Mixing
Find convergence interval of $f(x) =sum_{n=1}^∞ left(sinfrac{2}{n^2}right)x^n $
$begingroup$
$$f(x) =sum_{n=1}^∞ left(sinfrac{2}{n^2}right)x^n $$
so looks like what i have to do is use the formal
$$lim_{nto∞}left|frac{a_{n+1}}{a_n}right|=a$$ with $a_n=sinfrac{2}{n^2}$. And then $R=1/a$.
But I having trouble with the lim. Thank you for your help
sequences-and-series convergence
edited Jan 25 at 5:25
Chinnapparaj R
5,7432928
asked Jan 25 at 4:41
Vu Thanh PhanVu Thanh Phan
347
$endgroup$
add a comment |
$begingroup$
$$f(x) =sum_{n=1}^∞ left(sinfrac{2}{n^2}right)x^n $$
so looks like what i have to do is use the formal
$$lim_{nto∞}left|frac{a_{n+1}}{a_n}right|=a$$ with $a_n=sinfrac{2}{n^2}$. And then $R=1/a$.
But I having trouble with the lim. Thank you for your help
sequences-and-series convergence
edited Jan 25 at 5:25
Chinnapparaj R
5,7432928
asked Jan 25 at 4:41
Vu Thanh PhanVu Thanh Phan
347
$endgroup$
add a comment |
$begingroup$
$$f(x) =sum_{n=1}^∞ left(sinfrac{2}{n^2}right)x^n $$
so looks like what i have to do is use the formal
$$lim_{nto∞}left|frac{a_{n+1}}{a_n}right|=a$$ with $a_n=sinfrac{2}{n^2}$. And then $R=1/a$.
But I having trouble with the lim. Thank you for your help
sequences-and-series convergence
edited Jan 25 at 5:25
Chinnapparaj R
5,7432928
asked Jan 25 at 4:41
Vu Thanh PhanVu Thanh Phan
347
$endgroup$
$$f(x) =sum_{n=1}^∞ left(sinfrac{2}{n^2}right)x^n $$
so looks like what i have to do is use the formal
$$lim_{nto∞}left|frac{a_{n+1}}{a_n}right|=a$$ with $a_n=sinfrac{2}{n^2}$. And then $R=1/a$.
But I having trouble with the lim. Thank you for your help
sequences-and-series convergence
sequences-and-series convergence
edited Jan 25 at 5:25
Chinnapparaj R
5,7432928
asked Jan 25 at 4:41
Vu Thanh PhanVu Thanh Phan
347
edited Jan 25 at 5:25
Chinnapparaj R
5,7432928
asked Jan 25 at 4:41
Vu Thanh PhanVu Thanh Phan
347
edited Jan 25 at 5:25
Chinnapparaj R
5,7432928
edited Jan 25 at 5:25
Chinnapparaj R
5,7432928
edited Jan 25 at 5:25
Chinnapparaj R
5,7432928
5,7432928
asked Jan 25 at 4:41
Vu Thanh PhanVu Thanh Phan
347
asked Jan 25 at 4:41
Vu Thanh PhanVu Thanh Phan
347
asked Jan 25 at 4:41
Vu Thanh PhanVu Thanh Phan
347
347
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You may recall the fact that $sin' = cos$, which implies the standard limit
$$
lim_{xto 0} frac{sin x}{x} = lim_{xto 0} frac{sin x - sin 0}{x- 0} = sin' 0 = cos 0 = 1 tag{1}
$$
from which we have, as $1/n^2 to 0$ as $ntoinfty$,
$$
lim_{ntoinfty} frac{sin frac{2}{n^2}}{frac{2}{n^2}} = 1 tag{2}
$$
Therefore, we get, for $a_n stackrel{rm def}{=} sin frac{2}{n^2}$ (for all $ngeq 1$),
$$begin{align}
lim_{ntoinfty} frac{a_{n+1}}{a_n} &=
lim_{ntoinfty} frac{sin frac{2}{(n+1)^2}}{sin frac{2}{n^2}}
=
lim_{ntoinfty} frac{sin frac{2}{(n+1)^2}}{frac{2}{(n+1)^2}}cdotfrac{frac{2}{(n+1)^2}}{frac{2}{n^2}}cdot frac{frac{2}{n^2}}{sin frac{2}{n^2}}
\&=
lim_{ntoinfty} frac{sin frac{2}{(n+1)^2}}{frac{2}{(n+1)^2}}cdotlim_{ntoinfty} frac{n^2}{(n+1)^2}cdot lim_{ntoinfty} frac{frac{2}{n^2}}{sin frac{2}{n^2}}
end{align}$$
Now, all three limits in this last expression exist and are equal to $1$ (the first and last because of $(2)$). Therefore,
$$
boxed{lim_{ntoinfty} frac{a_{n+1}}{a_n} =
1}
$$
answered Jan 25 at 4:50
Clement C.Clement C.
