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Equivalent condition for convergent partial products.
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Is the folllowing argument correct? All cited theorems are presented below.
Proposition. Show, in general, that the sequence of partial products of the following series converges if
and only if $sum_n=1^inftya_n$ converges. (The inequality $1 +
xleq 3^x$ for positive $x$ will be useful in one direction.)
$$prod_n=1^infty(1+a_n) = (1+a_1)(1+a_2)(1+a_3)cdots,text text textwhere a_nge 0.$$
Proof. $(Rightarrow)$. Assume that the sequence $s_n = prod_k=1^n(1+a_k)$ of partial products converges, then by theorem $ textbf2.3.2$ there exists an $M>0$ such that $s_nleq M,forall ninmathbfN$. We now show that
$$sum_k=1^na_kleq prod_k=1^n(1+a_k),forall ninmathbfN$$
The base case is trivial. Now let $kinmathbfN$ and assume that $sum_j=1^ka_jleq prod_j=1^k(1+a_j)$, consequently $sum_j=1^k+1a_j = sum_j=1^ka_j +a_k+1leq prod_j=1^k(1+a_j)+a_k+1$.
Now assume that $a_k+1(1-prod_j=1^k(1+a_j))>0$, from hypothesis $a_k+1ge 0$, but then $prod_j=1^k(1+a_j)< 1$, a contradiction since by hypothesis $(1+a_n)ge 1,forall ninmathbfN$, thus $a_k+1leq a_k+1prod_j=1^k(1+a_j)$ but then by inductive hypothesis we have
$$sum_j=1^k+1a_j = sum_j=1^ka_j +a_k+1$$
$$leqprod_j=1^k(1+a_j)+a_k+1leq prod_j=1^k(1+a_j)+ a_k+1prod_j=1^k(1+a_j) = prod_j=1^k+1(1+a_j)$$
completing the induction. It is now apparent that $sum_k=1^na_kleq M,forall ninmathbfN$, and since $a_nge 0$ it follows that the corresponding sequence of partial sums is increasing, the series $sum_k=1^inftya_k$ is then convergent by theorem $textbf2.4.1$.
$(Leftarrow)$. Now assume that the sequence $b_n = sum_j=1^na_j$ of partial products is convergent, again by theorem $textbf2.3.2$ there exists a $M>0$ such that $b_nleq M,forall ninmathbfN$. We know that $(1+x)<3^x$ when $xge 0$ consequently
$$prod_j=1^k(1+a_j)<3^sum_j=1^ka_jleq 3^M$$
In summary then $prod_k=1^n(1+a_k)leq 3^M,forall ninmathbfN$, again since $a_nge 0,forall ninmathbfN$ it follows that corresponding sequence of partial products is increasing, appealing to theorem $textbf2.4.1$ implies that
$prod_n=1^infty(1+a_n)$ is convergent.
$blacksquare$
Note:
$(2.4.1):$ Every bounded montonic sequence converges.
$(2.3.2):$ Every convergent sequence is bounded.
real-analysis sequences-and-series proof-verification
asked Jul 29 at 20:18
Atif Farooq
2,7352824
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up vote
0
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Is the folllowing argument correct? All cited theorems are presented below.
Proposition. Show, in general, that the sequence of partial products of the following series converges if
and only if $sum_n=1^inftya_n$ converges. (The inequality $1 +
xleq 3^x$ for positive $x$ will be useful in one direction.)
$$prod_n=1^infty(1+a_n) = (1+a_1)(1+a_2)(1+a_3)cdots,text text textwhere a_nge 0.$$
Proof. $(Rightarrow)$. Assume that the sequence $s_n = prod_k=1^n(1+a_k)$ of partial products converges, then by theorem $ textbf2.3.2$ there exists an $M>0$ such that $s_nleq M,forall ninmathbfN$. We now show that
$$sum_k=1^na_kleq prod_k=1^n(1+a_k),forall ninmathbfN$$
The base case is trivial. Now let $kinmathbfN$ and assume that $sum_j=1^ka_jleq prod_j=1^k(1+a_j)$, consequently $sum_j=1^k+1a_j = sum_j=1^ka_j +a_k+1leq prod_j=1^k(1+a_j)+a_k+1$.
