upper bounds for partial sums of some cosine series

Question

I was wondering how one could show the following results:

1. $|\sum_{n=1}^N (-1)^n \cos^2 (n+1)| \leq 1$ for all $N\ge 1$.
2. There exists a real number $M$ so that $|\sum_{n=1}^N (-1)^n \cos(2(n+1))|\leq M$ and $|\sum_{n=1}^N \cos(2(n+1))|\leq M$ for all $N\ge 1$.

I know that $\cos (\theta) = \dfrac{e^{-i\theta} + e^{i\theta}}2$ for all $\theta,$ but I'm not sure how to prove the desired bounds since they involve $\cos$ taking integer values. And I know $\cos(2n) = 2\cos^2 n - 1$ for any real number $n$. But I can't seem to make much progress from here on.

Edit: For completeness, here's the computation for the case $N=2b$ (the case where $N=2b+1$ is similar):

We have $\begin{align} \sin(-1)\sum_{n=1}^b \sin(4n+1) &= \sin(-1) Im(\sum_{n=1}^b e^{i(4n+1)}) \\ &= \sin(-1)Im(\dfrac{e^{i(4b+5)} - e^{5i}}{e^{4i}-1})\\ &= \dfrac{\sin(-1)}{2-2\cos 4} Im(e^{i(4b+1)} - e^i - e^{i(4b+5)} + e^{5i})\\ &= \dfrac{\sin(-1)}{4\sin^2 2} (\sin(4b+1)-\sin(1) - \sin(4b+5) + \sin(5))\\ &= \dfrac{\sin(-1)}{4\sin^2 2} (2(\frac{1}2 (\sin 5 - \sin 1) - \frac{1}2 (\sin(4b+5) - \sin(4b+1))\\ &= \dfrac{\sin(-1)}{4\sin^2 2}(2\sin 2(\cos 3 - \cos (4b+3))\text{ using $\sin A \cos B = \frac{1}2 (\sin(A+B) - \sin(B-A))$}\\ &= \dfrac{\sin(-1)}{2\sin 2}(\cos 3 - \cos (4b+3))\\ &= \dfrac{\sin(-1)}{\sin 2} (\sin (2b)\sin(2b+3)) ( \end{align}$

Answer 1

As for $\displaystyle\sum_{n=1}^N \cos(2n+2)$,

this is just the real part of $\displaystyle\sum_{n=1}^N \exp(2in+2i)=\exp(2i)\cdot\exp(2i)\cdot\dfrac{\big(\exp(2i)\big)^n-1}{\exp(2i)-1}$ as is clear from the formula for geometric series.

How large can the modulus of it get? All values of $\big(\exp(2i)\big)^n$ must lie on the unit circle. The furthest that the unit circle gets from the point $1\in\mathbb{C}$ is $-1$. Draw a graph of the unit circle, if you're not convinced. Therefore, $\bigg|\big(\exp(2i)\big)^n-1\bigg|\leq2$. The other constants involved are fixed. Therefore, $\displaystyle\bigg|\sum_{n=1}^N \exp(2in+2i)\bigg|\leq2c$ for some fixed positive $c$. The real part of the sum would also have modulus $\leq2c$.

$\\$

The other sums in your question can also be simplified in a similar way.

$\displaystyle\sum_{n=1}^N (-1)^n \cos(2n+2)$ is the real part of $\displaystyle\sum_{n=1}^N (-1)^n\exp(2in+2i)$, which is also a geometric series.

$\displaystyle\sum_{n=1}^N (-1)^n \cos^2 (n+1)$ can be rewritten as $\displaystyle f(N)+\sum_{n=1}^{\lfloor N/2\rfloor}\bigg(\cos^2(2n+1)-\cos^2(2n)\bigg)$

where $f(N)$ is $0$ for even values of $N$, and $-\cos^2(N+1)$ for odd values of $N$.

Remember that $\cos^2(b)-\cos^2(a)=\sin(a+b)\sin(a-b)$.

Therefore, $\displaystyle\sum_{n=1}^N (-1)^n \cos^2 (n+1)=f(N)+\sin(-1)\sum_{n=1}^{\lfloor N/2\rfloor}\sin(4n+1)$.

Again, interpret the sum on the right as the imaginary part of a geometric series.

Split into cases.

If $N$ is even and $N=2b$,

then $\displaystyle\sum_{n=1}^N (-1)^n \cos^2 (n+1)=-\csc(2)\sin(1)\sin(2b)\sin(3 + 2 b)$.

See this link for confirmation.

You can graph this as a function of $b$, where $b$ ranges from $0$ to $\pi$. Notice that the function is periodic with period $\pi$, so you don't have to check beyond that interval. Indeed, the function is strictly between $1$ and $-1$.

If $N$ is odd and $N=2b+1$,

then $\displaystyle\sum_{n=1}^N (-1)^n \cos^2 (n+1)=-\cos^2(2b+2)-\csc(2)\sin(1)\sin(2b)\sin(3 + 2 b)$.

Again, graph this function and check.