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The problem:

I've hit what might be a dead-end. If it is true, I would like to show that for $c \in (0,1)$ and $1 \leq k \leq J$, the sum $$ \sum_{j=k}^J jc^{1-j-k} = \sum_{n=1}^{J+1-k} (k-n-1) c^{n} $$ has an upper bound that is independent of $k$. I tried the lazy approach of letting $J \to \infty$, but that doesn't seem to work. Any input here would be appreciated.

The context: This is for a PDE course where we are dabbling in functional analysis. I am trying to show that the spectrum of the right-shift operator on $\ell^2$ is $\{\lambda:|\lambda| \leq 1\}$, and this summation comes from the $|\lambda|>1$ case. We have not established the notion of an operator norm so I would prefer to avoid using it, though I am well aware it makes quick work of this part of the problem.

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  • $\begingroup$ Is $J > k{}{}$? Is $k > 0$? $\endgroup$ – Antonio Vargas Apr 16 '14 at 1:09
  • $\begingroup$ @AntonioVargas yes, $1 \leq k \leq J$ $\endgroup$ – Ben Grossmann Apr 16 '14 at 1:15
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If $0 < c < 1$ then $1/c > 1$, so if $1 \leq k \leq J$ we have

$$ \begin{align} \sum_{j=k}^J jc^{k - j - 1} &= \sum_{j=k}^J j \left(\frac{1}{c}\right)^{1+j-k} \\ &\leq \sum_{j=k}^J j \left(\frac{1}{c}\right)^{1+j-1} \\ &\leq \sum_{j=1}^J j \left(\frac{1}{c}\right)^{j}. \end{align} $$

The first inequality holds holds because the map $x \mapsto \left(\frac{1}{c}\right)^x$ is increasing.

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  • $\begingroup$ Thank you for your answer! Unfortunately, this doesn't happen to suit my needs since I need my upper bound to be finite as $J \to \infty$, sorry for not having that in the question statement. $\endgroup$ – Ben Grossmann Apr 16 '14 at 4:12
  • $\begingroup$ @Omnomnomnom, your quantity doesn't have a finite upper bound that holds for all $J,k$ with $1 \leq k \leq J$. The inequalities in my argument are tantamount to taking $k=1$. $\endgroup$ – Antonio Vargas Apr 16 '14 at 4:14
  • $\begingroup$ My apologies! The sign of the exponent was wrong. This certainly should change things $\endgroup$ – Ben Grossmann Apr 16 '14 at 4:20
  • $\begingroup$ Now take $k = J$ in the new version; it's still unbounded as $J \to \infty$. $\endgroup$ – Antonio Vargas Apr 16 '14 at 4:26
  • $\begingroup$ Well, that takes care of that then. Thank you. $\endgroup$ – Ben Grossmann Apr 16 '14 at 4:35

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