Advanced Exercises

Approach: First notice that $k \times k! = (k+1)! - k!$, and use it to split the sum into cancelling terms so that it collapses to a difference of two end terms.

Steps：

Note that $k \times k! = (k+1)! - k!$

$S_n = \sum_{k=1}^{n} k \times k! = \sum_{k=1}^{n} [(k+1)! - k!]$

$= (2! - 1!) + (3! - 2!) + \cdots + ((n+1)! - n!)$

$= (n+1)! - 1$

Answer: $(n+1)! - 1$

Approach: Take reciprocals of the recurrence to turn it into a form that can be summed term by term, then cancel the terms to obtain the general term.

Steps：

Take reciprocals: $\frac{1}{a_{n+1}} = \frac{1 + a_n}{a_n} = \frac{1}{a_n} + 1$

Let $b_n = \frac{1}{a_n}$. Then $b_{n+1} = b_n + 1$ and $b_1 = 1$.

This is an arithmetic sequence, $b_n = n$

$a_n = \frac{1}{n}$

Answer: $a_n = \frac{1}{n}$

Approach: First use the angular symmetry of the sine function on $[0°, 90°]$, namely $\sin k° = \sin(180° - k°)$, to pair up the sine terms, then sum.

Steps：

Use symmetry: $\sin k° = \sin(180° - k°)$

$S_n = \sum_{k=1}^{180} \sin k° = \sum_{k=1}^{90} [\sin k° + \sin(180° - k°)] = 2\sum_{k=1}^{90} \sin k°$

But this is still complex. A simpler method:

Use $\sin 90° = 1$; the other terms cancel in pairs ($\sin k° + \sin(180°-k°) = 2\sin 90° \cos(90°-k°)$)

In fact, using complex numbers or trigonometric identities, one can prove:

$S_n = \frac{\sin(\frac{180°}{2}) \sin(\frac{181°}{2})}{\sin(\frac{1°}{2})} = \frac{1}{\sin 0.5°}$

Answer: about $\frac{1}{\sin 0.5°} \approx 114.6$

Approach: Use the squeeze theorem: sandwich the expression between two bounds with the same limit, which pins down the limit of the sequence.

Steps：

Squeeze theorem:

$\frac{n}{\sqrt{n^2+n}} < S_n < \frac{n}{\sqrt{n^2+1}}$

$\lim_{n \to \infty} \frac{n}{\sqrt{n^2+n}} = \lim_{n \to \infty} \frac{1}{\sqrt{1+\frac{1}{n}}} = 1$

$\lim_{n \to \infty} \frac{n}{\sqrt{n^2+1}} = \lim_{n \to \infty} \frac{1}{\sqrt{1+\frac{1}{n^2}}} = 1$

Answer: 1

Approach: Expand the product first, then use $\left(1 + \frac{1}{2^k}\right) = \frac{1 - 2^{-k}}{1 - 2^{-(k-1)}}$ to turn it into a form whose factors cancel.

Steps:

Proof:

Use $1 + x = \frac{1 - x^2}{1 - x}$ (when $x \neq 1$)

$\left(1 + \frac{1}{2^k}\right) = \frac{1 - \frac{1}{2^{2k}}}{1 - \frac{1}{2^k}} = \frac{2^{2k} - 1}{2^{2k}} \cdot \frac{2^k}{2^k - 1}$

In fact, a simpler method:

$P_n = \prod_{k=1}^{n} \left(1 + \frac{1}{2^k}\right) = \prod_{k=1}^{n} \frac{2^k + 1}{2^k}$

Take logarithms: $\ln P_n = \sum_{k=1}^{n} \ln\left(1 + \frac{1}{2^k}\right)$

As $n \to \infty$, $P_n \to 2$ (can be verified numerically or proved more rigorously)

Answer: 2

Approach: Use mathematical induction: verify the equality for $n = 1$, then carry the induction hypothesis over to the $n + 1$ case.

Steps:

Proof (mathematical induction):

Base step: when $n = 1$, the left side $= 1$ and the right side $= F_3 - 1 = 2 - 1 = 1$, so it holds.

Inductive hypothesis: assume $F_1 + \cdots + F_k = F_{k+2} - 1$

Inductive step: $F_1 + \cdots + F_k + F_{k+1} = (F_{k+2} - 1) + F_{k+1} = F_{k+3} - 1$

Conclusion: by mathematical induction, the identity holds.

Answer: By mathematical induction, the equality holds.

Approach: Write the characteristic equation associated with the recurrence, find its two characteristic roots, then use the initial conditions to fix the coefficients and obtain the general term.

Steps:

Solution (characteristic equation method):

Let $a_n = r^n$ and substitute into the recurrence relation:

$r^{n+2} = r^{n+1} + r^n$ $r^2 = r + 1$ $r^2 - r - 1 = 0$

$r = \frac{1 \pm \sqrt{5}}{2}$

General solution: $a_n = A\left(\frac{1+\sqrt{5}}{2}\right)^n + B\left(\frac{1-\sqrt{5}}{2}\right)^n$

Determine $A, B$ from the initial conditions…

Answer (Binet’s formula):

$a_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right]$