Are these two naive upper bounds ok?

Question

EDIT:

Is my work ok? Here, I am trying to show a uniform bound for the sum of $cos(n)$

$$|\sum_{n=1}^{N} cos(n)|$$ $$=\big |\sum \frac{e^{in} + e^{-in}}{2}\big|$$ $$\le \sum |\frac{e^{in} + e^{-in}}{2}|$$ $$\le \sum |\frac{e^{in}}{2}| + \sum |\frac {e^{-in}}{2}|$$ $$= \frac{1}{2} \sum |(e^i)^n| + \frac{1}{2} \sum |(e^{-i})^n|$$ $$=\frac{1}{2} |\frac {e^i-e^{i(N+1)}}{1-e^i}| + \frac{1}{2} |\frac {e^{-i}-e^{-i(N+1)}}{1-e^{-i}}|$$ $$\le \frac{1}{2} \frac{1}{1-e^i}+ \frac{1}{2} \frac{1}{1-e^i} + \frac{1}{2}\frac{1}{1-e^{-i}} + \frac{1}{2}\frac{1}{1-e^{-i}}$$

Specifically, my concern is that I am not sure whether I am showing the upper bound of one of the factors correctly, in applying the Dirichlet Test.

For example, is this ok

$$\left|\sum_{n=1}^N \cos(n) \right| \le \sum_{n=1}^N \left|\cos(n)\right| \le \sum_{n=1}^N 1 \le N$$

And so $\sum \cos(n)$ has bounded partial sums for every positive integer $N$. My gut feeling is that this is not true, and seems like a naive "bound". But, it could be true, because $N$ is fixed, before giving the estimate.

A second example would be

$$\left|\int_1^R e^{ix}dx\right| \le \int_1^R \left|e^{ix}\right|dx \le \int_1^R 1 dx = R-1$$

which I claim is a "uniform bound" for any $R$ fixed. The bound certainly looks independent of $x$.

Are both of my examples ok?

Thanks,

Answer 1

While $$\left|\sum_{n=1}^{N}\cos(n)\right| \leq N$$ is certainly true, it is not likely to be of much use for the Dirichlet test, which requires a bound that is independent of $N$.

We can obtain a better bound by recognizing that $\cos(n) = (e^{in} + e^{-in})/2$, and for any nonzero integer $k$, we have $$\begin{aligned} \sum_{n=1}^{N} e^{ikn} &= e^{ik}\left(\sum_{n=0}^{N-1}e^{ikn}\right)\\ & = e^{ik}\left(\frac{e^{ikN} - 1}{e^{ik} - 1} \right) \\ &= e^{ik}\left( \frac{e^{ikN/2}}{e^{ik/2}}\right)\left( \frac{e^{ikN/2}-e^{-ikN/2}}{e^{ik/2}-e^{-ik/2}}\right) \\ &= e^{ik} \left(\frac{e^{ikN/2}}{e^{ik/2}}\right)\left(\frac{2i\sin(kN/2)}{2i\sin(k/2)}\right) \\ \end{aligned}$$ where we have used the fact that $e^{ix} - e^{-ix} = 2i\sin(x)$. Taking absolute values, we get $$\left|\sum_{n=1}^{N} e^{ikn}\right| = \left|\frac{\sin(kN/2)}{\sin(k/2)}\right|$$ Applying this result with $k=1$ and $k=-1$, we obtain $$\begin{aligned} \left|\sum_{n=1}^{N}\cos(n)\right| &= \frac{1}{2}\left|\sum_{n=1}^{N}(e^{in} + e^{-in})\right| \\ &\leq \frac{1}{2}\left|\sum_{n=1}^{N} e^{in} \right| + \frac{1}{2} \left|\sum_{n=1}^{N} e^{-in} \right|\\ &= \frac{1}{2}\left|\frac{\sin(N/2)}{\sin(1/2)}\right| + \frac{1}{2}\left|\frac{\sin(-N/2)}{\sin(-1/2)}\right| \\ &= \left|\frac{\sin(N/2)}{\sin(1/2)}\right| \\ &\leq \frac{1}{|\sin(1/2)|} \\ \end{aligned}$$

Answer 2

The first bound you have will not work for the Dirichlet test; we need a bound that is independent of $N$. The easiest approach is to explicitly calculate this sum by noting that $$ \cos(n) = \Re (\cos(1) + i\sin(1))^n $$ and apply the formula for the sum of a geometric series. Then, find an upper bound for the numerator.