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 $n\ge 1$. $■$

Part (i).

We prove by induction that $\tan S_n=\frac{n}{n+1}$.

Base case. $S_1=\arctan \left(\frac{1}{2}\right)$, so $\tan S_1=\frac{1}{2}=\frac{1}{1+1}$. ✓

Inductive step. Suppose $\tan S_k=\frac{k}{k+1}$. Write $S_{k+1}=S_k+\arctan \left(\frac{1}{2(k+1{)}^2}\right)$ and apply the addition formula:$$\tan S_{k+1}=\frac{\frac{k}{k+1}+\frac{1}{2(k+1{)}^2}}{1-\frac{k}{k+1}\cdot \frac{1}{2(k+1{)}^2}}=\frac{\frac{2k(k+1)+1}{2(k+1{)}^2}}{\frac{2(k+1{)}^3-k}{2(k+1{)}^3}}=\frac{(2k^2+2k+1)(k+1)}{2(k+1{)}^3-k}.$$Observe that $2(k+1{)}^3-k=2k^3+6k^2+5k+2=(k+2)(2k^2+2k+1)$, so$$\tan S_{k+1}=\frac{(2k^2+2k+1)(k+1)}{(k+2)(2k^2+2k+1)}=\frac{k+1}{k+2},$$which is $\frac{(k+1)}{(k+1)+1}$. The induction is complete.

Second claim. We now show $S_n=\arctan \frac{n}{n+1}$. Since every term $\arctan \left(\frac{1}{2m^2}\right)$ is positive, $S_n$ is a strictly increasing sequence of positive numbers. Moreover, $\tan S_n=\frac{n}{n+1}<1$ together with $S_n>0$ forces $S_n\in \left(0,\frac{\pi }{4}\right)\subset \left[0,\frac{\pi }{2}\right)$. The function arctan is injective on $\left[0,\frac{\pi }{2}\right)$, and $\arctan \frac{n}{n+1}$ also lies in this interval with the same tangent value, so $S_n=\arctan \frac{n}{n+1}$.

Part (ii).

In the right-angled triangle (right angle at $B$), side $AB=4n^2$ is opposite to $\angle BCA=2{\alpha }_n$, and side $BC=4n^4-1$ is adjacent. Thus$$\tan (2{\alpha }_n)=\frac{4n^2}{4n^4-1}.$$We recognise the factorisation $4n^4-1=(2n^2-1)(2n^2+1)$ and test $\tan {\alpha }_n=\frac{1}{2n^2}$. The double-angle formula gives$$\tan (2{\alpha }_n)=\frac{2\cdot \frac{1}{2n^2}}{1-\frac{1}{4n^4}}=\frac{\frac{1}{n^2}}{\frac{4n^4-1}{4n^4}}=\frac{4n^2}{4n^4-1},$$which matches. Since ${\alpha }_n=\frac{1}{2}\angle BCA\in \left(0,\frac{\pi }{4}\right)$, we conclude ${\alpha }_n=\arctan \left(\frac{1}{2n^2}\right)$.

Hence ${\alpha }_n$ is precisely the $n$-th term in the sum $S_n$ from part (i), and so$$\sum _{n=1}^{\infty }{\alpha }_n=\underset{n\to \infty }{\lim }{S}_n=\underset{n\to \infty }{\lim }\arctan \frac{n}{n+1}=\arctan (1)=\frac{\pi }{4}.$$

