Prove that the series $\sum_{n=1}^{\infty} x_n$ diverges in a Hilbert space $~X$

Question

Let $~X~$ be a Hilbert space and $~(e_n)_n~$ be an orthonormal basis for $~X.~$ Let us consider a sequence $~~(x_n)_n~~$ in $~~X~~$ given by $$x_n=e_n+\frac{e_n}{n+1},~~~~\text{ for all }~n \leq 1.$$ We need to prove that the series $~~\displaystyle \sum_{n=1}^{\infty} x_n~~$ diverges.
Now note that, $$\left\|\sum_{n=1}^{\infty} x_n\right\|\le \sum_{n \ge 1} \|x_n\| =\sum_{n \ge 1}\sqrt{1+\frac{2}{n+1}+\frac{1}{(n+1)^2}} \to \infty.$$ This proves that the given series $~~\displaystyle \sum_{n=1}^{\infty} x_n~~$ diverges.
Is my attempt is correct or anything I did wrong?
Thanks for your time.

Answer 1

Knowing that $\left\|\sum x_n\right\|$ is bounded above by infinity doesn’t show that it diverges. All real series are bounded above by $+\infty$ and below by $-\infty.$

A correct approach

$$x_n=\frac{n+2}{n+1}e_n$$ so $$\|x_n\|=\frac{n+2}{n+1}>1.$$

So if $S_n=\sum_{i=1}^n x_i,$ you have $$\|S_{n+1}-S_n\|=\|x_n\|>1$$

so $S_n$ is not Cauchy, and hence not convergent.

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Basically, like real and complex numbers and finite-dimensional vector spaces, if $\sum x_i$ converges, then $x_i\to 0.$

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If you really want to compute that $\left\|\sum_{n=1}^N x_i\right\|,$ you can, because $x_i\cdot x_j=0$ for $i\neq j.$

Then since $\|x_n\|>1$ for all $n,$ we have: $$\left\|\sum_{n=1}^N x_n\right\|^2=\sum_{n=1}^N \|x_n\|^2>N$$

So: $$\left\|\sum_{n=1}^Nx_n\right\|> \sqrt N.$$

This implies $\sum x_n$ diverges.

Answer 2

Your mistake has been pointed out in the comments. You cannot write an inequality for $\left\|\sum_{n=1}^{\infty} x_n\right\|$ when you dont even know that the infiníte sum makes sense.

$\|x_n\|=1+\frac 1 {n+1} \to 1$. $\sum x_n$ cannot converge unless $x_n \to 0$. This proves that the series diverges.