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Please show that, given that $\sum_{k\ge1}c_k^2=\infty$ and $c_k\rightarrow 0$, $$\lim_{n\rightarrow\infty}\prod_{k=1}^n\cos{tc_k}=0$$ for every $t\neq0$. (All variables here are real numbers.)

The above is a key step in the proof of the following result: If $\sum_{k\ge1}c_k^2=\infty$ then the sequence with random signs $\sum_{k\ge1}\pm c_k $ diverges with probability 1. (See p.31 of Marc Kac's lovely book Statistical Independence in Probability Analysis and Number Theory for details.) I am also wondering if one could prove the result without exploiting an infinite cosine product. Any hints or pointers are welcome. Thank you in advance!

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Note that $\cos x=1-\frac 12x^2+O(x^4)$ and hence $\ln \cos x =-\frac 12x^2+O(x^4)$ near $x=0$. Hence for $|x|$ small enough we have $\ln \cos x <-\frac 13 x^2$ (or $-ax^2$ for any specific $a$ with $0<a<\frac12$). Because $c_k\to 0$ we do have $|c_kt|$ small enough for almost all $k$, say for all $k>N$ (where $N$ depends on $t$). Then for $n>N$ $$\ln\prod_{k=1}^n\cos c_kt=\sum_{k=1}^n\ln\cos c_kt <\sum_{k=1}^N\left(\ln \cos c_kt+\frac{t^2}3c_k^2\right)-\frac{t^2}3\sum_{k=1}^nc_k^2.$$ As $\sum c_k^2$ diverges to $+\infty$, the log of the product diverges to $-\infty$, the original sequence of partial products itself therefore converges to $0$.

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