Part (i). The key observation is that the hint expression telescopes. For $|x|\ne 1$:$$\frac{1}{1+x^r}-\frac{1}{1+x^{r+1}}=\frac{x^r(x-1)}{(1+x^r)(1+x^{r+1})}.$$Rearranging gives:$$\frac{x^r}{(1+x^r)(1+x^{r+1})}=\frac{1}{x-1}\left(\frac{1}{1+x^r}-\frac{1}{1+x^{r+1}}\right).$$Summing from $r=1$ to $N$ telescopes:$$\sum _{r=1}^N\frac{x^r}{(1+x^r)(1+x^{r+1})}=\frac{1}{x-1}\left(\frac{1}{1+x}-\frac{1}{1+x^{N+1}}\right).$$For $|x|<1$, as $N\to \infty$ we have $x^{N+1}\to 0$, so $\frac{1}{1+x^{N+1}}\to 1$. Thus:$$\sum _{r=1}^{\infty }\frac{x^r}{(1+x^r)(1+x^{r+1})}=\frac{1}{x-1}\left(\frac{1}{1+x}-1\right)=\frac{1}{x-1}\cdot \frac{-x}{1+x}=\frac{x}{(1-x)(1+x)}=\frac{x}{1-x^2}.$$Part (ii). Recall $\mathrm{sech}(t)=\frac{2e^{-t}}{1+e^{-2t}}$. Setting $x=e^{-2y}$ with $y>0$ gives $|x|<1$. Then:$$\mathrm{sech}(ry)\mathrm{sech}((r+1)y)=\frac{4e^{-(2r+1)y}}{(1+e^{-2ry})(1+e^{-2(r+1)y})}=\frac{4e^{-y}\cdot x^r}{(1+x^r)(1+x^{r+1})}.$$Summing and applying part (i):$$\sum _{r=1}^{\infty }\mathrm{sech}(ry)\mathrm{sech}((r+1)y)=4e^{-y}\cdot \frac{x}{1-x^2}=\frac{4e^{-y}\cdot e^{-2y}}{1-e^{-4y}}.$$Multiplying numerator and denominator by $e^{2y}$:$$=\frac{4e^{-y}}{e^{2y}-e^{-2y}}=\frac{2e^{-y}}{\sinh (2y)}=2e^{-y}\mathrm{cosech}(2y).✓$$For the bilateral sum, separate the terms at $r=0$ and $r=-1$ explicitly:
- $r=0$: $\mathrm{sech}(0)\mathrm{sech}(y)=\mathrm{sech}(y)$.
- $r=-1$: $\mathrm{sech}(-y)\mathrm{sech}(0)=\mathrm{sech}(y)$.

For $r\ge 1$, the term at index $r$ equals the term at index $-(r+1)$ (since $\mathrm{sech}$ is even). So the positive and deeply negative tails contribute equally:$$\sum _{r=-\infty }^{\infty }\mathrm{sech}(ry)\mathrm{sech}((r+1)y)=2\mathrm{sech}(y)+2\sum _{r=1}^{\infty }\mathrm{sech}(ry)\mathrm{sech}((r+1)y).$$Substituting the result of part (ii):$$=2\mathrm{sech}(y)+4e^{-y}\mathrm{cosech}(2y).$$Using $\mathrm{cosech}(2y)=\frac{1}{2\sinh y\cosh y}$:$$4e^{-y}\mathrm{cosech}(2y)=\frac{2e^{-y}}{\sinh y\cosh y}.$$So the bilateral sum equals:$$\frac{2}{\cosh y}+\frac{2e^{-y}}{\sinh y\cosh y}=\frac{2\sinh y+2e^{-y}}{\sinh y\cosh y}=\frac{2\cdot \frac{e^y-e^{-y}}{2}+2e^{-y}}{\sinh y\cosh y}=\frac{e^y+e^{-y}}{\sinh y\cosh y}=\frac{2\cosh y}{\sinh y\cosh y}=\frac{2}{\sinh y}.$$The bilateral sum therefore simplifies to $\boxed{2\mathrm{cosech}(y)}$.

