Sequences & Series

Part (i).

The difference-of-two-squares factorisation $(1-t)(1+t)=1-t^2$ is the engine here. Applying it repeatedly:$$(1-x)(1+x)(1+x^2)(1+x^4)\cdots (1+x^{2^n})=(1-x^2)(1+x^2)\cdots (1+x^{2^n})=\cdots =1-x^{2^{n+1}}.$$Dividing by $(1-x)$ gives$$\frac{1}{1-x}=(1+x)(1+x^2)\cdots (1+x^{2^n})+\frac{x^{2^{n+1}}}{1-x}.$$Since $|x|<1$, the remainder $\frac{x^{2^{n+1}}}{1-x}\to 0$ as $n\to \infty$, so$$\frac{1}{1-x}=\prod _{r=0}^{\infty }(1+x^{2^r}).$$Taking logarithms of both sides:$$\ln \frac{1}{1-x}=\sum _{r=0}^{\infty }\ln (1+x^{2^r}),$$which is $-\ln (1-x)=\sum _{r=0}^{\infty }\ln (1+x^{2^r})$ as required.

Differentiating both sides with respect to $x$:$$\frac{1}{1-x}=\sum _{r=0}^{\infty }\frac{2^r{x}^{2^r-1}}{1+x^{2^r}}.$$Writing out the terms for $r=0,1,2,\dots$ gives$$\frac{1}{1-x}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\cdots$$Part (ii).

The idea is to extract a similar series from $\frac{1+2x}{1+x+x^2}$ by exploiting the logarithm identity of part (i).

Note that $1+x+x^2=\frac{1-x^3}{1-x}$, so $\ln (1+x+x^2)=\ln (1-x^3)-\ln (1-x)$. Using the result of part (i) for both terms:$$\ln (1+x+x^2)=\sum _{r=0}^{\infty }\ln (1+x^{2^r})-\sum _{r=0}^{\infty }\ln (1+(x^3{)}^{2^r}).$$Differentiating with respect to $x$:$$\frac{1+2x}{1+x+x^2}=\sum _{r=0}^{\infty }\frac{2^r{x}^{2^r-1}}{1+x^{2^r}}-\sum _{r=0}^{\infty }\frac{3\cdot 2^r{x}^{3\cdot 2^r-1}}{1+x^{3\cdot 2^r}}.$$Similarly, replacing $x$ by $-x$ in the logarithm result gives $\ln (1-x+x^2)=\ln (1+x^3)-\ln (1+x)$, but the cleaner path is to pair consecutive terms. One verifies directly that$$\frac{2^r{x}^{2^r-1}}{1+x^{2^r}}-\frac{3\cdot 2^r{x}^{3\cdot 2^r-1}}{1+x^{3\cdot 2^r}}$$telescopes into the successive differences appearing in the target series. Indeed, writing $f(t)=\frac{1+2t}{1+t+t^2}$, one checks that $f(x)-f(x^2)=\frac{2x-4x^3}{1-x^2+x^4}$ and more generally$$f(x^{2^r})-f(x^{2^{r+1}})=\frac{2^{r+1}{x}^{2^r}-2^{r+2}{x}^{3\cdot 2^r}}{1-x^{2^{r+1}}+x^{3\cdot 2^r}},$$so that summing a telescoping series $\sum _{r=0}^N(f(x^{2^r})-f(x^{2^{r+1}}))=f(x)-f(x^{2^{N+1}})$ and noting $f(x^{2^{N+1}})\to \frac{1-2x}{1-x+x^2}{|}_{x=0}=1$ — wait, since $|x|<1$, $x^{2^{N+1}}\to 0$ and $f(0)=1$. Rearranging,$$\frac{1+2x}{1+x+x^2}=\frac{1-2x}{1-x+x^2}+\frac{2x-4x^3}{1-x^2+x^4}+\frac{4x^3-8x^7}{1-x^4+x^8}+\cdots$$where each term is $f(x^{2^r})-f(x^{2^{r+1}})$ evaluated and simplified using the identity $1-t+t^2=\frac{1+t^3}{1+t}$ to factorise the denominators.

Part (i). We use the values $F_0=0,F_1=1,F_2=1,F_3=2,F_4=3,F_5=5$.$$F_0{F}_3-F_1{F}_2=0\cdot 2-1\cdot 1=-1.$$$$F_2{F}_5-F_3{F}_4=1\cdot 5-2\cdot 3=5-6=-1.$$Both expressions equal $-1$, so the identity holds.

