>>18
∫_{x=an+1 to bn +1} (1/x) dx
< S[a](a,b)
< ∫_{x=an to bn} (1/x) dx

という不等式で、積分区間をあわせて

∫_{x=an to bn} (1/(x+1)) dx
< S[a](a,b)
< ∫_{x=an to bn} (1/x) dx

を示せということだけどこれは
∫_{x=k to k+1} (1/(x+1)) dx
< 1/(k+1)
< ∫_{x=k to k+1} (1/x) dx

∫_{x=k to k+1} (1/(x+1)) dx
< ∫_{x=k to k+1} (1/(k+1)) dx
< ∫_{x=k to k+1} (1/x) dx

から出てくる