Let $p_1=2 Let $$S_n=\sum_{k_1=0}^{\infty}\sum_{k_2=0}^{\infty}\dots \sum_{k_n=0}^{\infty}\frac{1}{p_1^{k_1}p_2^{k_2}\dots p_n^{k_n}}$$ It is easy to check that $S_1=2,S_3=3$ and in general $S_n=\frac{1}{1-\frac{1}{p_1}}\frac{1}{1-\frac{1}{p_2}}\dots \frac{1}{1-\frac{1}{p_n}}$ Then if we can show that as $n\to\infty$, $S_n$ diverges to infinity we can say $\sum_{n=1}^{\infty} \frac{1}{n}$ also diverges to infinity. If we can show that $S_n\thicksim f(n)$, where $f(n)$ increases to infinity as $n\to\infty$, then we are done. But I have no idea about the following problem: $$T_n=\sum_{k_1\neq k_2\neq \dots \neq k_n}^{\infty}\frac{1}{p_1^{k_1}p_2^{k_2}\dots p_n^{k_n}}$$ Here $k_1\neq k_2\neq \dots\neq k_n$ means they are distinct($n!$ values).What is this $f(n)$?
Is there any other way to show $\sum_{k=1}^{\infty}\frac{1}{k}$ diverges using $S_n$?
How should we proceed to in this case?
Monday 23 June 2014
sequences and series - divergence of $sum_{k_1neq k_2neq dots neq k_n}^{infty}frac{1}{p_1^{k_1}p_2^{k_2}dots p_n^{k_n}}$
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