It is my calculations about the convergence of a series are correct?
Consider $0<q<p$ real numbers. I wana find for which values of $p$ the series $sumfrac{1}{p^n-q^n}$ is convergent.
First note that the following
$$
sumfrac{1}{p^n-q^n}=sumfrac{1}{p^n(1-(frac{q}{p})^n)}
$$
Now we have the inequalities
$$
0leqsqrt[n]{1-left(frac{q}{p}right)^n}leq 1
$$
So we have that $lim_{nrightarrowinfty}sqrt[n]{1-left(frac{q}{p}right)^n}=k$ where $0leq kleq 1$
Then, by the root criteria (if $kneq0$) the serie converge iff $frac{1}{pk}<1$ iff $1<pkleq p$.
My question: As you see, my process depend of $k$ because I assume that $kneq 0$ but I can't prove it. Can someone give me a hint?
calculus sequences-and-series convergence
asked Dec 1 at 23:14
Gödel
1,413319
add a comment |
Consider $0<q<p$ real numbers. I wana find for which values of $p$ the series $sumfrac{1}{p^n-q^n}$ is convergent.
First note that the following
$$
sumfrac{1}{p^n-q^n}=sumfrac{1}{p^n(1-(frac{q}{p})^n)}
$$
Now we have the inequalities
$$
0leqsqrt[n]{1-left(frac{q}{p}right)^n}leq 1
$$
So we have that $lim_{nrightarrowinfty}sqrt[n]{1-left(frac{q}{p}right)^n}=k$ where $0leq kleq 1$
Then, by the root criteria (if $kneq0$) the serie converge iff $frac{1}{pk}<1$ iff $1<pkleq p$.
My question: As you see, my process depend of $k$ because I assume that $kneq 0$ but I can't prove it. Can someone give me a hint?
calculus sequences-and-series convergence
asked Dec 1 at 23:14
Gödel
1,413319
add a comment |
Consider $0<q<p$ real numbers. I wana find for which values of $p$ the series $sumfrac{1}{p^n-q^n}$ is convergent.
First note that the following
$$
sumfrac{1}{p^n-q^n}=sumfrac{1}{p^n(1-(frac{q}{p})^n)}
$$
Now we have the inequalities
$$
0leqsqrt[n]{1-left(frac{q}{p}right)^n}leq 1
$$
So we have that $lim_{nrightarrowinfty}sqrt[n]{1-left(frac{q}{p}right)^n}=k$ where $0leq kleq 1$
Then, by the root criteria (if $kneq0$) the serie converge iff $frac{1}{pk}<1$ iff $1<pkleq p$.
My question: As you see, my process depend of $k$ because I assume that $kneq 0$ but I can't prove it. Can someone give me a hint?
calculus sequences-and-series convergence
asked Dec 1 at 23:14
Gödel
1,413319
Consider $0<q<p$ real numbers. I wana find for which values of $p$ the series $sumfrac{1}{p^n-q^n}$ is convergent.
First note that the following
$$
sumfrac{1}{p^n-q^n}=sumfrac{1}{p^n(1-(frac{q}{p})^n)}
$$
Now we have the inequalities
$$
0leqsqrt[n]{1-left(frac{q}{p}right)^n}leq 1
$$
So we have that $lim_{nrightarrowinfty}sqrt[n]{1-left(frac{q}{p}right)^n}=k$ where $0leq kleq 1$
Then, by the root criteria (if $kneq0$) the serie converge iff $frac{1}{pk}<1$ iff $1<pkleq p$.
