Assume that $(a_k)$ is a sequence of reals such that $$ \lim_{n\to \infty}\sum_{i=0}^n a_i\cos(n\pi b_i)\ =\ 0 $$ for some sequence $(b_k)$ with $1=b_0>b_1>b_2>\dots>0$.
Does it imply that all $a_k$'s are equal 0 ?
limit of a sum with cosines
0
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sequences-and-series
convergence
1 Answers
1
Trying again . . .
Let $b_0=1$, and let $b_1,b_2,...$ be any sequence of irrational numbers such that
$\qquad\qquad b_0>b_1>b_2>\dots>0$
Since $b_k$ is irrational when $k>0$, it follows that for all integers $k,n > 0$,
$\qquad\qquad\cos(n\pi b_k) \ne 0$.
Define the sequence $\,a_0,a_1,a_2,\ldots\;$ recursively by
$\qquad\qquad a_0 = 0$
$\qquad\qquad$for $n > 0$:
$\qquad\qquad\;\;\;\bullet\;\,$ If $\,\displaystyle{\sum_{k=0}^{n-1} a_k\cos(n\pi b_k)} \ne 0$, $\;\displaystyle{ a_n =-\frac{\sum_{k=0}^{n-1} a_k\cos(n\pi b_k)}{\cos(n\pi b_n)}}$
$\qquad\qquad\;\;\;\bullet\;\,$ If $\,\displaystyle{\sum_{k=0}^{n-1} a_k\cos(n\pi b_k)} = 0$, $\;\displaystyle{ a_n =\frac{1}{n\cos(n\pi b_n)}}$
The definition of $a_n$, as given above, guarantees that for $n > 0$,
$\qquad\qquad\displaystyle{\sum_{k=0}^{n} a_k\cos(n\pi b_k)} =\, 0\, \text{ or }\,1/n$
hence
$\qquad\qquad\displaystyle{\lim_{n\to \infty}\sum_{k=0}^n a_k\cos(n\pi b_k)=0}$
But the definition of $a_n$ also guarantees that
$\qquad\qquad\ a_n \ne 0,\,$ for all $n > 0$
so we have a counterexample.