A couple of difficulties in Tauber theory lecture notes of prof. Yum-Tong Siu Reading trough the lecture notes on Tauber theory of prof. Yum-Tong Siu I am a bit off right at the beginning. If someone could clarify the follow two steps in his proofs of Tauber's original 1897 theorem and the theorem of Hardy-Littlewood of 1914:
On p. 4 : We have assumed $n \geq N_0$ to be large enough so that the Tauber condition says $|na_n| < \epsilon$ (for arbitrary $\epsilon > 0$). How does it follow in the establishing of the second $\epsilon$-bound (for $|\sum_{n=0}^{N}a_n(1-x^n)|$) that
$$ \sum_{n=0}^{N}|na_n|(1-x) < N\epsilon(1-x)$$?
On  p. 5 : The very last line, why is $k+1 = \int_{t=0}^{1} t^k dt$? He continues using this throughout the remaining of the argument, so I guess it is not a typo. I am rather confused and a bit ashamed that I apparently do not understand calculus.
Both of these things seem flatly incorrect to me. But I am a noob and this document appears to have been written by an established professor (presumable, because his name is on it and I found it originally on his Harvard webpage). So I guess these ought to be typos but I do not see how to recover from them easily. Any thoughts?
 A: The estimate on p. 4 does not depend on the properties of $N_0$, only on the Tauberian condition $n|a_n|\to 0$ as $n\to\infty$. 
Let $b_n=n|a_n|$. Then $b_n\to 0$, and hence the averages ${1\over N}(b_1+\cdots+b_N)$ will $\to 0$ as well. So if $N$ is large enough, ${1\over N}(b_1+\cdots+b_N)<\epsilon$, or $\sum_{k=0}^Nn|a_n|<N\epsilon$.
A: *

*Write 
$$\sum_{n=0}^N \,\lvert n a_n \rvert= \sum_{n=0}^{N_0-1} \,\lvert n a_n \rvert + \sum_{n=N_0}^N \,\lvert n a_n \rvert.$$


The first sum is a constant $C$, and the second sum can be controlled with the Tauber condition, so we get
$$
\sum_{n=0}^N \,\lvert n a_n \rvert \leq C+N\epsilon.
$$
Note that this bound is a little weaker than what you're asking for. We can still conclude the argument on page 4 by using this weaker bound : we have
$$
\left\lvert \sum_{n=0}^N a_n(1-x^n)\right\rvert \leq (C+N\epsilon)(1-x)\leq C(1-x)+\epsilon.
$$
Recall that $C$ is a constant that only depends on $\epsilon$ : there exists $\delta'>0$ such that when $x>1-\delta'$, $C(1-x)<\epsilon$. Now when $x>\max(1-\delta,1-\delta')$, one has
$$\left\lvert \sum_{n=N+1}^\infty a_nx^n\right\rvert \leq \epsilon
$$
and
$$
\left\lvert \sum_{n=0}^N a_n(1-x^n)\right\rvert \leq 2\epsilon.
$$


*

*We have $\int_0^1 t^k \,\mathrm{d}t=[t^{k+1}/(k+1)]_0^1=1/(k+1)$. The argument on p.5 is still correct : from


$$\sum_{n=0}^\infty s_n(1-x^{k+1})x^n(x^n)^k \to A,
$$
multiplying both sides by $1/(k+1)$ yields
$$\frac{1-x^{k+1}}{(k+1)(1-x)}(1-x)\sum_{n=0}^\infty s_nx^n(x^n)^k \to A\int_0^1 t^k \,\mathrm{d}t.
$$
which is what is needed.
