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Let \begin{align} a_n= cos \left( \frac{\pi}{2} n\right)n^{- \frac{1}{2}n +\frac{1}{2} } \end{align}

and let \begin{align} c_{2n}= \sum_{m=0}^{2n} a_m a_{2n-m} \end{align}

Note that the above is an operation of discrete convolution

Question 1 : Can we establish that \begin{align} \lim_{n \to \infty } c_{2n}=0 \end{align}

Question 2: If the about is true can we also establish how fast does $c_{2n}$ go to zero by showing bounds on $|c_{2n}|$?

Thanks you.

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Using a Cauchy product :

$$\sum c_n = \left(\sum a_n\right)^2 = \left(\sum \frac{i^n +(-i)^n}{2} n^{- \frac{1}{2}n +\frac{1}{2} }\right)^2$$

which converges, thus $ c_n \longrightarrow 0$ .

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  • $\begingroup$ Thank you. Very nice proof. Two questions: How did you get a $(-1)^n$ inside the last sum? Also, any idea how to bound $|c_n|??? $\endgroup$ – Boby Jan 4 '17 at 15:09
  • $\begingroup$ Sorry there was a mistake, it's $\cos\left(n\frac{\pi}{2}\right) = \frac{i^n +(-i)^n}{2}$. For the bounding, let me think of it. $\endgroup$ – Blencer Jan 4 '17 at 15:17
  • $\begingroup$ Thanks. Let me know if you have something. $\endgroup$ – Boby Jan 4 '17 at 15:28

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