Let $langle a_nrangle$ be a sequence where $a_n = sumlimits_j=1^n(-1)^j2/n$ thenLet $(a_n)$ be a sequence of real numbers. Suppose each of the subsequences $(a_2n),(a_2n+1)$ and $(a_3n)$ converges to $p,q$ and $r$If $prodlimits_n(1+a_n)$ converges, does $sumlimits_nfraca_n1+a_n$ converge?Let $a_n rightarrow 0$. Show $sumlimits_n=0^infty (a_n - a_n+1) = a_0$.If $langlea_nrangle$ is convergent with limit l, how would I show that $langler_nrangle$ also converges?Convergence of infinite series for $sum a_n $ and $sum b_n$Let $a_n$ be a sequence of real numbers. which one of the following is always true?If a sequence diverges to infinity then so do all its subsequences. Then why does the sum for $frac1n$ diverge but not $frac1n^2$?Series with $sum a_n$ converges but $sum n a_n^2$ diverges, and $a_n$ is decreasingConvergence of $sumlimits_n=1^inftya_n$ implies convergence of $sumlimits_n=1^inftya_n^sigma_n$ where $sigma_n=fracnn+1$?$a_n$ is an infinite sequence with $sumlimits_n=1^infty a_nleq1$ and $0leq a_n<1$. Prove that $sumlimits_n=1^infty a_n/(a_n-1)$ converges?
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Let $langle a_nrangle$ be a sequence where $a_n = sumlimits_j=1^n(-1)^j2/n$ then
Let $(a_n)$ be a sequence of real numbers. Suppose each of the subsequences $(a_2n),(a_2n+1)$ and $(a_3n)$ converges to $p,q$ and $r$If $prodlimits_n(1+a_n)$ converges, does $sumlimits_nfraca_n1+a_n$ converge?Let $a_n rightarrow 0$. Show $sumlimits_n=0^infty (a_n - a_n+1) = a_0$.If $langlea_nrangle$ is convergent with limit l, how would I show that $langler_nrangle$ also converges?Convergence of infinite series for $sum a_n $ and $sum b_n$Let $a_n$ be a sequence of real numbers. which one of the following is always true?If a sequence diverges to infinity then so do all its subsequences. Then why does the sum for $frac1n$ diverge but not $frac1n^2$?Series with $sum a_n$ converges but $sum n a_n^2$ diverges, and $a_n$ is decreasingConvergence of $sumlimits_n=1^inftya_n$ implies convergence of $sumlimits_n=1^inftya_n^sigma_n$ where $sigma_n=fracnn+1$?$a_n$ is an infinite sequence with $sumlimits_n=1^infty a_nleq1$ and $0leq a_n<1$. Prove that $sumlimits_n=1^infty a_n/(a_n-1)$ converges?
$begingroup$
Let $langle a_nrangle$ be a sequence where $a_n = sumlimits_j=1^n(-1)^j2/n$
is oscillatory
diverges to $+infty$
diverges to $-infty$
converges to $0$
I think by taking subsequences $langle a_2nrangle =(2/n) +(2/n) +ldots +(2/n)$ ($n$-times) $= 2$ and $langle a_2n-1rangle = -2/n -2/n -ldots -2/n$ ($n$-times) $=-2$ which is cunverging to $2$ and $-2$ respectively. So that the sequence is oscillatory. I have doubt taking this type of subsequence is right or wrong.
real-analysis sequences-and-series
$endgroup$
add a comment |
$begingroup$
Let $langle a_nrangle$ be a sequence where $a_n = sumlimits_j=1^n(-1)^j2/n$
is oscillatory
diverges to $+infty$
diverges to $-infty$
converges to $0$
I think by taking subsequences $langle a_2nrangle =(2/n) +(2/n) +ldots +(2/n)$ ($n$-times) $= 2$ and $langle a_2n-1rangle = -2/n -2/n -ldots -2/n$ ($n$-times) $=-2$ which is cunverging to $2$ and $-2$ respectively. So that the sequence is oscillatory. I have doubt taking this type of subsequence is right or wrong.
real-analysis sequences-and-series
$endgroup$
add a comment |
$begingroup$
Let $langle a_nrangle$ be a sequence where $a_n = sumlimits_j=1^n(-1)^j2/n$
is oscillatory
diverges to $+infty$
diverges to $-infty$
converges to $0$
I think by taking subsequences $langle a_2nrangle =(2/n) +(2/n) +ldots +(2/n)$ ($n$-times) $= 2$ and $langle a_2n-1rangle = -2/n -2/n -ldots -2/n$ ($n$-times) $=-2$ which is cunverging to $2$ and $-2$ respectively. So that the sequence is oscillatory. I have doubt taking this type of subsequence is right or wrong.
