Exercise 8.1

Question

ewcommand\seqfin[2]{\left(#1_1, \ldots, #1_{#2}ight)}

ewcommand\seqinf[1]{\left(#1_1, #1_2, \ldotsight)}
\def{ho}
Let $\seqinf{b}$ be an infinite sequence of real numbers.
The sum $\sum_{k=1}^{n} b_k$ is defined by induction as follows:
\begin{align*}
\sum_{k=1}^{n} b_k &= b_1 \hspace{1cm} 	ext{for $n=1$,} \
\sum_{k=1}^{n} b_k &= \left(\sum_{k=1}^{n-1} b_kight) + b_n \hspace{1cm} 	ext{for $n > 1$.}
\end{align*}
Let $A$ be the set of real numbers; choose $$ so that Theorem~8.4 applies to define this sum rigorously.
We sometimes denote the sum $\sum_{k=1}^{n} b_k$ by the symbol $b_1 + b_2 + \cdots + b_n$.

Dan Whitman · Accepted Answer

ewcommand\seqfin[2]{\left(#1_1, \ldots, #1_{#2}ight)}

ewcommand\seqinf[1]{\left(#1_1, #1_2, \ldotsight)}
\def{ho}
\defest{estriction}

For a function $f: \left\{1, \ldots, might\} 	o A$, define $(f) = f(m) + b_{m+1}$.
For clarity, denote the sum function by $s : {{\mathbb{Z}}_+} 	o A$ so that $s(n) = \sum_{k=1}^{n} b_k$.
Then by Theorem~8.4 there is a unique $s : {{\mathbb{Z}}_+} 	o A$ such that
\begin{align*}
s(1) &= b_1\,, \
s(n) &= (s est \left\{1, \ldots, n-1ight\}) \hspace{1cm} 	ext{for $n>1$.}
\end{align*}
Then we clearly have that $\sum_{k=1}^{1} b_k = s(1) = b_1$ and
\begin{gather*}
\sum_{k=1}^{n} b_k = s(n) = (s est \left\{1, \ldots, n-1ight\}) = s(n-1) + b_{(n-1)+1} = \sum_{k=1}^{n-1} b_k + b_n
\end{gather*}
for $n>1$ as desired.

Exercise 8.1

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Exercise 8.1

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