1}{r_3}\right)^2=\\&=r_1[(...)^2-(...)^2]+r_2[(...)^2-(...)^2]+r_3[(...)^2]\\&=\left(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\right)+\left(\frac{1}{r_2}+\frac{1}{r_3}\right)+\left(\frac{1}{r_3}\right)\\&=\frac{1}{r_1}+\frac{2}{r_2}+\frac{3}{r_3}\end{align}$$
where the line with the $(...)^2$ is to give an idea on how to do the simplification, i.e. by opening the parentheses that have the differences of $r_i$'s. But you can do it in any way you like.