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REAL ANALYSIS

PARTIAL SUMMATION (ABEL SUMMATION)

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\begin{description}

\item[]
As part of the analogy existing between summation and integration,
partial summation corresponds to integration by parts.

If $u\leq v$ and $\ds s_m=\sum_{r=u}^ma_r$ then we have the
identity

\begin{equation}
\sum_{m=u}^v a_mb_m=b_{v+1}s_V+\sum_{m=u}^vs_m(b_m-b_{m+1})
\end{equation}

\item[Proof]

\begin{eqnarray*}
\sum_{m=u}^v a_mb_m&=&\sum_{m=u}^v(s_m-s_{m-1})b_m\\
&=&b_{v+1}s_v+\sum_{m=u}^vs_m(b_m-b_{m+1})
\end{eqnarray*}

with the convention that empty sums are zero.

\item[Abel's lemma]
With the above notation, suppose that $\{b_m\}$ is a positive
monotonic decreasing sequence, and that $|s_m|\leq M$ for all $m$.

Then

$$\left|\sum_{m=u}^va_mb_m\right|\leq Mb_v$$

\item[Proof]

\begin{eqnarray*}
\left|\sum_{m=u}^va_mb_m\right|&=&
\left|\sum_{m=u}^vs_m(b_m-b_{m+1})+s_vb_{v+1}\right|\\
&\leq&\sum_{m=u}^v|s_m|(b_m-b_{m+1})+|s_v|b_{v+1}\\
&\leq&M\left[\sum_{m=u}^v(b_m-b_{m+1})+b_{v+1}\right]\\ &=&Mb_0
\end{eqnarray*}

\item[Theorem 6 Dirichlet's test]
Suppose that $\phi_n$ is a monotonic decreasing sequence
converging to zero, and that $\sum a_n$ is a series with bounded
partial sums. Then $\ds\sum_{n=1}^\infty a_n\phi_n$ is convergent.

\item[Proof]
$\ds\left|\sum_{m=1}^n a_m\right|<K$ for all $n$

$$\left|\sum_{m=0}^v
a_m\right|=\left|\sum_1^va_m-\sum_1^{u-1}a_m\right|\leq\left|\sum_1^va_m\right|+\left|\sum_1^{u-1}a_m\right|<2K.$$

Given $\varepsilon>0,\ \exists$ a natural number
$N=N(\varepsilon)|phi_v<\frac{\varepsilon}{2K}$ for all $u\geq N$.
By Abel's Lemma, therefore,
$\ds\left|\sum_{m=0}^va_m\phi_m\right|\leq2K\left(\frac{\varepsilon}{2K}\right)=\varepsilon$
whenever $v\geq u\geq N\Rightarrow\sum a_n\phi_n$ converges by
general principle of convergence.

\item[Theorem 7 Abel's Test]
Suppose that $\phi_n$ is a monotonic sequence converging to a
finite limit. Let $\sum a_n$ be a convergent series. Then
$\ds\sum_{n=1}^\infty a_n\phi_n$ is convergent.

\item[Proof]

\begin{enumerate}

\item
Suppose $\phi_n$ is monotonic decreasing and $z\\phi_n\rightarrow
l$ as $m\to\infty$ therefore $\psi_n$ is decreasing and
$\psi_n=\phi_n-l\to 0$ as $n\to\infty$. Therefore by Direchlet's
test $\sum a_n\psi_n$ converges. Write

\begin{eqnarray*}
\Psi&=&\lim_{m\to\infty}\sum_{r=1}^ma_n(\phi_n-l)\\
&=&\lim_{m\to\infty}\sum_{n=1}^ma_n\phi_n-l\sum_1^\infty a_n
\end{eqnarray*}

Therefore $\sum_{a=1}^\infty a_n\phi_n=\Psi-l\sum_1^\infty a_n$.

\item
Suppose $\phi_n$ is monotonic increasing and $\phi_n\to L$ as
$m\to\infty$. Write $\psi'_n=l-\phi_n\ \psi'_n$ is increasing and
$\psi'_n\to0$. Therefore as before $\sum a_n\phi_n$ converges.

\end{enumerate}

\item[Theorem 8 Root Test]
The series $\sum u_n$ converges or diverges according as
$\overline{\lim}(u_n)^\frac{1}{n}$ is greater than or less than
one.

\item[Proof]

\begin{enumerate}

\item
Suppose
$\ds\overline{\lim}_{n\to\infty}(u_n)^\frac{1}{n}=\alpha<1$.
Choose $\beta|\alpha<\beta<1$. Take $\varepsilon=\beta-\alpha>0$.
>From the property of the upper limit, $\exists
m=m(\beta)|(u_n)^\frac{1}{n}<\beta$ for all $n\geq m$ so
$u_n<\beta^n$ for all $n\geq m$. Therefore $\sum u_n$ converges by
comparison with $\sum\beta^n$.

\item
Suppose $\overline{\lim}_{n\to\infty}(u_n)^\frac{1}{n}=\alpha>1$
then $(u_n)^\frac{1}{n}>1$ for an infinity of $n$ therefore
$u_n>1$ for an infinity of $n$ therefore $u_n\not\to0$ as
$n\to\infty$ therefore $\sum u_n$ diverges.

\end{enumerate}

\item[Theorem 9]
$\exists$ a number $R$ such that the power series $\sum a_nz^n$
converges absolutely for $|z|<R$ and diverges for $|z|>R$, and
$\ds R^{-1}=\overline{\lim}_{n\to\infty}|a_n|^\frac{1}{n}$, with
the appropriate conventions when RHS=0 or $+\infty$.

\item[Proof]

\begin{description}

\item[(i)]
if $|a_n|^\frac{1}{n}\to0$ as $n\to\infty\
|a_nz^n|^\frac{1}{n}=|a_n|^\frac{1}{n}|z|\to0$ as $n\to\infty$ for
all $z$ therefore by Root rest $\sum|a_nz^n|$ converges.

\item[(ii)]
If $\overline{\lim}|a_n|^\frac{1}{n}=\infty$ the power series does
not converge for $z\neq0$ since
$\overline{\lim}|a_nz^n|^\frac{1}{n}=\overline{\lim}|a_n|^\frac{1}{n}R=+\infty$.

\item[(iii)]
If $\overline{\lim}|a_n|^\frac{1}{n}$ is finite and non-zero, we
write it equal to $\frac{1}{R}\ R>0\ \
\overline{\lim}|a_nz^n|^\frac{1}{n}=\frac{|z|}{R}$. Hence by root
test, the series converges or diverges according as $|z|<R$ or
$|z|>R$. $R$ is called the radius of convergence.

$R^{-1}=\overline{\lim}|a_n|^\frac{1}{n}$ with, conventionally,
$R=0$ if RHS$=+\infty$ and $R=\infty$ is RHS=0.

\end{description}

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