real-analysis sequences-and-series
$endgroup$
Let $langle a_nrangle$ be a sequence where $a_n = sumlimits_j=1^n(-1)^j2/n$
is oscillatory
diverges to $+infty$
diverges to $-infty$
converges to $0$
I think by taking subsequences $langle a_2nrangle =(2/n) +(2/n) +ldots +(2/n)$ ($n$-times) $= 2$ and $langle a_2n-1rangle = -2/n -2/n -ldots -2/n$ ($n$-times) $=-2$ which is cunverging to $2$ and $-2$ respectively. So that the sequence is oscillatory. I have doubt taking this type of subsequence is right or wrong.
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Mar 13 at 10:18
rtybase
11.5k31534
11.5k31534
asked Mar 13 at 8:37
800123800123
73
73
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The correct answer is 4). Write down the sums $sumlimits_k=1^n (-1)^k$ and convince yourself that these are bounded. So if you multiply these by $frac 2n $ and let $n to infty$ you get $0$.
$endgroup$
add a comment |
$begingroup$
We have $a_n=frac2nsum_j=1^n(-1)^j=frac1n cdot (1-(-1)^n+1)$ and
$|a_n| le 2/n.$ for all $n$
Can you proceed ?
$endgroup$
$begingroup$
this means $-2/nleq a_nleq2/n$ then by sandwitch theorem $<a_n> converges to 0
$endgroup$
– 800123
Mar 13 at 11:04
add a comment |
$begingroup$
1)$a_n= (-2/n)$, $n$ is odd.
2) $a_n=0$, $n$ is even.
1')$lim_ k rightarrow infty a_2k+1 =0$, $k=0,1,2,..;$
2')$ lim_l rightarrow infty a_2l=0, l=1,2,3,...;$
Lemma:
The $2$ subsequences $a_2l, a_2k+1$,
$l=1,2,..,$ $k=0,1,2,..,$
converge to $0$, hence $a_n$ converges to $0$.
Can you prove the Lemma?
$endgroup$
add a comment |
$begingroup$
$$a_n = sumlimits_j=1^n(-1)^j2/n\to a_2k=0\,\a_2k+1=-2over 2k+1$$then this sequence oscillatorily converges to 0.
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The correct answer is 4). Write down the sums $sumlimits_k=1^n (-1)^k$ and convince yourself that these are bounded. So if you multiply these by $frac 2n $ and let $n to infty$ you get $0$.
$endgroup$
add a comment |
$begingroup$
The correct answer is 4). Write down the sums $sumlimits_k=1^n (-1)^k$ and convince yourself that these are bounded. So if you multiply these by $frac 2n $ and let $n to infty$ you get $0$.
$endgroup$
add a comment |
$begingroup$
The correct answer is 4). Write down the sums $sumlimits_k=1^n (-1)^k$ and convince yourself that these are bounded. So if you multiply these by $frac 2n $ and let $n to infty$ you get $0$.
$endgroup$
The correct answer is 4). Write down the sums $sumlimits_k=1^n (-1)^k$ and convince yourself that these are bounded. So if you multiply these by $frac 2n $ and let $n to infty$ you get $0$.
answered Mar 13 at 8:42
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
68.1k53068
add a comment |
add a comment |
$begingroup$
We have $a_n=frac2nsum_j=1^n(-1)^j=frac1n cdot (1-(-1)^n+1)$ and
$|a_n| le 2/n.$ for all $n$
Can you proceed ?
$endgroup$
$begingroup$
this means $-2/nleq a_nleq2/n$ then by sandwitch theorem $<a_n> converges to 0
$endgroup$
– 800123
Mar 13 at 11:04
add a comment |
$begingroup$
We have $a_n=frac2nsum_j=1^n(-1)^j=frac1n cdot (1-(-1)^n+1)$ and
$|a_n| le 2/n.$ for all $n$
Can you proceed ?
$endgroup$
$begingroup$
this means $-2/nleq a_nleq2/n$ then by sandwitch theorem $<a_n> converges to 0
$endgroup$
– 800123
Mar 13 at 11:04
add a comment |
$begingroup$
We have $a_n=frac2nsum_j=1^n(-1)^j=frac1n cdot (1-(-1)^n+1)$ and
$|a_n| le 2/n.$ for all $n$
Can you proceed ?
