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Question Number 184918 by Mastermind last updated on 14/Jan/23

$$\mathrm{Show}\mathrm{that}$$$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...\mathrm{is}$$$$\mathrm{not}\mathrm{convergent}$$$$$$$$\mathrm{Hi}$$

Commented by Frix last updated on 14/Jan/23

$$\frac{1}{5}\mathrm{is}\mathrm{missing}?$$

Commented by Mastermind last updated on 14/Jan/23

$$\mathrm{Yes},\mathrm{it}\mathrm{was}\mathrm{a}\mathrm{mistake}...\mathrm{it}\mathrm{has}\mathrm{been}$$$$\mathrm{corrected}\mathrm{tho}.$$

Answered by Frix last updated on 14/Jan/23

$$\underset{j=1}{\sum ^{\mathrm{\infty }}}\frac{1}{j}=\prod _{p\mathrm{is}\mathrm{prime}}\left(\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{p^k}\right)=\mathrm{\infty }$$$$\mathrm{The}\mathrm{sum}\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...\mathrm{is}\mathrm{the}$$$$\mathrm{same}\mathrm{as}\mathrm{the}\mathrm{product}\mathrm{of}\mathrm{these}\mathrm{sums}:$$$$(1+\frac{1}{2}+...+\frac{1}{2^n}+...)(1+\frac{1}{3}+...+\frac{1}{3^n}+...)(1+\frac{1}{5}+...+\frac{1}{5^n}+...)(1+\frac{1}{7}+...+\frac{1}{7^n}+...)...(1+\frac{1}{p}+...+\frac{1}{p^n}+...)...$$$$\mathrm{We}\mathrm{know}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{sums}\mathrm{is}>1\mathrm{and}\mathrm{we}$$$$\mathrm{know}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{primes}\mathrm{is}\mathrm{\infty }...$$

Answered by prakash jain last updated on 14/Jan/23

$$1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})$$$$+(\frac{1}{9}+\frac{1}{10}+....+\frac{1}{16})+...$$$$>1+\frac{1}{2}+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})+..$$$$=1+\frac{1}{2}\left(infinitetimes\right)$$$$\Rightarrow \mathrm{series}\mathrm{is}\mathrm{divergent}$$

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