My question: As you see, my process depend of $k$ because I assume that $kneq 0$ but I can't prove it. Can someone give me a hint?
calculus sequences-and-series convergence
calculus sequences-and-series convergence
asked Dec 1 at 23:14
Gödel
1,413319
asked Dec 1 at 23:14
Gödel
1,413319
asked Dec 1 at 23:14
Gödel
1,413319
asked Dec 1 at 23:14
Gödel
1,413319
asked Dec 1 at 23:14
Gödel
1,413319
1,413319
add a comment |
add a comment |
2 Answers
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$(frac q p)^{n} to 0$ because $q<p$. Hence $1-(frac q p)^{n} to 1$. Comparing the given series with $sum frac 1 {p^{n}}$ we see that the series converges iff $p>1$.
answered Dec 1 at 23:24
Kavi Rama Murthy
50.1k31854
add a comment |
You could better compute $k$, but it's not the best method. Note also that the simple fact that $0le a_nle1$ doesn't guarantee the existence of $lim_{ntoinfty}a_n$, so you don't really have $k$.
First method.
Note that
$$
lim_{ntoinfty}frac{dfrac{1}{p^n}}{dfrac{1}{p^n-q^n}}=
lim_{ntoinfty}left(1-left(dfrac{q}{p}right)^{!n}right)=1
$$
so the two series $sum_nfrac{1}{p^n-q^n}$ and $sum_{n}frac{1}{p^n}$ are either both convergent or both divergent.
Second method.
The ratio test, setting $r=q/p<1$,
$$
lim_{ntoinfty}frac{dfrac{1}{p^{n+1}-q^{n+1}}}{dfrac{1}{p^n-q^n}}=
lim_{ntoinfty}frac{p^{n}-q^n}{p^{n+1}-q^{n+1}}=
lim_{ntoinfty}frac{1-r^n}{p-qr^n}=frac{1}{p}
$$
would only leave the case $p=1$ to be determined. Can you do the case $p=1$?
answered Dec 1 at 23:58
egreg
178k1484201
add a comment |
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2 Answers
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2 Answers
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$(frac q p)^{n} to 0$ because $q<p$. Hence $1-(frac q p)^{n} to 1$. Comparing the given series with $sum frac 1 {p^{n}}$ we see that the series converges iff $p>1$.
answered Dec 1 at 23:24
Kavi Rama Murthy
50.1k31854
add a comment |
$(frac q p)^{n} to 0$ because $q<p$. Hence $1-(frac q p)^{n} to 1$. Comparing the given series with $sum frac 1 {p^{n}}$ we see that the series converges iff $p>1$.
answered Dec 1 at 23:24
Kavi Rama Murthy
50.1k31854
add a comment |
$(frac q p)^{n} to 0$ because $q<p$. Hence $1-(frac q p)^{n} to 1$. Comparing the given series with $sum frac 1 {p^{n}}$ we see that the series converges iff $p>1$.
answered Dec 1 at 23:24
Kavi Rama Murthy
50.1k31854
$(frac q p)^{n} to 0$ because $q<p$. Hence $1-(frac q p)^{n} to 1$. Comparing the given series with $sum frac 1 {p^{n}}$ we see that the series converges iff $p>1$.
answered Dec 1 at 23:24
Kavi Rama Murthy
50.1k31854
edited Dec 2 at 4:21
answered Dec 1 at 23:24
Kavi Rama Murthy
50.1k31854
answered Dec 1 at 23:24
Kavi Rama Murthy
50.1k31854
answered Dec 1 at 23:24
Kavi Rama Murthy
50.1k31854
50.1k31854
add a comment |
add a comment |
You could better compute $k$, but it's not the best method. Note also that the simple fact that $0le a_nle1$ doesn't guarantee the existence of $lim_{ntoinfty}a_n$, so you don't really have $k$.
First method.
Note that
$$
lim_{ntoinfty}frac{dfrac{1}{p^n}}{dfrac{1}{p^n-q^n}}=
lim_{ntoinfty}left(1-left(dfrac{q}{p}right)^{!n}right)=1
$$
so the two series $sum_nfrac{1}{p^n-q^n}$ and $sum_{n}frac{1}{p^n}$ are either both convergent or both divergent.
Second method.
