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| Title of object: |
partial summation |
| Canonical Name: |
PartialSummation |
| Type: |
Theorem |
| Created on: |
2006-12-17 17:09:19 |
| Modified on: |
2007-08-12 07:42:52 |
| Classification: |
msc:40D05, msc:40A05 |
Preamble:
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Content:
Partial summation is a way to estimate $\sum_1^{\infty}a_i b_i$ given sequences of complex numbers $a_i, b_i$ under certain circumstances.
\begin{thm}\emph{(Partial Summation Theorem)}\newline
Let $\{a_i\},\{b_i\}$ be sequences of complex numbers. Suppose the partial sums of the $a_i$ are bounded in magnitude by $h$, that $\sum_1^{\infty} |b_i-b_{i+1}|$ converges, and that $\lim_{i\to\infty} b_i=0$. Then $\sum_1^{\infty} a_i b_i$ converges, and
\[\left|\sum_1^{\infty}a_i b_i\right|\leq h\sum_1^{\infty}|b_i-b_{i+1}|\]
\end{thm}
\textbf{Proof.} We will show $\sum_m^n a_ib_i\to 0$ as $m,n\to\infty$.
Let $A_k=\sum_m^k a_i$. Then
\begin{eqnarray*}
&a_mb_m+\ldots+a_nb_n&=A_mb_m+(A_{m+1}-A_m)b_{m+1}+\ldots+(A_n-A_{n-1})b_n\\
&&=A_m(b_m-b_{m+1})+A_{m+1}(b_{m+1}-b_{m+2})+\ldots+A_{n-1}(b_{n-1}-b_n)+A_nb_n
\end{eqnarray*}
But by the triangle inequality, since $|\sum_1^m a_i|\leq h$ and $|\sum_1^n a_i|\leq h$, it follows that $|A_k|\leq 2h$ for each $m\leq k\leq n$.
Thus
\[\left|\sum_m^n a_ib_i\right|\leq 2h(|b_m-b_{m+1}|+\ldots+|b_n|)\]
by the triangle inequality. However, as $m,n\to \infty$, the right-hand \PMlinkescapetext{side} goes to zero by assumption. So by the Cauchy criterion for convergence, $\sum_1^{\infty}a_i b_i$ converges. Now,
\begin{eqnarray*}
&\left|\sum_1^n a_i b_i\right|&\leq \sum_1^n |a_i||b_i| \leq \sum_1^n h|b_i| =h\sum_1^n |b_i|\\
&&\leq h(|b_1-b_2|+\ldots+|b_{n-1}-b_n|+|b_n|) \leq h\sum_1^{\infty} |b_i-b_{i+1}|+h|b_n|
\end{eqnarray*}
But as $n\to\infty$, $h|b_n|\to 0$, and the result follows.
\begin{thm}\emph{(Partial Summation for Real Sequences)}\newline
Let $\{a_i\}$ be a sequence of complex numbers. Suppose the partial sums are bounded in magnitude by $h$. Let $\{b_i\}$ be a sequence of decreasing positive real numbers such that $\lim_{i\to\infty} b_i=0$. Then $\sum_1^{\infty} a_ib_i$ converges, and $|\sum_1^{\infty} a_ib_i|\leq hb_1$.
\end{thm}
\textbf{Proof.} Clear, since $|b_i-b_{i+1}|=b_i-b_{i+1}$.
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