# Using Summation By Parts on $\sum_{i=1}^{\infty} z^{i}\frac{1}{i}$

I'm trying to show that the power series $\sum_{k=1}^{\infty} z^{k}\frac{1}{k}$ converges for $z$ s.t. $|z| = 1$ (convergence on the unit circle), except for $z=1$ .

My exercise book says to do it using the summation by parts formula:
$\sum _{k=1}^{n} a_kb_k = A_nb_{n+1}- \sum _{k=1}^{n} A_k(b_{k+1}-b_k)$

I am having a hard time proceeding after substituting $e^{ki\theta}$, $\frac{1}{k}$ for $a_{k}$ and $b_{k}$ .

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Oh, but for $z = 1$ you have $\sum_k \frac{1}{k}$ which does not converge... –  dtldarek May 22 '12 at 23:26
I think you mean $|z|<1$ –  Pedro Tamaroff May 22 '12 at 23:32
@PeterTamaroff With $|z| < 1$ it is enough to go through absolute convergence. I think Mark means $|z| = 1, z \neq 1$, e.g. for $z = -1$ this converges conditionally, and should converge for other values too. –  dtldarek May 22 '12 at 23:34
Sorry, I made an edit to the question. It converges for all except z = 1. –  Mark May 23 '12 at 0:05

We have that $A_n=\sum_{k=1}^ne^{ik\theta}=\frac{e^{i\theta}-e^{i(n+1)\theta}}{1-e^{i\theta}}$, a simple geometric series. Notice that with $b_k=1/k$, the $A_nb_{n+1}$ term is bounded (for each fixed $\theta$) in absolute value for all $\theta\neq 2\pi m$, where $m\in\mathbb{Z}$. In particular, it's the denominator of $A_n$ that explodes when $\theta$ is a multiple of $2\pi$. That leaves the series. Again $A_n$ is bounded above in absolute value, and $\sum_{k=1}^\infty|b_{k+1}-b_k|=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}\ldots\rightarrow 1$, so that the series $\sum_{k=1}^\infty A_k(b_{k+1}-b_k)$ converges absolutely.

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To fill in the details, do you mean the following? In particular, the absolute value of the numerator of $A_{n}$: $|e^{i\theta} - e^{i(n+1)\theta}|$ is less than or equal to 2, since both terms are on the unit circle. So $|A_{n}|$ is bounded by some constant $C$, which means $A_{n}b_{n+1}$ approaches 0. On the other hand, $|A_{k}*(b_{k+1}-b_{k})|$ is less than $C|(b_{k+1}-b_{k})|$, so the summation also converges absolutely. –  Mark May 23 '12 at 1:07

This is an application of the Dirichlet's test also termed the generalized alternating test. The test goes as follows. If we have a sequence of positive real numbers, $\{a_n\}_{n=1}^{\infty}$ and a sequence of complex numbers $\{b_n\}_{n=1}^{\infty}$ such that

1. $a_n$ is a decreasing sequence i.e. $a_n > a_m$ whenever $m>n$.
2. $\lim_{n \rightarrow \infty} a_n = 0$.
3. $\left \lvert \displaystyle \sum_{k=0}^{n} b_k\right \rvert \leq M$, $\forall n \in \mathbb{Z}^+$, where $M$ is some constant independent of $n$

then the series $\displaystyle\sum_{n=0}^{\infty} a_n b_n$ converges.

The proof is by partial summation technique. Let $S_n = \displaystyle \sum_{k=0}^n a_k b_k$ and $B_n = \displaystyle \sum_{k=0}^n b_k$.

By partial summation, we have that $\displaystyle S_n = a_{n+1} B_n + \sum_{k=0}^n B_k (a_k - a_{k+1})$. Hence, $\displaystyle \lim_{n \rightarrow \infty} S_n = \lim_{n \rightarrow \infty} \left( a_{n+1} B_n + \sum_{k=0}^n B_k (a_k - a_{k+1}) \right) = 0 + \sum_{k=0}^{\infty} B_k (a_k - a_{k+1})$.

Hence, all we need to show now is that $\displaystyle \sum_{k=0}^{\infty} B_k (a_k - a_{k+1})$ converges.

We will show that $\displaystyle \sum_{k=0}^{\infty} B_k (a_k - a_{k+1})$ in fact converges absolutely. Note that

\begin{align} \left \lvert \sum_{k=0}^n B_k (a_k - a_{k+1}) \right \rvert & \leq \sum_{k=0}^n \left \lvert B_k (a_k - a_{k+1}) \right \rvert\\ & = \sum_{k=0}^n \left \lvert B_k \right \rvert \left \lvert (a_k - a_{k+1}) \right \rvert\\ & \leq M \sum_{k=0}^{n} \left( a_k - a_{k+1}\right)\\ & \leq M(a_0 - a_{n+1})\\ & \leq M a_0 \end{align}

Hence, we have that $\displaystyle \lim_{n \rightarrow \infty} S_n$ exists.

For your problem, $a_n = \dfrac1n$ and $b_n = \exp(in \theta)$. Clearly, $a_n$ and $b_n$ satisfy the assumptions for the Dirichlet test to be applicable. The only thing which might not be clear is why $\left \lvert \displaystyle \sum_{k=1}^{n}b_k \right \rvert$ bounded. But as Sam has shown $$\left \lvert \sum_{k=1}^{n}b_k \right \rvert = \left \lvert \dfrac{\exp(in \theta) - \exp(i (n+1) \theta)}{1 - \exp(i \theta)} \right \rvert \leq \dfrac{2}{\left \lvert 1 - \exp(i \theta) \right \rvert}$$ which is some finite real number independent of $n$. (Note that $\theta \neq 2 k \pi$).

Hence, $\displaystyle \sum_{k=1}^{\infty} \dfrac{z^{k}}{k}$ converges for $z$ s.t. $|z| = 1$, except for $z=1$

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