50.9k33992
$endgroup$
add a comment |
$begingroup$
Another approach.
$$a_n=sin left(frac{2}{n^2}right)implies R_n=frac{a_n}{a_{n+1}}=sin left(frac{2}{n^2}right) csc left(frac{2}{(n+1)^2}right)$$ Now, using Taylor series
$$R_n=frac{a_n}{a_{n+1}}=1+frac{2}{n}+Oleft(frac{1}{n^2}right)$$ and since $R=R_infty$, then ...
answered Jan 25 at 6:02
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
add a comment |
$begingroup$
$f(x) =sum_{n=1}^∞ left(sinfrac{2}{n^2}right)x^n
$
Since
$sin(x) approx x$
for small $x$,
$sinfrac{2}{n^2}
approx frac{2}{n^2}
$
so the sum behaves like
$sum_{n=1}^∞ frac{2x^n}{n^2}
$
and this converges for
$-1 lt x lt 1$
by comparison with
$sum x^n$
and converges for
$x = pm 1$
because it becomes
$sum frac{2}{n^2}
$
at $x=1$
and
$sum frac{2(-1)^n}{n^2}
$
at $x=-1$,
both of which converge.
answered Jan 25 at 6:10
marty cohenmarty cohen
74.4k549129
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You may recall the fact that $sin' = cos$, which implies the standard limit
$$
lim_{xto 0} frac{sin x}{x} = lim_{xto 0} frac{sin x - sin 0}{x- 0} = sin' 0 = cos 0 = 1 tag{1}
$$
from which we have, as $1/n^2 to 0$ as $ntoinfty$,
$$
lim_{ntoinfty} frac{sin frac{2}{n^2}}{frac{2}{n^2}} = 1 tag{2}
$$
Therefore, we get, for $a_n stackrel{rm def}{=} sin frac{2}{n^2}$ (for all $ngeq 1$),
$$begin{align}
lim_{ntoinfty} frac{a_{n+1}}{a_n} &=
lim_{ntoinfty} frac{sin frac{2}{(n+1)^2}}{sin frac{2}{n^2}}
=
lim_{ntoinfty} frac{sin frac{2}{(n+1)^2}}{frac{2}{(n+1)^2}}cdotfrac{frac{2}{(n+1)^2}}{frac{2}{n^2}}cdot frac{frac{2}{n^2}}{sin frac{2}{n^2}}
\&=
lim_{ntoinfty} frac{sin frac{2}{(n+1)^2}}{frac{2}{(n+1)^2}}cdotlim_{ntoinfty} frac{n^2}{(n+1)^2}cdot lim_{ntoinfty} frac{frac{2}{n^2}}{sin frac{2}{n^2}}
end{align}$$
Now, all three limits in this last expression exist and are equal to $1$ (the first and last because of $(2)$). Therefore,
$$
boxed{lim_{ntoinfty} frac{a_{n+1}}{a_n} =
1}
$$
answered Jan 25 at 4:50
Clement C.Clement C.
50.9k33992
$endgroup$
add a comment |
$begingroup$
You may recall the fact that $sin' = cos$, which implies the standard limit
$$
lim_{xto 0} frac{sin x}{x} = lim_{xto 0} frac{sin x - sin 0}{x- 0} = sin' 0 = cos 0 = 1 tag{1}
$$
from which we have, as $1/n^2 to 0$ as $ntoinfty$,
$$
lim_{ntoinfty} frac{sin frac{2}{n^2}}{frac{2}{n^2}} = 1 tag{2}
$$
Therefore, we get, for $a_n stackrel{rm def}{=} sin frac{2}{n^2}$ (for all $ngeq 1$),
$$begin{align}
lim_{ntoinfty} frac{a_{n+1}}{a_n} &=
lim_{ntoinfty} frac{sin frac{2}{(n+1)^2}}{sin frac{2}{n^2}}
=
lim_{ntoinfty} frac{sin frac{2}{(n+1)^2}}{frac{2}{(n+1)^2}}cdotfrac{frac{2}{(n+1)^2}}{frac{2}{n^2}}cdot frac{frac{2}{n^2}}{sin frac{2}{n^2}}
\&=
lim_{ntoinfty} frac{sin frac{2}{(n+1)^2}}{frac{2}{(n+1)^2}}cdotlim_{ntoinfty} frac{n^2}{(n+1)^2}cdot lim_{ntoinfty} frac{frac{2}{n^2}}{sin frac{2}{n^2}}
end{align}$$
Now, all three limits in this last expression exist and are equal to $1$ (the first and last because of $(2)$). Therefore,
$$
boxed{lim_{ntoinfty} frac{a_{n+1}}{a_n} =
1}
$$
answered Jan 25 at 4:50
Clement C.Clement C.