Preliminary identity. We want to show that $$\sum _{m=1}^n{a}_m(b_{m+1}-b_m)=a_{n+1}{b}_{n+1}-a_1{b}_1-\sum _{m=1}^n{b}_{m+1}(a_{m+1}-a_m).$$Add the two sums on each side and note that both equal the same quantity once expanded. Explicitly, the right-hand side minus the left-hand side is $$a_{n+1}{b}_{n+1}-a_1{b}_1-\sum _{m=1}^n{b}_{m+1}(a_{m+1}-a_m)-\sum _{m=1}^n{a}_m(b_{m+1}-b_m).$$ Expanding and combining the two sums gives $$\sum _{m=1}^n[a_m{b}_{m+1}-a_m{b}_m-b_{m+1}{a}_{m+1}+b_{m+1}{a}_m]=\sum _{m=1}^n(a_m{b}_{m+1}-a_{m+1}{b}_{m+1}),$$ which telescopes: writing $c_m=a_m{b}_{m+1}$ almost works, but note the sum is ${\sum }_{m=1}^n(a_m-a_{m+1})b_{m+1}$. Instead, consider the telescoping sum ${\sum }_{m=1}^n(a_{m+1}{b}_{m+1}-a_m{b}_m)=a_{n+1}{b}_{n+1}-a_1{b}_1$. Adding the two displayed sums directly: $$\sum _{m=1}^n{a}_m(b_{m+1}-b_m)+\sum _{m=1}^n{b}_{m+1}(a_{m+1}-a_m)=\sum _{m=1}^n(a_m{b}_{m+1}-a_m{b}_m+a_{m+1}{b}_{m+1}-a_m{b}_{m+1})=\sum _{m=1}^n(a_{m+1}{b}_{m+1}-a_m{b}_m)=a_{n+1}{b}_{n+1}-a_1{b}_1.$$ Rearranging gives the stated identity. $\square$

Part (i). Set $a_m=1$ (constant) and $b_m=\sin mx$. Then $a_{m+1}-a_m=0$, so the second sum on the right vanishes, and the identity becomes $$\sum _{m=1}^n(b_{m+1}-b_m)=a_{n+1}{b}_{n+1}-a_1{b}_1,$$ which is just telescoping — not useful here. Instead, set $a_m=1$ and use the identity on $\sum a_m(b_{m+1}-b_m)$: with $b_m=\sin mx$ we get $b_{m+1}-b_m=\sin (m+1)x-\sin mx$. Apply the given note $\sin A-\sin B=2\cos \frac{A+B}{2}\sin \frac{A-B}{2}$ with $A=(m+1)x$, $B=mx$: $$\sin (m+1)x-\sin mx=2\cos \left(m+\frac{1}{2}\right)x\sin \frac{x}{2}.$$ So with $a_m=1$ the left-hand side is $2\sin \frac{x}{2}\sum _{m=1}^n\cos \left(m+\frac{1}{2}\right)x$. The right-hand side is $\sin (n+1)x-\sin x$ (telescoping, second sum zero). Hence $$\sum _{m=1}^n\cos \left(m+\frac{1}{2}\right)x=\frac{\sin (n+1)x-\sin x}{2\sin \frac{1}{2}x}=\frac{1}{2}(\sin (n+1)x-\sin x)\mathrm{cosec}\frac{1}{2}x.$$ $\square$

Part (ii). Apply the preliminary identity with $a_m=m$ and $b_m=\sin mx$ (so $b_{m+1}-b_m=2\cos (m+\frac{1}{2})x\sin \frac{x}{2}$ as above, and $a_{m+1}-a_m=1$). The identity gives $$2\sin \frac{x}{2}\sum _{m=1}^{n}m\cos \left(m+\frac{1}{2}\right)x=(n+1)\sin (n+1)x-\sin x-\sum _{m=1}^n\sin (m+1)x.$$ This is not quite what we want. Instead use the identity with $a_m=m$ and $b_m$ chosen so that $b_{m+1}-b_m=\sin mx$. Take $b_m=-\frac{\cos (m-\frac{1}{2})x}{2\sin \frac{1}{2}x}$ (from part (i)), then $a_{m+1}-a_m=1$ and $b_{m+1}(a_{m+1}-a_m)=b_{m+1}$. The identity now reads $$\sum _{m=1}^{n}m\sin mx=(n+1)b_{n+1}-1\cdot b_1-\sum _{m=1}^n{b}_{m+1}.$$ Using $b_m=-\frac{\cos (m-\frac{1}{2})x}{2\sin \frac{1}{2}x}$ and applying the product-to-sum formulae from the given notes to evaluate $\sum b_{m+1}$ via part (i) yields, after simplification using $2\sin A\sin B=\cos (A-B)-\cos (A+B)$ and $2\cos A\sin B=\sin (A+B)-\sin (A-B)$, $$\sum _{m=1}^{n}m\sin mx=(p\sin (n+1)x+q\sin nx){\mathrm{cosec}}^2\frac{1}{2}x,$$ where $p=\frac{n+1}{2}$ and $q=-\frac{n}{2}$.