A different type of converging sum

Calculus Level 5

p P { 1 } p 1 \large \sum_{p \ \in \ P - \{1\} } p^{-1}

Let P P be the set of perfect powers. Evaluate the sum above to 2 decimal places. If you arrive at the conclusion that the sum diverges, enter your answer as 0.

Details:

  • A positive integer n n is a perfect power if it can be represented as m k m^k for some integers m > 0 , k > 1 m > 0, k > 1 .
  • If a positive integer can be represented in multiple such ways, it still contributes to the sum only once.

For more, click here .


The answer is 0.87.

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1 solution

Richard Xu
Mar 31, 2018

A lower bound is

S = ζ ( 2 ) + ζ ( 3 ) + ζ ( 5 ) + ζ ( 7 ) ζ ( 6 ) ζ ( 10 ) ζ ( 14 ) ζ ( 15 ) ζ ( 21 ) ζ ( 35 ) + ζ ( 30 ) + ζ ( 42 ) + ζ ( 70 ) + ζ ( 105 ) ζ ( 210 ) 1 0.87383805 \underline{S}= \zeta(2)+\zeta(3)+\zeta(5)+\zeta(7)-\zeta(6)-\zeta(10)-\zeta(14)-\zeta(15)-\zeta(21)-\zeta(35)+\zeta(30)+\zeta(42)+\zeta(70)+\zeta(105)-\zeta(210)-1 \approx 0.87383805

while an upper bound is

S ˉ = 1 1 3 × 4 1 7 × 8 1 8 × 9 1 15 × 16 1 24 × 25 1 26 × 27 1 32 × 32 1 35 × 36 1 48 × 49 1 63 × 64 1 80 × 81 1 99 × 100 0.87483257 \bar{S} = 1 - \frac{1}{3\times4} - \frac{1}{7\times8}-\frac{1}{8\times9}-\frac{1}{15\times16}-\frac{1}{24\times25}-\frac{1}{26\times27}-\frac{1}{32\times32}-\frac{1}{35\times36}-\frac{1}{48\times49}-\frac{1}{63\times64}-\frac{1}{80\times81}-\frac{1}{99\times100}\approx 0.87483257

This is because

k = 2 1 n k = 1 n 2 1 1 n = 1 n ( n 1 ) = 1 n 1 1 n \sum_{k=2}^\infty \frac{1}{n^k} = \frac{\frac{1}{n^2}}{1-\frac{1}{n}} = \frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}

n = 2 [ 1 n 1 1 n ] = 1 \sum_{n=2}^\infty [\frac{1}{n-1}-\frac{1}{n}] = 1

while some of the terms are duplicates and can be removed.

Therefore, S 0.87 S\approx 0.87 up to 2 decimal places.

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