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Question Number 3546 by prakash jain last updated on 15/Dec/15

$$\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{p_i},p_{i}is{i}^{th}primediverges.$$$$\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{s_i},s_{i}is{i}^{th}wholesquareconverges.$$$$\mathrm{Why}?$$$$\mathrm{Even}\mathrm{though}\mathrm{it}\mathrm{appears}\mathrm{that}\mathrm{we}\mathrm{have}\mathrm{more}$$$$\mathrm{whole}\mathrm{squares}\mathrm{than}\mathrm{primes}.$$

Commented by Yozzii last updated on 15/Dec/15

$$Yes.$$

Commented by prakash jain last updated on 15/Dec/15

$$\mathrm{Thanks}.\mathrm{It}\mathrm{answers}\mathrm{the}\mathrm{question}\mathrm{that}$$$$\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{inverse}\mathrm{of}\mathrm{prime}\mathrm{number}\mathrm{can}$$$$\mathrm{be}\mathrm{divergent}\mathrm{if}\frac{1}{n^2}\mathrm{is}\mathrm{convergent}.$$

Commented by Filup last updated on 15/Dec/15

$$\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{s_i}=\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\sqrt{s_i}}\frac{1}{\sqrt{s_i}}\sqrt{s_i}\in \mathbb{Z}$$$$p_{i}hasnofactors$$

Commented by Yozzii last updated on 15/Dec/15

$$\frac{1}{3}>\frac{1}{9}\frac{1}{7}>\frac{1}{25}$$$$\frac{1}{2}>\frac{1}{4}\frac{1}{5}>\frac{1}{16}$$$$\therefore \frac{1}{p_i}>\frac{1}{s_i}\mathrm{\forall }i\in \mathbb{N}$$$$\therefore \underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{p_i}>\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{s_i}-\frac{1}{1}$$$$1+\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{p_i}>\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{s_i}$$$$s_i=i^{2}i\in {\mathbb{Z}}^{+}$$$$\therefore \underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{s_i}=\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{i^2}.pserieswithp=2.$$$$Sincep>1,theseriesisconvergent.$$

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