an = 1/(2n+1)^2
= 1/(4n^2 + 4n + 1)
令bn = 1/(4n^2 + 4n) 则 an < bn
bn = 1/4 * 1/n * 1/(n+1)
= 1/4 * [1/n - 1/(n+1)]
设数列bn前n项和为Tn，
则Tn = b1 + b2 + ... + bn
= 1/4 * (1 - 1/2) + 1/4 * (1/2 - 1/3) + ... 1/4 * [1/n - 1/(n+1)]
= 1/4 * [1 - 1/(n+1)]
< 1／4
由于an < bn，所以Sn < Tn < 1/4
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