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# example of summation by parts

Proposition. The series $\displaystyle\sum_{{n=1}}^{\infty}\frac{\sin{n\varphi}}{n}$ and $\displaystyle\sum_{{n=1}}^{\infty}\frac{\cos{n\varphi}}{n}$ converge for every complex value $\varphi$ which is not an even multiple of $\pi$.

Proof. Let $\varepsilon$ be an arbitrary positive number. One uses the identities

$\displaystyle\sin{\varphi}+\sin{2\varphi}+\ldots+\sin{n\varphi}=\frac{\sin(n+% \frac{1}{2})\varphi-\sin\frac{\varphi}{2}}{2\sin\frac{\varphi}{2}},$ | (1) |

$\displaystyle\cos{\varphi}+\cos{2\varphi}+\ldots+\cos{n\varphi}=\frac{-\cos(n+% \frac{1}{2})\varphi+\cos\frac{\varphi}{2}}{2\sin\frac{\varphi}{2}},$ | (2) |

proved in the entry “example of telescoping sum”. These give the estimates

$|\sin{\varphi}+\sin{2\varphi}+\ldots+\sin{n\varphi}|\leqq\frac{2}{2|\sin\frac{% \varphi}{2}|}\,:=\,K_{\varphi},$ |

$|\cos{\varphi}+\cos{2\varphi}+\ldots+\cos{n\varphi}|\leqq\frac{2}{2|\sin\frac{% \varphi}{2}|}\,:=\,K_{\varphi}$ |

for any $n=1,\,2,\,3,\,\ldots$. We want to apply to the series $\sum_{{n=1}}^{\infty}\frac{\cos{n\varphi}}{n}$ the Cauchy general convergence criterion for series. Let us use here the short notation

$\cos{N\varphi}+\cos{(N\!+\!1)\varphi}+\ldots+\cos{(N\!+\!p)\varphi}:=S_{{N,N+p% }}\quad(p=0,\,1,\,2,\,\ldots).$ |

Then, utilizing Abel’s summation by parts, we obtain

$\left|\sum_{{n=N}}^{{N+P}}\frac{\cos{n\varphi}}{n}\right|=\left|\sum_{{p=0}}^{% {P}}\frac{1}{N\!+\!p}\cos{(N+p)\varphi}\right|=\left|\sum_{{p=0}}^{{P-1}}\left% (\frac{1}{N\!+\!p}-\frac{1}{N\!+\!p\!+\!1}\right)S_{{N,N+p}}+\frac{1}{N\!+\!P}% S_{{N,N+P}}\right|\leqq$ |

$\leqq\sum_{{p=0}}^{{P-1}}\left(\frac{1}{N\!+\!p}-\frac{1}{N\!+\!p\!+\!1}\right% )|S_{{N,N+P}}|+\frac{1}{N+P}|S_{{N,N+P}}|<$ |

$<\sum_{{p=0}}^{{P-1}}\left(\frac{1}{N\!+\!p}-\frac{1}{N\!+\!p\!+\!1}\right)% \cdot 2K_{\varphi}+\frac{1}{N\!+\!P}\cdot 2K_{\varphi}\,=\,\frac{1}{N}\cdot 2K% _{\varphi};$ |

the last form is gotten by telescoping the preceding sum and before that by using the identity

$S_{{N,N+p}}=[\cos\varphi+\cos 2\varphi+\ldots+\cos(N\!+\!p)\varphi]-[\cos% \varphi+\cos 2\varphi+\ldots+\cos(N\!-\!1)\varphi].$ |

Thus we see that

$\left|\sum_{{n=N}}^{{N+P}}\frac{\cos{n\varphi}}{n}\right|<\frac{2K_{\varphi}}{% N}<\varepsilon$ |

for all natural numbers $P$ as soon as $N>\frac{2K_{\varphi}}{\varepsilon}$. According to the Cauchy criterion, the latter series is convergent for the mentioned values of $\varphi$. The former series is handled similarly.

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## Comments

## \sum_n \sin{nz}/n and \sum_n \cos{nz}/n

These series converge quite slowly and I don't know other ways to demonstrate their convergence than the one in "example of summation by parts". Are there other ways?

## Re: \sum_n \sin{nz}/n and \sum_n \cos{nz}/n

Hmm, I might be totally wrong, but I seems like I always taken the convergence of these series for granted...

Isn't one of these series the Fourier series for the sawtooth blade function? The sawtooth blade function is of bounded variation, so Dirichlet's theorem applies.

Now, the proof of Dirichlet's theorem may use the second mean-value theorem for integrals, so I suppose the summation-by-parts argument is actually hidden there.

// Steve

## Re: \sum_n \sin{nz}/n and \sum_n \cos{nz}/n

You mean the Dirichlet\'s convergence test? I have not known it, but now I found it in PM. Indeed, lieven uses the summation-by-parts in the proof.

Jussi

## Re: \sum_n \sin{nz}/n and \sum_n \cos{nz}/n

> You mean the Dirichlet\'s convergence test? I have not known

> it, but now I found it in PM. Indeed, lieven uses the

> summation-by-parts in the proof.

No, I mean the famous Dirichlet theorem on convergence of Fourier series

(http://planetmath.org/encyclopedia/DirichletConditions.html)

which you surely must know?

Although that theorem you cite (which I also did not know previously)

does have a tangential connection.

// Steve

## Re: \sum_n \sin{nz}/n and \sum_n \cos{nz}/n

> No, I mean the famous Dirichlet theorem on convergence of Fourier series (http://planetmath.org/encyclopedia/DirichletConditions.html)

which you surely must know?

Yes, I know it (although I didn't remember its name).

Jussi