[EX] A nice and simple series in \( \displaystyle \mathbb{C} \)

[Paolo90]

Exercise. Consider the series

\[
\sum_{n=1}^{\infty} \frac{i^{n}}{n}
\]

Does it converge? In any case, prove your conjecture.

[gugo82]

Addendum: If the previous series converges, evaluate its sum.

[fu^2]

we should try to split the sum:

\( \displaystyle \sum_{{{n}\in{\mathbb{{{N}}}}}}{\frac{{{{i}}^{{n}}}}{{{n}}}}=\sum_{{{n}\in{\mathbb{{{N}}}}}}{\frac{{{{i}}^{{{2}{n}}}}}{{{2}{n}}}}+\sum_{{{n}\in{\mathbb{{{N}}}}}}{\frac{{{{i}}^{{{2}{n}+{1}}}}}{{{2}{n}+{1}}}}=\sum_{{{n}\in{\mathbb{{{N}}}}}}{\frac{{{{\left(-\right)}}^{{{n}}}}}{{{2}{n}}}}+{i}\sum_{{{n}\in{\mathbb{{{N}}}}}}{\frac{{{{\left(-\right)}}^{{{n}}}}}{{{2}{n}+{1}}}}={\frac{{{1}}}{{{2}}}}{\log{{\left\lbrace{2}\right\rbrace}}}+{i}\text{arctan}{\left({1}\right)} \)

since \( \displaystyle {\log{{\left\lbrace{2}\right\rbrace}}}=\sum_{{{n}\in{\mathbb{{{N}}}}}}{\frac{{{{\left(-\right)}}^{{n}}}}{{{n}}}} \) and \( \displaystyle \text{arctan}{\left({x}\right)}=\sum_{{{n}\in{\mathbb{{{N}}}}}}{{\left(-\right)}}^{{n}}{\frac{{{{x}}^{{{2}{n}+{1}}}}}{{{2}{n}+{1}}}} \)


it's ok?

[fu^2]

It's turn of that the core of this exercise regards the fact that \( \displaystyle \sum_{{{n}\in{\mathbb{{{N}}}}}}{\frac{{{{\left(-\right)}}^{{n}}}}{{{n}}}} \) converges.

It's well konw that the sum \( \displaystyle \sum_{{{i}\in{\mathbb{{{N}}}}}}{\frac{{{1}}}{{{i}}}} \) has the opposite behavior.

Then, it's natural ask himself ("chiedersi"?) how many "oscillations" are necessary to make converge this series.

Thus, let \( \displaystyle {\left\lbrace\xi_{{n}}\right\rbrace}_{{{n}\in{\mathbb{{{N}}}}}} \) be a sequence of IID r.v., with \( \displaystyle {\mathbb{{{P}}}}{\left(\xi_{{n}}={1}\right)}={1}-{\mathbb{{{P}}}}{\left(\xi_{{n}}=-{1}\right)}={p} \), with \( \displaystyle {p}\in{\left[{0},{1}\right]} \) and consider the series

\( \displaystyle \sum_{{{n}\in{\mathbb{{{N}}}}}}{\frac{{{{\left(-\right)}}^{{\xi_{{n}}}}}}{{{n}}}} \)

and I repeat the Paolo90's question:

Does it converge? In any case, prove your conjecture.

[gugo82]

Exercise. Consider the series
\[
\sum_{n=1}^{\infty} \frac{i^{n}}{n}
\]
Does it converge? In any case, prove your conjecture.

Addendum: If the previous series converges, evaluate its sum.

A solution using a power series argument.

Testo nascosto, fai click qui per vederlo

Let us consider the power series:
\[
\tag{1} \sum_{n=1}^\infty \frac{1}{n}\ z^n
\]
It is clear that the radius of convergence of (1) equals \(1\); for a well-known (to those who know it well) theorem of Picard the region of convergence of (1) is either the whole of \(\overline{D}(0;1)\) or \(\overline{D}(0;1)\setminus \{1\}\); but the series clearly does not converge in \(1\), therefore \(\sum_{n=1}^\infty \frac{1}{n}\ z^n\) converges in \(\overline{D}(0;1)\setminus \{1\}\).
In particular, the numerical series \(\sum_{n=1}^\infty \frac{1}{n}\ \imath^n\) converges for \(\imath \in \overline{D}(0;1)\setminus \{1\}\).

