We prove Theorem 1.1 by induction on *n*. It is easy to check that Theorem 1.1 holds for 4 ≤ *n* ≤ 12. Thus, we always assume that *n* ≥ 13 in the following proof. Suppose that Theorem 1.1 holds for *n* = *m*, namely,

$$\begin{array}{}{\displaystyle \frac{{B}_{m}(i)}{{B}_{m}(i+1)}<\frac{{B}_{m}(i+1)}{{B}_{m}(i+2)},\phantom{\rule{1em}{0ex}}1\le i\le {m}^{2}-2.}\end{array}$$(20)

We proceed to show that Theorem 1.1 holds for *n* = *m* + 1, that is,

$$\begin{array}{}{\displaystyle \frac{{B}_{m+1}(k)}{{B}_{m+1}(k+1)}<\frac{{B}_{m+1}(k+1)}{{B}_{m+1}(k+2)},\phantom{\rule{1em}{0ex}}1\le k\le (m+1{)}^{2}-2.}\end{array}$$(21)

Employing (11), (15) and (20), we see that (21) holds for 1 ≤ *k* ≤ 2*m*. It follows from (11), (16) and (20) that (21) is true for the case *k* = 2*m* + 1. In view of (11), (17) and (20), we find that (21) holds for 2*m* + 2 ≤ *k* ≤ *m*^{2} – 2. From (11), (18) and (20), we deduce that (21) is true for the case *k* = *m*^{2} – 1. By (11), (19) and (20), we can prove that (21) holds for *m*^{2} ≤ *k* ≤ *m*^{2} + *m* – 3 and *m*^{2} + *m* + 1 ≤ *k* ≤ (*m* + 1)^{2} – 2.

Now, special attentions should be paid to three cases *k* = *m*^{2} + *m* – 2, *k* = *m*^{2} + *m* – 1 and *k* = *m*^{2} + *m*.

By (12) and (13), it is easy to check that for *m* ≥ 4,

$$\begin{array}{}{\displaystyle}& \frac{{B}_{m}({m}^{2}-2)+{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})-m-3}{{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})+1}-\frac{{B}_{m}({m}^{2}-3)}{{B}_{m}({m}^{2}-2)}\\ & \phantom{\rule{2em}{0ex}}=\frac{{m}^{4}-13{m}^{2}+72m-132}{6(m-3)({m}^{2}-m+2)}>0.\end{array}$$(22)

In view of (20) and (22),

$$\begin{array}{}{\displaystyle \frac{{B}_{m}({m}^{2}-m-3)}{{B}_{m}({m}^{2}-m-2)}<\frac{{B}_{m}({m}^{2}-3)}{{B}_{m}({m}^{2}-2)}<\frac{{B}_{m}({m}^{2}-2)+{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})-m-3}{{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})+1}.}\end{array}$$(23)

From (11), it is easy to prove that for *m* ≥ 4,

$$\begin{array}{}{\displaystyle {B}_{m}({m}^{2}-m-2)>{B}_{m}({m}^{2}-m-1)>\cdots >{B}_{m}({m}^{2}-1)>{B}_{m}({m}^{2}).}\end{array}$$(24)

Thanks to (23) and (24),

$$\begin{array}{}{\displaystyle}& {B}_{m}({m}^{2}-m-3)({B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})+(-1{)}^{m+1})+(-1{)}^{m+1}\sum _{i=0}^{m+2}{B}_{m}({m}^{2}-i)\\ & \phantom{\rule{1em}{0ex}}<{B}_{m}({m}^{2}-m-3)({B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})+1)+(m+3){B}_{m}({m}^{2}-m-2)\\ & \phantom{\rule{1em}{0ex}}<{B}_{m}({m}^{2}-m-2)({B}_{m}({m}^{2}-2)+{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})).\end{array}$$(25)

By (20),

$$\begin{array}{}{\displaystyle {B}_{m}({m}^{2}-m-3)\sum _{i=2}^{m+1}{B}_{m}({m}^{2}-i)<{B}_{m}({m}^{2}-m-2)\sum _{i=3}^{m+2}{B}_{m}({m}^{2}-i).}\end{array}$$(26)

Combining (25) and (26) yields

$$\begin{array}{}{\displaystyle \frac{\sum _{i={m}^{2}-m-3}^{{m}^{2}}{B}_{m}(i)}{\sum _{i={m}^{2}-m-2}^{{m}^{2}}{B}_{m}(i)}<\frac{\sum _{i={m}^{2}-m-2}^{{m}^{2}}{B}_{m}(i)}{\sum _{i={m}^{2}-m-1}^{{m}^{2}}{B}_{m}(i)+(-1{)}^{m+1}},}\end{array}$$(27)

which can be rewritten as

$$\begin{array}{}{\displaystyle \frac{{B}_{m+1}({m}^{2}+m-2)}{{B}_{m+1}({m}^{2}+m-1)}<\frac{{B}_{m+1}({m}^{2}+m-1)}{{B}_{m+1}({m}^{2}+m)}.}\end{array}$$(28)

Therefore, (21) holds for the case *k* = *m*^{2} + *m* – 2.

