For large NNN, which series has the smallest tail sum ∑n=N∞an\sum_{n=N}^{\infty} a_n∑n=N∞an?
∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^2}∑n=1∞n21 has tail approximately 1N\frac{1}{N}N1
∑n=1∞12n\sum_{n=1}^{\infty} \frac{1}{2^n}∑n=1∞2n1 has tail approximately 12N−1\frac{1}{2^{N-1}}2N−11 (exponentially small)
∑n=1∞1nlnn\sum_{n=1}^{\infty} \frac{1}{n \ln n}∑n=1∞nlnn1 has tail approximately lnlnN\ln \ln NlnlnN
∑n=1∞1n3/4\sum_{n=1}^{\infty} \frac{1}{n^{3/4}}∑n=1∞n3/41 has tail approximately 1N1/4\frac{1}{N^{1/4}}N1/41