# i-1-1-p-i-p-i-is-i-th-prime-diverges-i-1-1-s-i-s-i-is-i-th-whole-square-converges-Why-Even-though-it-appears-that-we-have-more-whole-squares-than-primes-

Question Number 3546 by prakash jain last updated on 15/Dec/15
$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{p}_{{i}} },\:{p}_{{i}} \:{is}\:{i}^{{th}} \:{prime}\:{diverges}. \\$$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{s}_{{i}} },\:{s}_{{i}} \:{is}\:{i}^{{th}} \:{whole}\:{square}\:{converges}. \\$$$$\mathrm{Why}? \\$$$$\mathrm{Even}\:\mathrm{though}\:\mathrm{it}\:\mathrm{appears}\:\mathrm{that}\:\mathrm{we}\:\mathrm{have}\:\mathrm{more} \\$$$$\mathrm{whole}\:\mathrm{squares}\:\mathrm{than}\:\mathrm{primes}. \\$$
Commented by Yozzii last updated on 15/Dec/15
$${Yes}. \\$$
Commented by prakash jain last updated on 15/Dec/15
$$\mathrm{Thanks}.\:\mathrm{It}\:\mathrm{answers}\:\mathrm{the}\:\mathrm{question}\:\mathrm{that} \\$$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{inverse}\:\mathrm{of}\:\mathrm{prime}\:\mathrm{number}\:\mathrm{can} \\$$$$\mathrm{be}\:\mathrm{divergent}\:\mathrm{if}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{convergent}. \\$$
Commented by Filup last updated on 15/Dec/15
$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{s}_{{i}} }=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{s}_{{i}} }}\:\frac{\mathrm{1}}{\:\sqrt{{s}_{{i}} }}\:\:\:\:\:\:\:\:\:\:\:\sqrt{{s}_{{i}} }\in\mathbb{Z} \\$$$${p}_{{i}} \:{has}\:{no}\:{factors} \\$$
Commented by Yozzii last updated on 15/Dec/15
$$\frac{\mathrm{1}}{\mathrm{3}}>\frac{\mathrm{1}}{\mathrm{9}}\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{7}}>\frac{\mathrm{1}}{\mathrm{25}} \\$$$$\frac{\mathrm{1}}{\mathrm{2}}>\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{5}}>\frac{\mathrm{1}}{\mathrm{16}} \\$$$$\therefore\:\frac{\mathrm{1}}{{p}_{{i}} }>\frac{\mathrm{1}}{{s}_{{i}} }\:\forall{i}\in\mathbb{N} \\$$$$\therefore\:\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{p}_{{i}} }>\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{s}_{{i}} }−\frac{\mathrm{1}}{\mathrm{1}} \\$$$$\mathrm{1}+\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{p}_{{i}} }>\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{s}_{{i}} } \\$$$${s}_{{i}} ={i}^{\mathrm{2}} \:\:\:{i}\in\mathbb{Z}^{+} \: \\$$$$\therefore\:\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{s}_{{i}} }=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{i}^{\mathrm{2}} }.\:{p}\:{series}\:{with}\:{p}=\mathrm{2}. \\$$$${Since}\:{p}>\mathrm{1},\:{the}\:{series}\:{is}\:{convergent}. \\$$