Now, let \(f(z)\) be the sum of \(\sum_{n=1}^\infty \frac{1}{n}\ z^n\); differentiating both \(f(z)\) and its series expansion one gets:
\[
f^\prime (z) = \sum_{n=0}^\infty z^n = \frac{1}{1-z}
\]
hence \(f(z)\) is an antiderivative of \(\frac{1}{1-z}\), in particular, the one whose real restriction takes the value \(0\) in \(0\) i.e.:
\[
f(z) = -\log (1-z)\; .
\]
Therefore:
\[
\sum_{n=1}^\infty \frac{1}{n}\ \imath^n =-\log (1-\imath)=-\frac{1}{2}\ \ln 2 +\frac{\pi}{4}\ \imath \; .
\]

[dissonance]

This is an answer to the original question by Paolo.

Testo nascosto, fai click qui per vederlo

Let \(S_n=\sum_1^n i^n / n\). We claim that this sequence is Cauchy. To see this fix \(n < m\) and observe that

\[S_m-S_n= \frac{i^{n+1}}{n+1} + \frac{i^{n+2}}{n+2}+\frac{i^{n+1}}{n+1} +\ldots+ \frac{i^{m}}{m} \]

We picture this sum in the complex plane as follows:

thus inferring the bound
\[\lvert S_m-S_n \lvert^2 \le \left\lvert \frac{i}{n+1}- \frac{1}{n+2}\right\rvert^2=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}\]
and the required Cauchy condition. So the given series converges. /////

[dissonance]

it's ok?

The mathematics is ok but English is not: you should say

is it ok?

It's turn of that ...

Turns out that...

Then, it's natural ask himself ("chiedersi"?)...

Sounds weird. I would say: we may naturally ask ourselves...
but I don't find it very satisfactory either. I hope someone may help us on this.

to make converge this series.

Wrong use of "to make". I would say: so that this series converges.

Thus, let \( \displaystyle {\left\lbrace\xi_{{n}}\right\rbrace}_{{{n}\in{\mathbb{{{N}}}}}} \) be a sequence of IID r.v., with \( \displaystyle {\mathbb{{{P}}}}{\left(\xi_{{n}}={1}\right)}={1}-{\mathbb{{{P}}}}{\left(\xi_{{n}}=-{1}\right)}={p} \), with \( \displaystyle {p}\in{\left[{0},{1}\right]} \) and consider the series

\( \displaystyle \sum_{{{n}\in{\mathbb{{{N}}}}}}{\frac{{{{\left(-\right)}}^{{\xi_{{n}}}}}}{{{n}}}} \)

and I repeat the Paolo90's question:

Does it converge? In any case, prove your conjecture.

And here we come to mathematics! Unfortunately I have no idea on how to approach this problem.

[gugo82]

Thus, let \( \displaystyle {\left\lbrace\xi_{{n}}\right\rbrace}_{{{n}\in{\mathbb{{{N}}}}}} \) be a sequence of IID r.v., with \( \displaystyle {\mathbb{{{P}}}}{\left(\xi_{{n}}={1}\right)}={1}-{\mathbb{{{P}}}}{\left(\xi_{{n}}=-{1}\right)}={p} \), with \( \displaystyle {p}\in{\left[{0},{1}\right]} \) and consider the series

\( \displaystyle \sum_{{{n}\in{\mathbb{{{N}}}}}}{\frac{{{{\left(-\right)}}^{{\xi_{{n}}}}}}{{{n}}}} \)

and I repeat the Paolo90's question:

Does it converge? In any case, prove your conjecture.

I think there are some kind of typos here...
In fact, if \(\xi_n=\pm 1\) with probability \(1\), then \((-1)^{\xi_n}=-1\) with the same probability making the series \(\sum (-1)^{\xi_n}/n\) diverge surely.

Thus, I think you should correct the statement in one of the following way: either:

[...] let \( \displaystyle {\left\lbrace\xi_{{n}}\right\rbrace}_{{{n}\in{\mathbb{{{N}}}}}} \) be a sequence of i.i.d. RVs, with \(\color{maroon}{\mathbb{P}(\xi_n=1)=p=1-\mathbb{P}(\xi_n=0)}\) and \( \color{maroon}{ p\in [0,1]}\) [...]

or:

[...] let \( \displaystyle {\left\lbrace\xi_{{n}}\right\rbrace}_{{{n}\in{\mathbb{{{N}}}}}} \) be a sequence of i.i.d. RVs, with \( \displaystyle {\mathbb{{{P}}}}{\left(\xi_{{n}}={1}\right)}={p}={1}-{\mathbb{{{P}}}}{\left(\xi_{{n}}=-{1}\right)} \), with \( \displaystyle {p}\in{\left[{0},{1}\right]} \) and consider the series
\[
\color{maroon}{\sum_{n\in\mathbb{N}}\frac{\xi_n}{n}}
\] [...]