The ratio test, setting $r=q/p<1$,
$$
lim_{ntoinfty}frac{dfrac{1}{p^{n+1}-q^{n+1}}}{dfrac{1}{p^n-q^n}}=
lim_{ntoinfty}frac{p^{n}-q^n}{p^{n+1}-q^{n+1}}=
lim_{ntoinfty}frac{1-r^n}{p-qr^n}=frac{1}{p}
$$
would only leave the case $p=1$ to be determined. Can you do the case $p=1$?
answered Dec 1 at 23:58
egreg
178k1484201
add a comment |
You could better compute $k$, but it's not the best method. Note also that the simple fact that $0le a_nle1$ doesn't guarantee the existence of $lim_{ntoinfty}a_n$, so you don't really have $k$.
First method.
Note that
$$
lim_{ntoinfty}frac{dfrac{1}{p^n}}{dfrac{1}{p^n-q^n}}=
lim_{ntoinfty}left(1-left(dfrac{q}{p}right)^{!n}right)=1
$$
so the two series $sum_nfrac{1}{p^n-q^n}$ and $sum_{n}frac{1}{p^n}$ are either both convergent or both divergent.
Second method.
The ratio test, setting $r=q/p<1$,
$$
lim_{ntoinfty}frac{dfrac{1}{p^{n+1}-q^{n+1}}}{dfrac{1}{p^n-q^n}}=
lim_{ntoinfty}frac{p^{n}-q^n}{p^{n+1}-q^{n+1}}=
lim_{ntoinfty}frac{1-r^n}{p-qr^n}=frac{1}{p}
$$
would only leave the case $p=1$ to be determined. Can you do the case $p=1$?
answered Dec 1 at 23:58
egreg
178k1484201
add a comment |
You could better compute $k$, but it's not the best method. Note also that the simple fact that $0le a_nle1$ doesn't guarantee the existence of $lim_{ntoinfty}a_n$, so you don't really have $k$.
First method.
Note that
$$
lim_{ntoinfty}frac{dfrac{1}{p^n}}{dfrac{1}{p^n-q^n}}=
lim_{ntoinfty}left(1-left(dfrac{q}{p}right)^{!n}right)=1
$$
so the two series $sum_nfrac{1}{p^n-q^n}$ and $sum_{n}frac{1}{p^n}$ are either both convergent or both divergent.
Second method.
The ratio test, setting $r=q/p<1$,
$$
lim_{ntoinfty}frac{dfrac{1}{p^{n+1}-q^{n+1}}}{dfrac{1}{p^n-q^n}}=
lim_{ntoinfty}frac{p^{n}-q^n}{p^{n+1}-q^{n+1}}=
lim_{ntoinfty}frac{1-r^n}{p-qr^n}=frac{1}{p}
$$
would only leave the case $p=1$ to be determined. Can you do the case $p=1$?
answered Dec 1 at 23:58
egreg
178k1484201
You could better compute $k$, but it's not the best method. Note also that the simple fact that $0le a_nle1$ doesn't guarantee the existence of $lim_{ntoinfty}a_n$, so you don't really have $k$.
First method.
Note that
$$
lim_{ntoinfty}frac{dfrac{1}{p^n}}{dfrac{1}{p^n-q^n}}=
lim_{ntoinfty}left(1-left(dfrac{q}{p}right)^{!n}right)=1
$$
so the two series $sum_nfrac{1}{p^n-q^n}$ and $sum_{n}frac{1}{p^n}$ are either both convergent or both divergent.
Second method.
The ratio test, setting $r=q/p<1$,
$$
lim_{ntoinfty}frac{dfrac{1}{p^{n+1}-q^{n+1}}}{dfrac{1}{p^n-q^n}}=
lim_{ntoinfty}frac{p^{n}-q^n}{p^{n+1}-q^{n+1}}=
lim_{ntoinfty}frac{1-r^n}{p-qr^n}=frac{1}{p}
$$
would only leave the case $p=1$ to be determined. Can you do the case $p=1$?
answered Dec 1 at 23:58
egreg
178k1484201
answered Dec 1 at 23:58
egreg
178k1484201
answered Dec 1 at 23:58
egreg
178k1484201
answered Dec 1 at 23:58
egreg
178k1484201
178k1484201
add a comment |
add a comment |
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