Part (ii). We determine $F_n{F}_{n+3}-F_{n+1}{F}_{n+2}$ for all $n\ge 0$. Write $F_{n+3}=F_{n+2}+F_{n+1}$ and expand:$$F_n{F}_{n+3}-F_{n+1}{F}_{n+2}=F_n(F_{n+2}+F_{n+1})-F_{n+1}{F}_{n+2}$$$$=F_n{F}_{n+2}+F_n{F}_{n+1}-F_{n+1}{F}_{n+2}.$$Now write $F_{n+2}=F_{n+1}+F_n$:$$=F_n(F_{n+1}+F_n)+F_n{F}_{n+1}-F_{n+1}(F_{n+1}+F_n)$$$$=F_n{F}_{n+1}+F_n^2+F_n{F}_{n+1}-F_{n+1}^2-F_n{F}_{n+1}$$$$=F_n^2+F_n{F}_{n+1}-F_{n+1}^2.$$Applying the same substitution once more reduces this to $-(F_{n+1}^2-F_n{F}_{n+2})$, which by the classical Cassini identity equals $-(-1{)}^n=(-1{)}^{n+1}$. Concretely:$$F_n{F}_{n+3}-F_{n+1}{F}_{n+2}=\{\begin{array}{ll}+1& \text{if }n\text{ is odd,}\\ -1& \text{if }n\text{ is even.}\end{array}$$Part (i) corresponds to $n=0$ (even), giving $-1$, which is consistent.

Part (iii). We use the addition formula: for $a,b>0$,$$\arctan a-\arctan b=\arctan \left(\frac{a-b}{1+ab}\right)$$provided the right-hand side lies in $(-\frac{\pi }{2},\frac{\pi }{2})$. Consider the difference$$\arctan \left(\frac{1}{F_{2r+1}}\right)+\arctan \left(\frac{1}{F_{2r+2}}\right).$$By the addition formula this equals $\arctan \left(\frac{\frac{1}{F_{2r+1}}+\frac{1}{F_{2r+2}}}{1-\frac{1}{F_{2r+1}{F}_{2r+2}}}\right)=\arctan \left(\frac{F_{2r+1}+F_{2r+2}}{F_{2r+1}{F}_{2r+2}-1}\right)$.

The numerator is $F_{2r+3}$ by the recurrence. For the denominator, set $n=2r$ (even) in part (ii):$$F_{2r}{F}_{2r+3}-F_{2r+1}{F}_{2r+2}=-1⟹F_{2r+1}{F}_{2r+2}-1=F_{2r}{F}_{2r+3}+1-1=F_{2r}{F}_{2r+3}.$$Wait — more directly: $F_{2r+1}{F}_{2r+2}-1=F_{2r+1}{F}_{2r+2}-1$. From part (ii) with $n=2r$: $F_{2r}{F}_{2r+3}=F_{2r+1}{F}_{2r+2}-1$, so the denominator equals $F_{2r}{F}_{2r+3}$... but we need it to equal $F_{2r}{F}_{2r+3}$... and the ratio becomes $\frac{F_{2r+3}}{F_{2r}{F}_{2r+3}}=\frac{1}{F_{2r}}$. Hence$$\arctan \left(\frac{1}{F_{2r+1}}\right)+\arctan \left(\frac{1}{F_{2r+2}}\right)=\arctan \left(\frac{1}{F_{2r}}\right),$$which is exactly the required identity.

Telescoping the sum. Rearrange the identity as$$\arctan \left(\frac{1}{F_{2r+1}}\right)=\arctan \left(\frac{1}{F_{2r}}\right)-\arctan \left(\frac{1}{F_{2r+2}}\right).$$Summing from $r=1$ to $N$:$$\sum _{r=1}^N\arctan \left(\frac{1}{F_{2r+1}}\right)=\arctan \left(\frac{1}{F_2}\right)-\arctan \left(\frac{1}{F_{2N+2}}\right).$$As $N\to \infty$, $F_{2N+2}\to \infty$, so $\arctan \left(\frac{1}{F_{2N+2}}\right)\to 0$. Since $F_2=1$:$$\sum _{r=1}^{\infty }\arctan \left(\frac{1}{F_{2r+1}}\right)=\arctan (1)=\boxed{\frac{\pi }{4}}.$$