Now assume that $a_k+1(1-prod_j=1^k(1+a_j))>0$, from hypothesis $a_k+1ge 0$, but then $prod_j=1^k(1+a_j)< 1$, a contradiction since by hypothesis $(1+a_n)ge 1,forall ninmathbfN$, thus $a_k+1leq a_k+1prod_j=1^k(1+a_j)$ but then by inductive hypothesis we have
$$sum_j=1^k+1a_j = sum_j=1^ka_j +a_k+1$$
$$leqprod_j=1^k(1+a_j)+a_k+1leq prod_j=1^k(1+a_j)+ a_k+1prod_j=1^k(1+a_j) = prod_j=1^k+1(1+a_j)$$
completing the induction. It is now apparent that $sum_k=1^na_kleq M,forall ninmathbfN$, and since $a_nge 0$ it follows that the corresponding sequence of partial sums is increasing, the series $sum_k=1^inftya_k$ is then convergent by theorem $textbf2.4.1$.
$(Leftarrow)$. Now assume that the sequence $b_n = sum_j=1^na_j$ of partial products is convergent, again by theorem $textbf2.3.2$ there exists a $M>0$ such that $b_nleq M,forall ninmathbfN$. We know that $(1+x)<3^x$ when $xge 0$ consequently
$$prod_j=1^k(1+a_j)<3^sum_j=1^ka_jleq 3^M$$
In summary then $prod_k=1^n(1+a_k)leq 3^M,forall ninmathbfN$, again since $a_nge 0,forall ninmathbfN$ it follows that corresponding sequence of partial products is increasing, appealing to theorem $textbf2.4.1$ implies that
$prod_n=1^infty(1+a_n)$ is convergent.
$blacksquare$
Note:
$(2.4.1):$ Every bounded montonic sequence converges.
$(2.3.2):$ Every convergent sequence is bounded.
real-analysis sequences-and-series proof-verification
asked Jul 29 at 20:18
Atif Farooq
2,7352824
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is the folllowing argument correct? All cited theorems are presented below.
Proposition. Show, in general, that the sequence of partial products of the following series converges if
and only if $sum_n=1^inftya_n$ converges. (The inequality $1 +
xleq 3^x$ for positive $x$ will be useful in one direction.)
$$prod_n=1^infty(1+a_n) = (1+a_1)(1+a_2)(1+a_3)cdots,text text textwhere a_nge 0.$$
Proof. $(Rightarrow)$. Assume that the sequence $s_n = prod_k=1^n(1+a_k)$ of partial products converges, then by theorem $ textbf2.3.2$ there exists an $M>0$ such that $s_nleq M,forall ninmathbfN$. We now show that
$$sum_k=1^na_kleq prod_k=1^n(1+a_k),forall ninmathbfN$$
The base case is trivial. Now let $kinmathbfN$ and assume that $sum_j=1^ka_jleq prod_j=1^k(1+a_j)$, consequently $sum_j=1^k+1a_j = sum_j=1^ka_j +a_k+1leq prod_j=1^k(1+a_j)+a_k+1$.
Now assume that $a_k+1(1-prod_j=1^k(1+a_j))>0$, from hypothesis $a_k+1ge 0$, but then $prod_j=1^k(1+a_j)< 1$, a contradiction since by hypothesis $(1+a_n)ge 1,forall ninmathbfN$, thus $a_k+1leq a_k+1prod_j=1^k(1+a_j)$ but then by inductive hypothesis we have
$$sum_j=1^k+1a_j = sum_j=1^ka_j +a_k+1$$
$$leqprod_j=1^k(1+a_j)+a_k+1leq prod_j=1^k(1+a_j)+ a_k+1prod_j=1^k(1+a_j) = prod_j=1^k+1(1+a_j)$$
completing the induction. It is now apparent that $sum_k=1^na_kleq M,forall ninmathbfN$, and since $a_nge 0$ it follows that the corresponding sequence of partial sums is increasing, the series $sum_k=1^inftya_k$ is then convergent by theorem $textbf2.4.1$.
