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Let M = 10^10, and let J be the number of primes < M, i.e., J = pi(M) = 455052511; then prime(J+1) = 10000000019.
Since prime(J+1) > M+2 and prime(k+1) - prime(k) >= 2 for all k > 1, it follows that, for all k > J,
prime(k) > M + 2*(k - J)
and thus
k/prime(k)^4 < k/(M + 2*(k - J))^4
so
Sum_{k>J} k/prime(k)^4 < Sum_{k>J} k/(M + 2*(k - J))^4
and it can be shown that the sum on the right-hand side is a value < 5*10^-22.
Summing the values of k/prime(k)^4 for all k <= J to obtain
Sum_{k=1..J} k/prime(k)^4 = 0.0944418581965049421841...
yields a lower bound on the infinite sum, and since the infinite sum is
Sum_{k>=1} k/prime(k)^4 = Sum_{k=1..J} k/prime(k)^4 + Sum_{k>J} k/prime(k)^4,
it must be less than
Sum_{k=1..J} k/prime(k)^4 + Sum_{k>J} k/(M + 2*(k - J))^4,
which is less than
0.0944418581965049421842 + 5*10^-22 = 0.0944418581965049421847,
which thus provides an upper bound on the infinite sum. (End)
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