Exercise 2.7.9

Question

[Ratio Test]
  Given a series $\sum_{n=1}^{\infty} a_{n}$ with $a_{n} 
eq 0$, the Ratio Test states that if $\left(a_{n}ight)$ satisfies
  $$
  \lim \left|\frac{a_{n+1}}{a_{n}}ight|=r<1
  $$
  then the series converges absolutely.

\begin{enumerate}[label=(\alph*)] 
  \item Let $r^{\prime}$ satisfy $r<r^{\prime}<1$. Explain why there exists an $N$ such that $n \geq N$ implies $\left|a_{n+1}ight| \leq\left|a_{n}ight| r^{\prime}$.
  \item Why does $\left|a_{N}ight| \sum\left(r^{\prime}ight)^{n}$ converge?
  \item Now, show that $\sum\left|a_{n}ight|$ converges, and conclude that $\sum a_{n}$ converges.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item We are given $$ \left|\frac{a_{n+1}}{a_n} - r ight| < \epsilon $$ Since $1 > r' > r$ we can set $\epsilon = r' - r$ meaning the neighborhood $$ \frac{a_{n+1}}{a_n} \in (r-\epsilon, r+\epsilon) = (2r-r', r') $$ Is all less then $r'$ meaning $$ \left|\frac{a_{n+1}}{a_n} ight| \le r' \implies |a_{n+1}| \le r'|a_n| $$ \item Let $N$ be large enough that for $n > N$ we have $|a_{n+1}| \le |a_n|r'$. Applying this multiple times gives $|a_n| \le (r')^{n-N} |a_N|$ which gives $$ |a_N| + |a_{N+1}| + \dots + |a_n| \le |a_N| + r'|a_N| + \dots + (r')^{n-N}|a_N| $$ Factoring out $|a_N|$ and writing with sums gives $$ \sum_{k=N}^n |a_k| \le |a_N|\sum_{k=0}^{n-1} (r')^k $$ Which converges as $n o \infty$ since $|r'| < 1$ and $|a_N|$ is constant. Implying $\sum_{k=N}^n |a_k|$ converges and thus $\sum_{k=1}^n |a_k|$ also converges since we only omitted finitely many terms. \item See (b) \end{enumerate}

Exercise 2.7.9

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Exercise 2.7.9

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