Possible troubles in Shelah's book "Cardinal Arithmetic" I found some possible troubles in Observation 5.3(7) in the Chapter II of the Shelah's book "Cardinal Arithmetic" (page 86). For convenience, I quote the result and the proof in the book here (together with Observation 5.3(4) that is used in the proof of 5.3(7)):

Observation 5.3
(4) If $\lambda > \kappa \left( \geq \theta > \sigma \right)$, $\sigma$ regular then
  $$
\operatorname{cov} \left( \lambda , \kappa , \theta , \sigma \right) =
\sum_{\mu \in \left[ \kappa , \lambda \right]}
\operatorname{cov} \left( \mu , \mu , \theta , \sigma \right) .
$$
(7) If $\lambda \geq \kappa \geq \theta > \sigma = \operatorname{cf} (\sigma)$,
  $\operatorname{cf} (\kappa) \geq \theta$, $\lambda_{0} = \lambda$,
  $$
\lambda_{n+1} = \sup
\left\lbrace
\operatorname{cov} \left( \mu , \mu , \tau^{+} , \tau \right) :
\kappa \leq \mu \leq \lambda_{n} , \operatorname{cf} (\mu) = \tau \in \left[ \sigma, \theta \right)
\right\rbrace
$$
  then
  $$
\operatorname{cov} \left( \lambda , \kappa , \theta , \sigma \right) \leq
\bigcup_{n < \omega} \lambda_{n} .
$$
Proof:
  7) Let $\chi$ be regular large enough, by induction on $n$ choose $N_{n} \prec \left( H(\chi) , \in \right)$
  of cardinality $\lambda_{n}$ such that
  $$
\left\lbrace N_{0}, \ldots , N_{n-1}, \lambda , \kappa , \theta , \sigma \right\rbrace \cup
\left( \lambda_{n} + 1 \right) \subseteq N_{n} ,
$$
  and
  $$
\mathcal{P}_{n} =
\left\lbrace A \in N_{n} : \left| A \right| < \kappa , A \subseteq \lambda \right\rbrace
$$
  and $\mathcal{P}_{\omega} = \bigcup_{n < \omega} \mathcal{P}_{n}$.
  Suppose $X \subseteq \lambda$, $\left| X \right| < \theta$ and for no
  $\mathcal{P} \subseteq \mathcal{P}_{\omega}$, $\left| \mathcal{P} \right| < \sigma$ is
  $X \subseteq \bigcup_{A \in \mathcal{P}} A$; let $I$ be the $\sigma$-complete ideal on $X$
  generated by $\left\lbrace X \cap A : A \in \mathcal{P}_{\omega} \right\rbrace$,
  so $X \notin I$. Let
  $$
\theta_{n} = \min
\left\lbrace
\left| \mathcal{P} \right| : \mathcal{P} \subseteq \mathcal{P}_{n} ,
\bigcup_{A \in \mathcal{P}} A \cap X \notin I
\right\rbrace ;
$$
  now $\theta_{n} \leq \left| X \right| < \theta$ and
  $\operatorname{cf} \left( \theta_{n} \right) \geq \sigma$ and $\theta_{n+1} < \theta_{n}$
  (use 5.3(4) applied to
  $\operatorname{cov} \left( \lambda_{n} , \kappa , \theta_{n} , \theta_{n} \right)$), contradiction.
  $\square$