50.9k33992
$endgroup$
add a comment |
$begingroup$
You may recall the fact that $sin' = cos$, which implies the standard limit
$$
lim_{xto 0} frac{sin x}{x} = lim_{xto 0} frac{sin x - sin 0}{x- 0} = sin' 0 = cos 0 = 1 tag{1}
$$
from which we have, as $1/n^2 to 0$ as $ntoinfty$,
$$
lim_{ntoinfty} frac{sin frac{2}{n^2}}{frac{2}{n^2}} = 1 tag{2}
$$
Therefore, we get, for $a_n stackrel{rm def}{=} sin frac{2}{n^2}$ (for all $ngeq 1$),
$$begin{align}
lim_{ntoinfty} frac{a_{n+1}}{a_n} &=
lim_{ntoinfty} frac{sin frac{2}{(n+1)^2}}{sin frac{2}{n^2}}
=
lim_{ntoinfty} frac{sin frac{2}{(n+1)^2}}{frac{2}{(n+1)^2}}cdotfrac{frac{2}{(n+1)^2}}{frac{2}{n^2}}cdot frac{frac{2}{n^2}}{sin frac{2}{n^2}}
\&=
lim_{ntoinfty} frac{sin frac{2}{(n+1)^2}}{frac{2}{(n+1)^2}}cdotlim_{ntoinfty} frac{n^2}{(n+1)^2}cdot lim_{ntoinfty} frac{frac{2}{n^2}}{sin frac{2}{n^2}}
end{align}$$
Now, all three limits in this last expression exist and are equal to $1$ (the first and last because of $(2)$). Therefore,
$$
boxed{lim_{ntoinfty} frac{a_{n+1}}{a_n} =
1}
$$
answered Jan 25 at 4:50
Clement C.Clement C.
50.9k33992
$endgroup$
You may recall the fact that $sin' = cos$, which implies the standard limit
$$
lim_{xto 0} frac{sin x}{x} = lim_{xto 0} frac{sin x - sin 0}{x- 0} = sin' 0 = cos 0 = 1 tag{1}
$$
from which we have, as $1/n^2 to 0$ as $ntoinfty$,
$$
lim_{ntoinfty} frac{sin frac{2}{n^2}}{frac{2}{n^2}} = 1 tag{2}
$$
Therefore, we get, for $a_n stackrel{rm def}{=} sin frac{2}{n^2}$ (for all $ngeq 1$),
$$begin{align}
lim_{ntoinfty} frac{a_{n+1}}{a_n} &=
lim_{ntoinfty} frac{sin frac{2}{(n+1)^2}}{sin frac{2}{n^2}}
=
lim_{ntoinfty} frac{sin frac{2}{(n+1)^2}}{frac{2}{(n+1)^2}}cdotfrac{frac{2}{(n+1)^2}}{frac{2}{n^2}}cdot frac{frac{2}{n^2}}{sin frac{2}{n^2}}
\&=
lim_{ntoinfty} frac{sin frac{2}{(n+1)^2}}{frac{2}{(n+1)^2}}cdotlim_{ntoinfty} frac{n^2}{(n+1)^2}cdot lim_{ntoinfty} frac{frac{2}{n^2}}{sin frac{2}{n^2}}
end{align}$$
Now, all three limits in this last expression exist and are equal to $1$ (the first and last because of $(2)$). Therefore,
$$
boxed{lim_{ntoinfty} frac{a_{n+1}}{a_n} =
1}
$$
answered Jan 25 at 4:50
Clement C.Clement C.
50.9k33992
answered Jan 25 at 4:50
Clement C.Clement C.
50.9k33992
answered Jan 25 at 4:50
Clement C.Clement C.
50.9k33992
answered Jan 25 at 4:50
Clement C.Clement C.
50.9k33992
50.9k33992
add a comment |
add a comment |
$begingroup$
Another approach.
$$a_n=sin left(frac{2}{n^2}right)implies R_n=frac{a_n}{a_{n+1}}=sin left(frac{2}{n^2}right) csc left(frac{2}{(n+1)^2}right)$$ Now, using Taylor series
$$R_n=frac{a_n}{a_{n+1}}=1+frac{2}{n}+Oleft(frac{1}{n^2}right)$$ and since $R=R_infty$, then ...
answered Jan 25 at 6:02
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
add a comment |
$begingroup$
Another approach.
