Mixing
$sum_{ n = j+1}^infty a_nw^{n-1-j} = O(1/r^j)$
$begingroup$
Let $a_n$ be a sequence of complex numbers. Moreover we suppose that the radisu of convergence of the power serie :$sum_{n = 0}^infty a_nx^n$ is $R$. $w$ is a complex number in ${z in mathbb{C} mid mid z mid < R }$, and $r in ]mid w mid, R [$.
Now prove that :
$$lim_{j to infty} sum_{n = j+1}^infty a_nw^{n-1-j} = O(1/r^j)$$
When trying to solve this exercises it seems that we even have $o(1/r^j)$ instead of $O(1/r^j)$ but I might be wrong.
We have :
$$r^j cdot sum_{n = j+1}^infty a_nw^{n-1-j} leq frac{r^j}{r^{j+1}} cdot sum_{ n = j+1}^infty mid a_n mid mid r^{n} mid $$
Hence : since
$lim_{j to infty} sum_{n = j+1}^infty mid a_n mid mid r^n mid = 0$ we deduce that we have : $sum_{n = j+1}^infty a_nw^{n-1-j} = o(1/r^j)$
real-analysis calculus integration sequences-and-series power-series
asked Feb 1 at 7:25
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
add a comment |
$begingroup$
Let $a_n$ be a sequence of complex numbers. Moreover we suppose that the radisu of convergence of the power serie :$sum_{n = 0}^infty a_nx^n$ is $R$. $w$ is a complex number in ${z in mathbb{C} mid mid z mid < R }$, and $r in ]mid w mid, R [$.
Now prove that :
$$lim_{j to infty} sum_{n = j+1}^infty a_nw^{n-1-j} = O(1/r^j)$$
When trying to solve this exercises it seems that we even have $o(1/r^j)$ instead of $O(1/r^j)$ but I might be wrong.
We have :
$$r^j cdot sum_{n = j+1}^infty a_nw^{n-1-j} leq frac{r^j}{r^{j+1}} cdot sum_{ n = j+1}^infty mid a_n mid mid r^{n} mid $$
Hence : since
$lim_{j to infty} sum_{n = j+1}^infty mid a_n mid mid r^n mid = 0$ we deduce that we have : $sum_{n = j+1}^infty a_nw^{n-1-j} = o(1/r^j)$
real-analysis calculus integration sequences-and-series power-series
asked Feb 1 at 7:25
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
add a comment |
$begingroup$
Let $a_n$ be a sequence of complex numbers. Moreover we suppose that the radisu of convergence of the power serie :$sum_{n = 0}^infty a_nx^n$ is $R$. $w$ is a complex number in ${z in mathbb{C} mid mid z mid < R }$, and $r in ]mid w mid, R [$.
Now prove that :
$$lim_{j to infty} sum_{n = j+1}^infty a_nw^{n-1-j} = O(1/r^j)$$
When trying to solve this exercises it seems that we even have $o(1/r^j)$ instead of $O(1/r^j)$ but I might be wrong.
We have :
$$r^j cdot sum_{n = j+1}^infty a_nw^{n-1-j} leq frac{r^j}{r^{j+1}} cdot sum_{ n = j+1}^infty mid a_n mid mid r^{n} mid $$
Hence : since
$lim_{j to infty} sum_{n = j+1}^infty mid a_n mid mid r^n mid = 0$ we deduce that we have : $sum_{n = j+1}^infty a_nw^{n-1-j} = o(1/r^j)$
real-analysis calculus integration sequences-and-series power-series
asked Feb 1 at 7:25
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
Let $a_n$ be a sequence of complex numbers. Moreover we suppose that the radisu of convergence of the power serie :$sum_{n = 0}^infty a_nx^n$ is $R$. $w$ is a complex number in ${z in mathbb{C} mid mid z mid < R }$, and $r in ]mid w mid, R [$.
Now prove that :
$$lim_{j to infty} sum_{n = j+1}^infty a_nw^{n-1-j} = O(1/r^j)$$
When trying to solve this exercises it seems that we even have $o(1/r^j)$ instead of $O(1/r^j)$ but I might be wrong.
We have :
$$r^j cdot sum_{n = j+1}^infty a_nw^{n-1-j} leq frac{r^j}{r^{j+1}} cdot sum_{ n = j+1}^infty mid a_n mid mid r^{n} mid $$
Hence : since
$lim_{j to infty} sum_{n = j+1}^infty mid a_n mid mid r^n mid = 0$ we deduce that we have : $sum_{n = j+1}^infty a_nw^{n-1-j} = o(1/r^j)$
real-analysis calculus integration sequences-and-series power-series
real-analysis calculus integration sequences-and-series power-series
asked Feb 1 at 7:25
dghkgfzyukzdghkgfzyukz
16612
asked Feb 1 at 7:25
dghkgfzyukzdghkgfzyukz
16612
asked Feb 1 at 7:25
dghkgfzyukzdghkgfzyukz
16612
asked Feb 1 at 7:25
dghkgfzyukzdghkgfzyukz
16612
asked Feb 1 at 7:25
dghkgfzyukzdghkgfzyukz
16612
16612
add a comment |
add a comment |
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