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Weyl's Criterion states that $a_1,a_2,\dots,$ will be equidistributed modulo $1$ if an only if $$\lim_{N\rightarrow \infty}\frac{1}{N}\sum_{n\leq N} e^{2\pi i a_n m}=0$$ for every $m$. While this is not a quantitative result, it suggests that we should examine the Fourier coefficients $$\sum_{n\leq N}e^{2\pi i m \sqrt{n} }.$$ The Erdos-Turan inequality ...


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\begin{align*} \sum\limits_{n=4}^{\infty} \frac{2^n + 8^n}{10^n} &= \sum\limits_{n=4}^{\infty } \frac{2^n}{10^n}+\sum\limits_{n=4}^{\infty } \frac{8^n}{10^n} \\ &=\sum\limits_{n=4}^{\infty } \left(\frac{1}{5}\right)^n+\sum\limits_{n=4}^{\infty } \left(\frac{4}{5}\right)^n \\ ...


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$$\sum\limits_{n=4}^{\infty } \frac{2^n + 8^n}{10^n} \\ = \sum\limits_{n=4}^{\infty }\left( \frac{1}{5^n}+\frac{4^n}{5^n}\right)\\=\frac{1}{1-\frac{1}{5}}-1-\frac{1}{5}-\frac{1}{25}-\frac{1}{125}+\frac{1}{1-\frac{4}{5}}-1-\frac{4}{5}-\frac{16}{25}-\frac{64}{125}\\=0.002+2.048\\=2.05$$


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Hint: you can write $\sum\limits_{n=4}^{n= \infty } \frac{2^n + 8^n}{10^n} = \sum\limits_{n=4}^{n= \infty } \frac{2^n }{10^n} + \sum\limits_{n=4}^{n= \infty } \frac{8^n }{10^n} $ because the two series on the right hand side converge. And then you have two geometric series, which you can probably solve.


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Hint: Let $x=\ln(t)$ and $\ln(b_i)$ be non-negative integers, then $$f(\ln(t))=a_1t^{\ln(b_1)}+a_2t^{\ln(b_2)}\cdots+a_Nt^{\ln(b_N)}$$ is a polynomial in $t$. You can easily construct a polynomial of $N$ terms having $N-1$ non-negative distinct roots. This establishes a lower bound on the maximum number of roots.



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