lim(n->∞)an²/an
=lim(n->∞)an
=0
所以
∑an²是弱级数，强级数∑an收敛，其必收敛。
an+1/n²≥2√an· 1/n
∑an收敛
∑1/n²收敛
所以
∑√an/n 收敛。
数列下极限 > 0,则自N项后各项均大于某正数c.
可得a_N·u_N > a_(N+1)·u_(N+1)+c·u_(N+1)
a_(N+1)·u_(N+1) > a_(N+2)·u_(N+2)+c·u_(N+2)
a_(N+2)·u_(N+2) > a_(N+3)·u_(N+3)+c·u_(N+3)
...
相加得a_N·u_N > a_M·u_M+c(u_(N+1)+u_(N+2)+u_(N+3)+...+u_M).
又∵a_M,u_M > 0,c > 0,∴u_(N+1)+u_(N+2)+u_(N+3)+...+u_M < a_N·u_N/c.
正项级数∑{1≤n≤∞} u_n ≤ a_N·u_N/c+∑{1≤n≤N} u_n有界,故收敛.
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