$(Leftarrow)$. Now assume that the sequence $b_n = sum_j=1^na_j$ of partial products is convergent, again by theorem $textbf2.3.2$ there exists a $M>0$ such that $b_nleq M,forall ninmathbfN$. We know that $(1+x)<3^x$ when $xge 0$ consequently
$$prod_j=1^k(1+a_j)<3^sum_j=1^ka_jleq 3^M$$
In summary then $prod_k=1^n(1+a_k)leq 3^M,forall ninmathbfN$, again since $a_nge 0,forall ninmathbfN$ it follows that corresponding sequence of partial products is increasing, appealing to theorem $textbf2.4.1$ implies that
$prod_n=1^infty(1+a_n)$ is convergent.
$blacksquare$
Note:
$(2.4.1):$ Every bounded montonic sequence converges.
$(2.3.2):$ Every convergent sequence is bounded.
real-analysis sequences-and-series proof-verification
asked Jul 29 at 20:18
Atif Farooq
2,7352824
Is the folllowing argument correct? All cited theorems are presented below.
Proposition. Show, in general, that the sequence of partial products of the following series converges if
and only if $sum_n=1^inftya_n$ converges. (The inequality $1 +
xleq 3^x$ for positive $x$ will be useful in one direction.)
$$prod_n=1^infty(1+a_n) = (1+a_1)(1+a_2)(1+a_3)cdots,text text textwhere a_nge 0.$$
Proof. $(Rightarrow)$. Assume that the sequence $s_n = prod_k=1^n(1+a_k)$ of partial products converges, then by theorem $ textbf2.3.2$ there exists an $M>0$ such that $s_nleq M,forall ninmathbfN$. We now show that
$$sum_k=1^na_kleq prod_k=1^n(1+a_k),forall ninmathbfN$$
The base case is trivial. Now let $kinmathbfN$ and assume that $sum_j=1^ka_jleq prod_j=1^k(1+a_j)$, consequently $sum_j=1^k+1a_j = sum_j=1^ka_j +a_k+1leq prod_j=1^k(1+a_j)+a_k+1$.
Now assume that $a_k+1(1-prod_j=1^k(1+a_j))>0$, from hypothesis $a_k+1ge 0$, but then $prod_j=1^k(1+a_j)< 1$, a contradiction since by hypothesis $(1+a_n)ge 1,forall ninmathbfN$, thus $a_k+1leq a_k+1prod_j=1^k(1+a_j)$ but then by inductive hypothesis we have
$$sum_j=1^k+1a_j = sum_j=1^ka_j +a_k+1$$
$$leqprod_j=1^k(1+a_j)+a_k+1leq prod_j=1^k(1+a_j)+ a_k+1prod_j=1^k(1+a_j) = prod_j=1^k+1(1+a_j)$$
completing the induction. It is now apparent that $sum_k=1^na_kleq M,forall ninmathbfN$, and since $a_nge 0$ it follows that the corresponding sequence of partial sums is increasing, the series $sum_k=1^inftya_k$ is then convergent by theorem $textbf2.4.1$.
$(Leftarrow)$. Now assume that the sequence $b_n = sum_j=1^na_j$ of partial products is convergent, again by theorem $textbf2.3.2$ there exists a $M>0$ such that $b_nleq M,forall ninmathbfN$. We know that $(1+x)<3^x$ when $xge 0$ consequently
$$prod_j=1^k(1+a_j)<3^sum_j=1^ka_jleq 3^M$$
In summary then $prod_k=1^n(1+a_k)leq 3^M,forall ninmathbfN$, again since $a_nge 0,forall ninmathbfN$ it follows that corresponding sequence of partial products is increasing, appealing to theorem $textbf2.4.1$ implies that
$prod_n=1^infty(1+a_n)$ is convergent.
$blacksquare$
Note:
$(2.4.1):$ Every bounded montonic sequence converges.
$(2.3.2):$ Every convergent sequence is bounded.
real-analysis sequences-and-series proof-verification
asked Jul 29 at 20:18
Atif Farooq
2,7352824
asked Jul 29 at 20:18
Atif Farooq
2,7352824
asked Jul 29 at 20:18
Atif Farooq
2,7352824
asked Jul 29 at 20:18
Atif Farooq
2,7352824
2,7352824
add a comment |Â
add a comment |Â
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