$$a_n=sin left(frac{2}{n^2}right)implies R_n=frac{a_n}{a_{n+1}}=sin left(frac{2}{n^2}right) csc left(frac{2}{(n+1)^2}right)$$ Now, using Taylor series
$$R_n=frac{a_n}{a_{n+1}}=1+frac{2}{n}+Oleft(frac{1}{n^2}right)$$ and since $R=R_infty$, then ...
answered Jan 25 at 6:02
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
add a comment |
$begingroup$
Another approach.
$$a_n=sin left(frac{2}{n^2}right)implies R_n=frac{a_n}{a_{n+1}}=sin left(frac{2}{n^2}right) csc left(frac{2}{(n+1)^2}right)$$ Now, using Taylor series
$$R_n=frac{a_n}{a_{n+1}}=1+frac{2}{n}+Oleft(frac{1}{n^2}right)$$ and since $R=R_infty$, then ...
answered Jan 25 at 6:02
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
Another approach.
$$a_n=sin left(frac{2}{n^2}right)implies R_n=frac{a_n}{a_{n+1}}=sin left(frac{2}{n^2}right) csc left(frac{2}{(n+1)^2}right)$$ Now, using Taylor series
$$R_n=frac{a_n}{a_{n+1}}=1+frac{2}{n}+Oleft(frac{1}{n^2}right)$$ and since $R=R_infty$, then ...
answered Jan 25 at 6:02
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 25 at 6:02
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 25 at 6:02
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 25 at 6:02
Claude LeiboviciClaude Leibovici
124k1158135
124k1158135
add a comment |
add a comment |
$begingroup$
$f(x) =sum_{n=1}^∞ left(sinfrac{2}{n^2}right)x^n
$
Since
$sin(x) approx x$
for small $x$,
$sinfrac{2}{n^2}
approx frac{2}{n^2}
$
so the sum behaves like
$sum_{n=1}^∞ frac{2x^n}{n^2}
$
and this converges for
$-1 lt x lt 1$
by comparison with
$sum x^n$
and converges for
$x = pm 1$
because it becomes
$sum frac{2}{n^2}
$
at $x=1$
and
$sum frac{2(-1)^n}{n^2}
$
at $x=-1$,
both of which converge.
answered Jan 25 at 6:10
marty cohenmarty cohen
74.4k549129
$endgroup$
add a comment |
$begingroup$
$f(x) =sum_{n=1}^∞ left(sinfrac{2}{n^2}right)x^n
$
Since
$sin(x) approx x$
for small $x$,
$sinfrac{2}{n^2}
approx frac{2}{n^2}
$
so the sum behaves like
$sum_{n=1}^∞ frac{2x^n}{n^2}
$
and this converges for
$-1 lt x lt 1$
by comparison with
$sum x^n$
and converges for
$x = pm 1$
because it becomes
$sum frac{2}{n^2}
$
at $x=1$
and
$sum frac{2(-1)^n}{n^2}
$
at $x=-1$,
both of which converge.
answered Jan 25 at 6:10
marty cohenmarty cohen
74.4k549129
$endgroup$
add a comment |
$begingroup$
$f(x) =sum_{n=1}^∞ left(sinfrac{2}{n^2}right)x^n
$
Since
$sin(x) approx x$
for small $x$,
$sinfrac{2}{n^2}
approx frac{2}{n^2}
$
so the sum behaves like
$sum_{n=1}^∞ frac{2x^n}{n^2}
$
and this converges for
$-1 lt x lt 1$
by comparison with
$sum x^n$
and converges for
$x = pm 1$
because it becomes
$sum frac{2}{n^2}
$
at $x=1$
and
$sum frac{2(-1)^n}{n^2}
$
at $x=-1$,
both of which converge.
answered Jan 25 at 6:10
marty cohenmarty cohen
74.4k549129
$endgroup$
$f(x) =sum_{n=1}^∞ left(sinfrac{2}{n^2}right)x^n
$
Since
$sin(x) approx x$
for small $x$,
$sinfrac{2}{n^2}
approx frac{2}{n^2}
$
so the sum behaves like
$sum_{n=1}^∞ frac{2x^n}{n^2}
$
and this converges for
$-1 lt x lt 1$
by comparison with
$sum x^n$
and converges for
$x = pm 1$
because it becomes
$sum frac{2}{n^2}
$
at $x=1$
and
$sum frac{2(-1)^n}{n^2}
$
at $x=-1$,
both of which converge.
answered Jan 25 at 6:10
marty cohenmarty cohen
74.4k549129
answered Jan 25 at 6:10
marty cohenmarty cohen
74.4k549129
answered Jan 25 at 6:10
marty cohenmarty cohen
74.4k549129
answered Jan 25 at 6:10
marty cohenmarty cohen
74.4k549129
74.4k549129
add a comment |
add a comment |
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