putnam_like_set3_b4. There is a mistake that does not impact the answer and the idea of this solution. Last step in grading_scheme.md should be as follows.
\bs[3]
Let $N$ be an integer in $[N_n,N_{n+1}]$ i.e. $N=N_n+m$. Then
[
b_N=\frac{b_{N_n}\cdot N_n^\alpha +1+2+\ldots+m}{(N_n+m)^{\alpha}}
]
and we estimate
[
b_{N_n}\cdot\frac{N_n^{\alpha}}{N_{n+1}^{\alpha}}<b_N<b_{N_{n+1}}\cdot\frac{N_{n+1}^{\alpha}}{N_{n}^{\alpha}}.
]
Since $\lim_{n\to\infty} \frac{N_n}{N_{n+1}}=1$, by the squeeze theorem we get $\lim_{N\to\infty}b_N=\beta$.
\es
putnam_like_set3_b4. There is a mistake that does not impact the answer and the idea of this solution. Last step in grading_scheme.md should be as follows.
\bs[3]$N$ be an integer in $[N_n,N_{n+1}]$ i.e. $N=N_n+m$ . Then$\lim_{n\to\infty} \frac{N_n}{N_{n+1}}=1$ , by the squeeze theorem we get $\lim_{N\to\infty}b_N=\beta$ .
Let
[
b_N=\frac{b_{N_n}\cdot N_n^\alpha +1+2+\ldots+m}{(N_n+m)^{\alpha}}
]
and we estimate
[
b_{N_n}\cdot\frac{N_n^{\alpha}}{N_{n+1}^{\alpha}}<b_N<b_{N_{n+1}}\cdot\frac{N_{n+1}^{\alpha}}{N_{n}^{\alpha}}.
]
Since
\es