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 = 1}{\sum^{\infty}} \frac{1}{p_{i}}, p_{i} i s i^{t h} p r i m e d i v e r g e s .

\underset{i = 1}{\sum^{\infty}} \frac{1}{s_{i}}, s_{i} i s i^{t h} w h o l e s q u a r e c o n v e r g e s .

Why ?

Even though it appears that we have more

whole squares than primes .

Commented by Yozzii last updated on 15/Dec/15

Y e s .

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

Thanks . It answers the question that

the sum of inverse of prime number can

be divergent if \frac{1}{n^{2}} is convergent .

Commented by Filup last updated on 15/Dec/15

\underset{i = 1}{\sum^{\infty}} \frac{1}{s_{i}} = \underset{i = 1}{\sum^{\infty}} \frac{1}{\sqrt{s_{i}}} \frac{1}{\sqrt{s_{i}}} \sqrt{s_{i}} \in Z

p_{i} h a s n o f a c t o r s

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}

∴ \frac{1}{p_{i}} > \frac{1}{s_{i}} \forall i \in N

∴ \underset{i = 1}{\sum^{\infty}} \frac{1}{p_{i}} > \underset{i = 1}{\sum^{\infty}} \frac{1}{s_{i}} - \frac{1}{1}

1 + \underset{i = 1}{\sum^{\infty}} \frac{1}{p_{i}} > \underset{i = 1}{\sum^{\infty}} \frac{1}{s_{i}}

s_{i} = i^{2} i \in Z^{+}

∴ \underset{i = 1}{\sum^{\infty}} \frac{1}{s_{i}} = \underset{i = 1}{\sum^{\infty}} \frac{1}{i^{2}} . p s e r i e s w i t h p = 2 .

S i n c e p > 1, t h e s e r i e s i s c o n v e r g e n t .

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