First, it is easy to show that $\theta_{n} \leq \left| X \right| < \theta$,
$\operatorname{cf} \left( \theta_{n} \right) \geq \sigma$ and
$\theta_{n+1} \leq \theta_{n}$. To show that $\theta_{n+1} < \theta_{n}$,
we must apply Observation 5.3(4) to
$\operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right)$
($\operatorname{cov} \left( \lambda_{n} , \kappa , \theta_{n} , \theta_{n} \right)$ is an "uninteresting" covering number: $\operatorname{cov} \left( \lambda_{n} , \kappa , \theta_{n} , \theta_{n} \right) \leq \lambda_{n}$ - consider the family ${[\lambda_{n}]}^1$).
Now, the major difficulty is that we need
$\operatorname{cf} (\theta_{n}) = \theta_{n}$
to use 5.3(4), but I don't see how to prove this.
Supposing $\operatorname{cf} (\theta_{n}) = \theta_{n}$, I wrote a detailed proof for 5.3(7): applying 5.3(4), we can show that
$$
\operatorname{cov}
\left( \left| \mathcal{P}_{n} \right| , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq
\operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq \lambda_{n+1} = \left| N_{n+1} \right| ,
$$
and then use this to "cover" the set in
${[\mathcal{P}_{n}]}^{\theta_{n}}$ that testifies the definition of
$\theta_{n}$, with a set in ${[\mathcal{P}_{n+1}]}^{< \theta_{n}}$.
My questions are: (answers specific to the case $\sigma = \aleph_0$ are welcome too)
1) Is possible to prove that $\operatorname{cf} (\theta_{n}) = \theta_{n}$?
2) Is possible to prove that
$\operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq \lambda_{n+1}$
when $\operatorname{cf} (\theta_{n}) < \theta_{n}$?
3) Is possible to prove that the sequence
$(\theta_{n})$
is not eventually constant?
Some observations:
i) This is my proof that
$\operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq \lambda_{n+1}$,
if $\operatorname{cf} (\theta_{n}) = \theta_{n}$:
$$
\operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) =
\sum_{\mu \in \left[ \kappa , \lambda_{n} \right]}
\operatorname{cov} \left( \mu , \mu , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq
$$
$$
\lambda_{n} \cdot \sup
\left\lbrace
\operatorname{cov} \left( \mu , \mu , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) :
\kappa \leq \mu \leq \lambda_{n}
\right\rbrace .
$$
Now, it is easy to show that
$\operatorname{cov} \left( \mu , \mu , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) = \operatorname{cf} (\mu)$
when $\operatorname{cf} (\mu) \neq \theta_{n}$. Thus,
$$
\sup
\left\lbrace
\operatorname{cov} \left( \mu , \mu , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) :
\kappa \leq \mu \leq \lambda_{n}
\right\rbrace \leq
$$
$$
\lambda_{n} + \sup
\left\lbrace
\operatorname{cov} \left( \mu , \mu , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) :
\kappa \leq \mu \leq \lambda_{n} , \operatorname{cf} (\mu) = \theta_{n}
\right\rbrace =
$$
$$
\lambda_{n} + \sup
\left\lbrace
\operatorname{cov} \left( \mu , \mu , {\left( \operatorname{cf} (\mu) \right)}^{+} , \operatorname{cf} (\mu) \right) :
\kappa \leq \mu \leq \lambda_{n} , \operatorname{cf} (\mu) = \theta_{n}
\right\rbrace \leq
$$
$$
\lambda_{n} + \sup
\left\lbrace
\operatorname{cov} \left( \mu , \mu , {\left( \operatorname{cf} (\mu) \right)}^{+} , \operatorname{cf} (\mu) \right) :
\kappa \leq \mu \leq \lambda_{n} , \sigma \leq \operatorname{cf} (\mu) < \theta
\right\rbrace =
$$
$$
\lambda_{n} + \lambda_{n+1} = \lambda_{n+1},
$$
hence
$$
\operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq
\lambda_{n} \cdot \lambda_{n+1} = \lambda_{n+1} .
$$
$\square$
ii) It is more convenient to define
$$
\lambda_{n+1} = \lambda_{n} + \sup
\left\lbrace
\operatorname{cov} \left( \mu , \mu , {\left( \operatorname{cf} (\mu) \right)}^{+} , \operatorname{cf} (\mu) \right) :
\kappa \leq \mu \leq \lambda_{n} , \sigma \leq \operatorname{cf} (\mu) < \theta
\right\rbrace .
$$
iii) If $\eta$ is any cardinal with
$\sigma \leq \operatorname{cf} (\eta) = \eta < \theta$,
then we can show that
$$
\operatorname{cov} \left( \lambda_{n} , \kappa , \eta^+ , \eta \right) \leq
\lambda_{n+1}
$$
(same argument of my observation (i)).
iv) The proof works when
$\sigma < \theta \leq \aleph_{\sigma}$,
since $\sigma \leq \operatorname{cf} (\xi) \leq \xi < \aleph_{\sigma}$ implies
$\operatorname{cf} (\xi) = \xi$.
v) When $\sigma = \aleph_0$ (the case that interests me): if we define
$$
\theta_{\omega} = \min
\left\lbrace
\left| \mathcal{P} \right| : \mathcal{P} \subseteq \mathcal{P}_{\omega} ,
\bigcup_{A \in \mathcal{P}} A \cap X \notin I
\right\rbrace ,
$$
then we can show that $\theta_{\omega}$ is regular. If the sequence $(\theta_{n})$
is eventually constant, then there is $k \in \omega$
such that $\theta_{n} = \theta_{k}$ for any $n \geq k$, and
$$
\aleph_0 = \sigma \leq \operatorname{cf} (\theta_{\omega}) = \theta_{\omega}
\leq \operatorname{cf} (\theta_{k}) < \theta_{k} < \theta .
$$
A more elaborated argument shows that
$\operatorname{cf} (\theta_{k}) = \aleph_0$.
Hence,
$$
\aleph_0 = \sigma = \operatorname{cf} (\theta_{\omega}) = \theta_{\omega}
= \operatorname{cf} (\theta_{k}) < \theta_{k} < \theta .
$$
vi) Considering observation (iv), everything works for
$$
\operatorname{cov}
\left( \aleph_{\omega + \omega} , \aleph_{\omega + 1} , \aleph_{\omega} , \aleph_0 \right) .
$$
Does the same occur with
$$
\operatorname{cov}
\left( \aleph_{\omega + \omega} , \aleph_{\omega + 1} , \aleph_{\omega + 1} , \aleph_0 \right) ?
$$
 A: I found a way to prove in detail a slightly weaker version of the Observation 5.3(7) cited above:

Definition
  Given an ordinal $\alpha$, and any cardinals
  $\mu$, $\eta$, $\theta$ and $\sigma$,
  with $\eta \geq \theta$, define
  $$
\operatorname{s}_{\mu , \eta , \theta , \sigma} =
\sup
\left\lbrace
\operatorname{cov} \left( \nu , \nu , {\left( \operatorname{cf} (\nu) \right)}^{+} , \operatorname{cf} (\nu) \right) :
\eta \leq \nu \leq \mu , \sigma \leq \operatorname{cf} (\nu) < \theta
\right\rbrace
$$
  and
  $$
\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \alpha \right) =
\left\lbrace
\begin{array}{ll}
\mu , & \textrm{if } \alpha = 0 ; \\
\sup \left\lbrace
\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \beta \right) : \beta < \alpha
\right\rbrace , & \textrm{if } \alpha \textrm{ is a limit ordinal;} \\
\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \beta \right) +
\operatorname{s}_{\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \beta \right) , \eta , \theta , \sigma} ,
& \textrm{if } \alpha = \beta + 1 .
\end{array}
\right.
$$
Proposition
  Let $\mu$, $\eta$, $\theta$ and $\sigma$ be cardinals such that
  $$
\mu \geq \eta = \operatorname{cf} (\eta) \geq \theta > \sigma = \operatorname{cf} (\sigma) \geq \aleph_{0} .
$$
  Then,
  $$
\operatorname{cov} \left( \mu , \eta , \theta , \sigma \right) \leq
\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \omega^{2} \right) .
$$

I needed this proposition in the following paper, where I included, for the reader's convenience, this proposition as Proposition 3.17, with a detailed proof.

A.M.E. Levi. Reflection for lindelöf degree and inaccessible cardinals,
  Acta Mathematica Hungarica 144(1) (2014), 182-195.
  doi:10.1007/s10474-014-0442-0

A manuscript version can be viewed here.
When $\sigma = \aleph_{0}$, we show that
$$
\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \omega \right) <
\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \omega + 1 \right) \leq
\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \omega^{2} \right) ;
$$
but when $\sigma > \aleph_{0}$, we prove that
$$
\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \omega \right) =
\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \alpha \right)
$$
for every ordinal $\alpha > \omega$. Hence, we can prove the following result, which is essentially the book's one, with the additional hypotheses
$\kappa = \operatorname{cf} (\kappa)$ and
$\sigma > \aleph_{0}$:
Proposition
If
$\lambda \geq \kappa = \operatorname{cf} (\kappa) \geq \theta > \sigma =
\operatorname{cf} (\sigma) > \aleph_{0}$,
$\lambda_{0} = \lambda$,
$$
\lambda_{n+1} = \lambda_{n} +
\sup
\left\lbrace
\operatorname{cov} \left( \mu , \mu , {\left( \operatorname{cf} (\mu) \right)}^{+} , \operatorname{cf} (\mu) \right) :
\kappa \leq \mu \leq \lambda_{n} , \sigma \leq \operatorname{cf} (\mu) < \theta
\right\rbrace ,
$$
then
$$
\operatorname{cov} \left( \lambda , \kappa , \theta , \sigma \right) \leq
\bigcup_{n < \omega} \lambda_{n} .
$$
