Define a sequence $a_1=0,\ \frac{1}{1-a_{n+1}}-\frac{1}{1-a_n}=2n+1\ (n=1,\ 2,\ 3,\ \cdots)$.

(1) Find $a_n$.

(2) Let ${b_k=\sqrt{\frac{k+1}{k}}\ (1-\sqrt{a_{k+1}}})$ for $k=1,\ 2,\ 3,\ \cdots$.

Prove that $\sum_{k=1}^n b_k<\sqrt{2}-1$ for each $n$.

Last Edited
### Part (1): Finding \(a_n\)

Given the recurrence relation:
\[
\frac{1}{1-a_{n+1}} - \frac{1}{1-a_n} = 2n + 1 \quad (n = 1, 2, 3, \ldots)
\]

We start by summing both sides from \(n = 1\) to \(n = m\):
\[
\sum_{n=1}^m \left( \frac{1}{1-a_{n+1}} - \frac{1}{1-a_n} \right) = \sum_{n=1}^m (2n + 1)
\]

The left-hand side is a telescoping series:
\[
\frac{1}{1-a_{m+1}} - \frac{1}{1-a_1}
\]

Given \(a_1 = 0\), we have:
\[
\frac{1}{1-a_1} = \frac{1}{1-0} = 1
\]

Thus, the left-hand side becomes:
\[
\frac{1}{1-a_{m+1}} - 1
\]

The right-hand side is the sum of the first \(m\) odd numbers:
\[
\sum_{n=1}^m (2n + 1) = 1 + 3 + 5 + \cdots + (2m + 1)
\]

The sum of the first \(m\) odd numbers is:
\[
\sum_{n=1}^m (2n + 1) = m^2 + m
\]

Equating both sides, we get:
\[
\frac{1}{1-a_{m+1}} - 1 = m^2 + m
\]

Solving for \(\frac{1}{1-a_{m+1}}\):
\[
\frac{1}{1-a_{m+1}} = m^2 + m + 1
\]

Thus:
\[
1 - a_{m+1} = \frac{1}{m^2 + m + 1}
\]

So:
\[
a_{m+1} = 1 - \frac{1}{m^2 + m + 1}
\]

To find \(a_n\), we replace \(m+1\) with \(n\):
\[
a_n = 1 - \frac{1}{(n-1)^2 + (n-1) + 1} = 1 - \frac{1}{n^2}
\]

### Part (2): Proving the Inequality

Given:
\[
b_k = \sqrt{\frac{k+1}{k}} \left(1 - \sqrt{a_{k+1}}\right)
\]

From part (1), we have:
\[
a_{k+1} = 1 - \frac{1}{(k+1)^2}
\]

Thus:
\[
\sqrt{a_{k+1}} = \sqrt{1 - \frac{1}{(k+1)^2}} = \sqrt{\frac{(k+1)^2 - 1}{(k+1)^2}} = \sqrt{\frac{k(k+2)}{(k+1)^2}} = \frac{\sqrt{k(k+2)}}{k+1}
\]

So:
\[
1 - \sqrt{a_{k+1}} = 1 - \frac{\sqrt{k(k+2)}}{k+1}
\]

Therefore:
\[
b_k = \sqrt{\frac{k+1}{k}} \left(1 - \frac{\sqrt{k(k+2)}}{k+1}\right) = \sqrt{\frac{k+1}{k}} - \sqrt{\frac{k+2}{k+1}}
\]

Summing \(b_k\) from \(k=1\) to \(k=n\):
\[
\sum_{k=1}^n b_k = \sum_{k=1}^n \left( \sqrt{\frac{k+1}{k}} - \sqrt{\frac{k+2}{k+1}} \right)
\]

This is a telescoping series, so most terms cancel out:
\[
\sum_{k=1}^n b_k = \sqrt{\frac{2}{1}} - \sqrt{\frac{n+2}{n+1}} = \sqrt{2} - \sqrt{\frac{n+2}{n+1}}
\]

Since \(\sqrt{\frac{n+2}{n+1}} < 1\) for all \(n\):
\[
\sqrt{2} - \sqrt{\frac{n+2}{n+1}} < \sqrt{2} - 1
\]

Thus:
\[
\sum_{k=1}^n b_k < \sqrt{2} - 1
\]

\(\blacksquare\)

The final answer is \( \boxed{ a_n = 1 - \frac{1}{n^2} } \)