Revisiting a lemma

We’ve discussed before the “trigonometric harmonic” series $\sum_{n=1}^\infty\cos n/n$. In particular, we showed that the series converges (conditionally). The argument involved the partial sums of the sequence $\{\cos n\}_{n=1}^\infty$, and we denoted these by $C(m)$. The closed-form expression we found for $C(m)$ involved the quantity $\cos m-\cos(m+1)$; in this post we show that this expression can also be written in the alternative form $2\sin(1/2)\sin(m+1/2)$.

The proof is a straightforward application of the complex exponential expression for sine and cosine:

2\sin(1/2)\sin(m+1/2) &= 2\frac{e^{i/2}-e^{-i/2}}{2i}\cdot\frac{e^{i(m+1/2)}-e^{-i(m+1/2)}}{2i} \\
&= \frac{e^{i(m+1)}-e^{im}-e^{-im}+e^{-i(m+1)}}{-2} \\
&= \cos m-\cos(m+1)

Applying this with $m:=0$ yields $2\sin(1/2)^2=1-\cos 1$, and we have

C(m) &= \frac{\cos m-\cos(m+1)}{2(1-\cos 1)}-\frac{1}{2} \\
&= \frac{2\sin(1/2)\sin(m+1/2)}{4\sin(1/2)^2}-\frac{1}{2} \\
&= \frac{\sin(m+1/2)}{2\sin(1/2)}-\frac{1}{2} .

Written in this form, the naive upper bound on $\lvert C(m)\rvert$ becomes

\[ \frac{1}{2\sin(1/2)}+\frac{1}{2} \approx 1.54 , \]

which is actually slightly better than the previous upper bound we gave of

\[ \frac{1}{1-\cos1}+\frac{1}{2} \approx 2.68 . \]

In fact, it is easy to see that our new upper bound the best possible, since it is reached at for example $m:=(3\pi-1)/2$.