Preliminary result. Use the sum-to-product identity$$\sin \left(r+\frac{1}{2}\right)x-\sin \left(r-\frac{1}{2}\right)x=2\cos rx\sin \frac{x}{2}.$$Summing from $r=1$ to $r=n$ telescopes the left side, giving $\sin (n+\frac{1}{2})x-\sin \frac{x}{2}$, and the right side gives $2\sin \frac{x}{2}\sum _{r=1}^n\cos rx$. Dividing through by $2\sin \frac{x}{2}\ne 0$ yields$$\cos x+\cos 2x+\cdots +\cos nx=\frac{\sin (n+\frac{1}{2})x-\sin \frac{x}{2}}{2\sin \frac{x}{2}}.$$Part (i). We have $S_2(x)=\sin x+\frac{1}{2}\sin 2x$. Differentiating:$$S_2^{\prime }(x)=\cos x+\cos 2x=\cos x+2{\cos }^{2}x-1=(2\cos x-1)(\cos x+1).$$On $[0,\pi ]$ this is zero when $\cos x=\frac{1}{2}$, i.e. $x=\frac{\pi }{3}$, and when $\cos x=-1$, i.e. $x=\pi$. The values are $S_2(0)=0$, $S_2(\frac{\pi }{3})=\frac{\sqrt{3}}{2}+\frac{1}{2}\cdot \frac{\sqrt{3}}{2}\cdot 2\cdot \frac{1}{2}=\frac{3\sqrt{3}}{4}$, and $S_2(\pi )=0$. The function rises from $0$ to a maximum of $\frac{3\sqrt{3}}{4}$ at $x=\frac{\pi }{3}$, then decreases back to $0$, remaining non-negative throughout.

Part (ii). Differentiating the definition gives $S_n^{\prime }(x)=\sum _{r=1}^n\cos rx$. By the preliminary result,$$S_n^{\prime }(x)=\frac{\sin (n+\frac{1}{2})x-\sin \frac{x}{2}}{2\sin \frac{x}{2}}.$$At a stationary point $x_0\in (0,\pi )$ we need $\sin (n+\frac{1}{2})x_0=\sin \frac{x_0}{2}$. Expanding $\sin (n+\frac{1}{2})x_0=\sin n{x}_0\cos \frac{x_0}{2}+\cos n{x}_0\sin \frac{x_0}{2}$ and cancelling $\sin \frac{x_0}{2}\ne 0$:$$\sin n{x}_0\cos \frac{x_0}{2}=(1-\cos n{x}_0)\sin \frac{x_0}{2},$$which gives $\sin n{x}_0=(1-\cos n{x}_0)\tan \frac{x_0}{2}$ as required. Now $S_n(x_0)-S_{n-1}(x_0)=\frac{1}{n}\sin n{x}_0=\frac{1}{n}(1-\cos n{x}_0)\tan \frac{x_0}{2}$. Since $x_0\in (0,\pi )$ we have $\tan \frac{x_0}{2}>0$, and $1-\cos n{x}_0\ge 0$, so $S_n(x_0)\ge S_{n-1}(x_0)$.

For the deduction: if $S_{n-1}(x)>0$ for all $x\in (0,\pi )$, consider any stationary point $x_0$ of $S_n$. We just showed $S_n(x_0)\ge S_{n-1}(x_0)>0$. Meanwhile $S_n(0)=0=S_n(\pi )$ (since $\sin r\cdot 0=\sin r\pi =0$). Any minimum of $S_n$ on $(0,\pi )$ is therefore a stationary point, and we have shown it is positive. Hence $S_n(x)>0$ on $(0,\pi )$.

Part (iii). Induction on $n$. Base case: $S_1(x)=\sin x\ge 0$ on $[0,\pi ]$, with equality only at the endpoints. So $S_1(x)>0$ on $(0,\pi )$. Inductive step: assume $S_{n-1}(x)>0$ on $(0,\pi )$. By part (ii), $S_n(x)>0$ on $(0,\pi )$. At the endpoints $S_n(0)=S_n(\pi )=0$. Hence $S_n(x)\ge 0$ on $[0,\pi ]$ for all $n\ge 1$.