求教一道高数题

2025-05-09 15:32:22
推荐回答(1个)
回答1:

直接Taylor展开,取有限项:
sin(1/n) ~ 1/n - (1/n)^3/3!
ln(1-1/n) ~ -1/n - (1/n)^2/2
sin(1/n) - kln(1-1/n) = (1/n)(1+k) - (1/n)^3/3! + k[1/n + (1/n)^2/2]
因为调和级数{1/n}发散,若要此级数收敛,则 1+k = 0。得, k = -1