Based on (12) and (13), we deduce that for *m* ≥ 13,

$$\begin{array}{}{\displaystyle}& \frac{{B}_{m}({m}^{2}-2)+{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})-2(m+2)}{{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})}-\frac{{B}_{m}({m}^{2}-3)}{{B}_{m}({m}^{2}-2)}\\ & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}=\frac{{m}^{4}-12{m}^{3}-13{m}^{2}+36m-36}{6m({m}^{2}-m+2)}>0.\end{array}$$(29)

By (20) and (29),

$$\begin{array}{}{\displaystyle \frac{{B}_{m}({m}^{2}-m-2)}{{B}_{m}({m}^{2}-m-1)}<\frac{{B}_{m}({m}^{2}-3)}{{B}_{m}({m}^{2}-2)}<\frac{{B}_{m}({m}^{2}-2)+{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})-2(m+2)}{{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})}.}\end{array}$$(30)

It follows from (24) and (30) that

$$\begin{array}{}{\displaystyle}& {B}_{m}({m}^{2}-m-2)({B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2}))\\ & \phantom{\rule{2em}{0ex}}<{B}_{m}({m}^{2}-m-1)({B}_{m}({m}^{2}-2)+{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})-2(m+2))\\ & \phantom{\rule{2em}{0ex}}<{B}_{m}({m}^{2}-m-1)({B}_{m}({m}^{2}-2)+{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2}))-2\sum _{i=0}^{m+1}{B}_{m}({m}^{2}-i)\\ & \phantom{\rule{2em}{0ex}}\le {B}_{m}({m}^{2}-m-1)({B}_{m}({m}^{2}-2)+{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2}))\\ & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}+2\times (-1{)}^{m+1}\sum _{i=0}^{m+1}{B}_{m}({m}^{2}-i).\end{array}$$(31)

In view of (20),

$$\begin{array}{}{\displaystyle {B}_{m}({m}^{2}-m-2)\sum _{i=2}^{m}{B}_{m}({m}^{2}-i)<{B}_{m}({m}^{2}-m-1)\sum _{i=3}^{m+1}{B}_{m}({m}^{2}-i).}\end{array}$$(32)

It follows from (31) and (32) that

$$\begin{array}{}{\displaystyle \frac{\sum _{i={m}^{2}-m-2}^{{m}^{2}}{B}_{m}(i)}{\sum _{i={m}^{2}-m-1}^{{m}^{2}}{B}_{m}(i)+(-1{)}^{m+1}}<\frac{\sum _{i={m}^{2}-m-1}^{{m}^{2}}{B}_{m}(i)+(-1{)}^{m+1}}{\sum _{i={m}^{2}-m}^{{m}^{2}}{B}_{m}(i)}.}\end{array}$$(33)

By (11), we can rewrite (33) as follows

$$\begin{array}{}{\displaystyle \frac{{B}_{m+1}({m}^{2}+m-1)}{{B}_{m+1}({m}^{2}+m)}<\frac{{B}_{m+1}({m}^{2}+m)}{{B}_{m+1}({m}^{2}+m+1)},}\end{array}$$(34)

which implies that (21) holds for the case *k* = *m*^{2} + *m* – 1.

In view of (12) and (13), we see that for *m* ≥ 4,

$$\begin{array}{}{\displaystyle}& \frac{{B}_{m}({m}^{2}-2)+{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})-m}{{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})}-\frac{{B}_{m}({m}^{2}-3)}{{B}_{m}({m}^{2}-2)}\\ & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}=\frac{(m+1)({m}^{3}-7{m}^{2}+12m+12)}{6m({m}^{2}-m+2)}>0.\end{array}$$(35)

By (20) and (35), we find that for *m* ≥ 4,

$$\begin{array}{}{\displaystyle \frac{{B}_{m}({m}^{2}-m-1)}{{B}_{m}({m}^{2}-m)}<\frac{{B}_{m}({m}^{2}-3)}{{B}_{m}({m}^{2}-2)}<\frac{{B}_{m}({m}^{2}-2)+{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})-m}{{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})}.}\end{array}$$(36)

It follows from (24) and (36) that

$$\begin{array}{}{\displaystyle}& {B}_{m}({m}^{2}-m-1)({B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2}))+(-1{)}^{m+1}\sum _{i=0}^{m-1}{B}_{m}({m}^{2}-i)\\ & \phantom{\rule{2em}{0ex}}<{B}_{m}({m}^{2}-m-1)({B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2}))+m{B}_{m}({m}^{2}-m)\\ & \phantom{\rule{2em}{0ex}}\le {B}_{m}({m}^{2}-m)({B}_{m}({m}^{2}-2)+{B}_{m}({m}^{2}-1)+{B}_{m}({m}^{2})).\end{array}$$(37)

By (20),

$$\begin{array}{}{\displaystyle {B}_{m}({m}^{2}-m-1)\sum _{i=2}^{m-1}{B}_{m}({m}^{2}-i)<{B}_{m}({m}^{2}-m)\sum _{i=3}^{m}{B}_{m}({m}^{2}-i).}\end{array}$$(38)

In view of (37) and (38), we can prove that

$$\begin{array}{}{\displaystyle \frac{\sum _{i={m}^{2}-m-1}^{{m}^{2}}{B}_{m}(i)+(-1{)}^{m+1}}{\sum _{i={m}^{2}-m}^{{m}^{2}}{B}_{m}(i)}<\frac{\sum _{i={m}^{2}-m}^{{m}^{2}}{B}_{m}(i)}{\sum _{i={m}^{2}-m+1}^{{m}^{2}}{B}_{m}(i)}.}\end{array}$$(39)

By (11), we can rewrite (39) as follows

$$\begin{array}{}{\displaystyle \frac{{B}_{m+1}({m}^{2}+m)}{{B}_{m+1}({m}^{2}+m+1)}<\frac{{B}_{m+1}({m}^{2}+m+1)}{{B}_{m+1}({m}^{2}+m+2)},}\end{array}$$(40)

which implies that (21) holds for the case *k* = *m*^{2} + *m*. This completes the proof.

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