$endgroup$
We have $a_n=frac2nsum_j=1^n(-1)^j=frac1n cdot (1-(-1)^n+1)$ and
$|a_n| le 2/n.$ for all $n$
Can you proceed ?
edited Mar 13 at 9:11
answered Mar 13 at 8:43
FredFred
48.5k1849
48.5k1849
$begingroup$
this means $-2/nleq a_nleq2/n$ then by sandwitch theorem $<a_n> converges to 0
$endgroup$
– 800123
Mar 13 at 11:04
add a comment |
$begingroup$
this means $-2/nleq a_nleq2/n$ then by sandwitch theorem $<a_n> converges to 0
$endgroup$
– 800123
Mar 13 at 11:04
$begingroup$
this means $-2/nleq a_nleq2/n$ then by sandwitch theorem $<a_n> converges to 0
$endgroup$
– 800123
Mar 13 at 11:04
$begingroup$
this means $-2/nleq a_nleq2/n$ then by sandwitch theorem $<a_n> converges to 0
$endgroup$
– 800123
Mar 13 at 11:04
add a comment |
$begingroup$
1)$a_n= (-2/n)$, $n$ is odd.
2) $a_n=0$, $n$ is even.
1')$lim_ k rightarrow infty a_2k+1 =0$, $k=0,1,2,..;$
2')$ lim_l rightarrow infty a_2l=0, l=1,2,3,...;$
Lemma:
The $2$ subsequences $a_2l, a_2k+1$,
$l=1,2,..,$ $k=0,1,2,..,$
converge to $0$, hence $a_n$ converges to $0$.
Can you prove the Lemma?
$endgroup$
add a comment |
$begingroup$
1)$a_n= (-2/n)$, $n$ is odd.
2) $a_n=0$, $n$ is even.
1')$lim_ k rightarrow infty a_2k+1 =0$, $k=0,1,2,..;$
2')$ lim_l rightarrow infty a_2l=0, l=1,2,3,...;$
Lemma:
The $2$ subsequences $a_2l, a_2k+1$,
$l=1,2,..,$ $k=0,1,2,..,$
converge to $0$, hence $a_n$ converges to $0$.
Can you prove the Lemma?
$endgroup$
add a comment |
$begingroup$
1)$a_n= (-2/n)$, $n$ is odd.
2) $a_n=0$, $n$ is even.
1')$lim_ k rightarrow infty a_2k+1 =0$, $k=0,1,2,..;$
2')$ lim_l rightarrow infty a_2l=0, l=1,2,3,...;$
Lemma:
The $2$ subsequences $a_2l, a_2k+1$,
$l=1,2,..,$ $k=0,1,2,..,$
converge to $0$, hence $a_n$ converges to $0$.
Can you prove the Lemma?
$endgroup$
1)$a_n= (-2/n)$, $n$ is odd.
2) $a_n=0$, $n$ is even.
1')$lim_ k rightarrow infty a_2k+1 =0$, $k=0,1,2,..;$
2')$ lim_l rightarrow infty a_2l=0, l=1,2,3,...;$
Lemma:
The $2$ subsequences $a_2l, a_2k+1$,
$l=1,2,..,$ $k=0,1,2,..,$
converge to $0$, hence $a_n$ converges to $0$.
Can you prove the Lemma?
answered Mar 13 at 9:02
Peter SzilasPeter Szilas
11.6k2822
11.6k2822
add a comment |
add a comment |
$begingroup$
$$a_n = sumlimits_j=1^n(-1)^j2/n\to a_2k=0\,\a_2k+1=-2over 2k+1$$then this sequence oscillatorily converges to 0.
$endgroup$
add a comment |
$begingroup$
$$a_n = sumlimits_j=1^n(-1)^j2/n\to a_2k=0\,\a_2k+1=-2over 2k+1$$then this sequence oscillatorily converges to 0.
$endgroup$
add a comment |
$begingroup$
$$a_n = sumlimits_j=1^n(-1)^j2/n\to a_2k=0\,\a_2k+1=-2over 2k+1$$then this sequence oscillatorily converges to 0.
$endgroup$
$$a_n = sumlimits_j=1^n(-1)^j2/n\to a_2k=0\,\a_2k+1=-2over 2k+1$$then this sequence oscillatorily converges to 0.
answered Mar 13 at 11:58
Mostafa AyazMostafa Ayaz
16.5k3939
16.5k3939
add a comment |
add a comment |
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