Moreover, what do you mean when you ask "does the series converge"?
Are solvers supposed to look for the values of \(p\) which make the series converge? Or should they evaluate the probability the series converge for a given \(p\) (though I can't say if it's actually possible)? Or you mean something else, perhaps?
Please, clarify that.
Finally, if you already solved the problem, consider also to give some hints.


BTW, is this problem taken from a book?

[fu^2]

@dissonance: thank you! your correction are alway appreciated
Now we came back to the exercise:

I write the correct statment (I have change one of the hypothesis: We start to study the classical harmonic series with random signs: in this case \( \displaystyle {p}={\frac{{{1}}}{{{2}}}} \)).

Thus, let \( \displaystyle {\left\lbrace\xi_{{n}}\right\rbrace}_{{{n}\in{\mathbb{{{N}}}}}} \) be a sequence of IID RVs, with \( \displaystyle {\mathbb{{{P}}}}{\left(\xi_{{n}}={1}\right)}={1}-{\mathbb{{{P}}}}{\left(\xi_{{n}}=-{1}\right)}={p} \) and \( \displaystyle {p}={\frac{{{1}}}{{{2}}}} \). I consider the series

\( \displaystyle \sum_{{{n}\in{\mathbb{{{N}}}}}}{\frac{{\xi_{{n}}}}{{{n}}}} \)

and I "repeat" (a.s. ) the Paolo90's question:

Does it converge \( \displaystyle {\mathbb{{{P}}}} \) - a.s.?
And in \( \displaystyle {{L}}^{{1}}{\left({\mathbb{{{P}}}}\right)} \)?

I have done a typical error deriving from the use of copy&paste technique added at the stupidity of the writer.

BTW, is this problem taken from a book?

No, I take this problem from my exercise book of the last corse of Probability.
This is a nice application of the Martingale theory (in fact this ex. is strictly probabilistic).

Another way to approach this problem is using the "Kolmogorov's three series theorem", but in this case is not clear the solution... Using this theorem is equivalent to use the dark magic .

If none want to prove to provide the solution, i will write the result

Testo nascosto, fai click qui per vederlo

Hint: this problem is equivalent to show the following result (that i write in its general form):

Let \( \displaystyle {M}={\left\lbrace{M}_{{N}}\right\rbrace}_{{{N}\in{\mathbb{{{N}}}}}} \) be a martingale in \( \displaystyle {{L}}^{{2}} \). Then \( \displaystyle {\left\lbrace\lt{M}\gt_{{\infty}}\lt+\infty\right\rbrace}\subset{\left\lbrace\exists\lim_{{{n}\to+\infty}}{M}_{{n}}\in{\left(-\infty,+\infty\right)}\right\rbrace} \)

in our problem, you can assume that \( \displaystyle {M}_{{N}}={\sum_{{{n}={1}}}^{{N}}}{a}_{{n}}{X}_{{n}} \) and \( \displaystyle \lt{M}\gt_{{N}}={\sum_{{{n}={1}}}^{{N}}}{{a}_{{n}}^{{2}}} \), with \( \displaystyle {a}_{{n}}={{n}}^{{-{1}}} \) and \( \displaystyle {X}_{{i}} \) are the sequence of RVs IID with \( \displaystyle {\mathbb{{{E}}}}{\left({X}_{{i}}\right)}={0} \).

But if you don't know something about the martingale theory, this problem can be very hard...

Testo nascosto, fai click qui per vederlo

if someone want to devote to the black magic: http://en.wikipedia.org/wiki/Kolmogorov ... es_theorem

The next step is study this problem with \( \displaystyle {p}\in{\left[{0},{1}\right]} \).

[j18eos]

Testo nascosto, fai click qui per vederlo

...that I write in its general form...

But no-one know that "io" in English is I!

Sorry me for my bad English!