Mixing
Proving that $lim_{ntoinfty}s_n=0$
Prove: If $lim_{ntoinfty}s_n=s$ and ${s_n}$ has a subsequence ${s_{n_k}}$ such that $(-1)^ks_{n_k}geq0$ then $s=0$.
Proof. Since ${s_n}$ converges to $s$ then so does its subsequence ${s_{n_k}}$ . Again, if it is convergent then $liminf=limsup$, but for $k$ even the subsequence assumes the form $s_{n_{2k}}$, instead for odd the form $-s_{n_{2k+1}}$. For these to be equal, for large $n$, both limits $s$ and $-s$ must be equal to $0$. Since $liminf=limsup=0$ then $s=0$.
real-analysis sequences-and-series proof-verification
asked Nov 20 '18 at 14:43
DMH16
578217
add a comment |
Prove: If $lim_{ntoinfty}s_n=s$ and ${s_n}$ has a subsequence ${s_{n_k}}$ such that $(-1)^ks_{n_k}geq0$ then $s=0$.
Proof. Since ${s_n}$ converges to $s$ then so does its subsequence ${s_{n_k}}$ . Again, if it is convergent then $liminf=limsup$, but for $k$ even the subsequence assumes the form $s_{n_{2k}}$, instead for odd the form $-s_{n_{2k+1}}$. For these to be equal, for large $n$, both limits $s$ and $-s$ must be equal to $0$. Since $liminf=limsup=0$ then $s=0$.
real-analysis sequences-and-series proof-verification
asked Nov 20 '18 at 14:43
DMH16
578217
"For these to be equal" -> the sequences are not equal. Their limits (respectively $s$ and $-s$) have to be.
– Clement C.
Nov 20 '18 at 14:45
Sorry, that is what I intended to write
– DMH16
Nov 20 '18 at 14:46
add a comment |
Prove: If $lim_{ntoinfty}s_n=s$ and ${s_n}$ has a subsequence ${s_{n_k}}$ such that $(-1)^ks_{n_k}geq0$ then $s=0$.
Proof. Since ${s_n}$ converges to $s$ then so does its subsequence ${s_{n_k}}$ . Again, if it is convergent then $liminf=limsup$, but for $k$ even the subsequence assumes the form $s_{n_{2k}}$, instead for odd the form $-s_{n_{2k+1}}$. For these to be equal, for large $n$, both limits $s$ and $-s$ must be equal to $0$. Since $liminf=limsup=0$ then $s=0$.
real-analysis sequences-and-series proof-verification
asked Nov 20 '18 at 14:43
DMH16
578217
Prove: If $lim_{ntoinfty}s_n=s$ and ${s_n}$ has a subsequence ${s_{n_k}}$ such that $(-1)^ks_{n_k}geq0$ then $s=0$.
Proof. Since ${s_n}$ converges to $s$ then so does its subsequence ${s_{n_k}}$ . Again, if it is convergent then $liminf=limsup$, but for $k$ even the subsequence assumes the form $s_{n_{2k}}$, instead for odd the form $-s_{n_{2k+1}}$. For these to be equal, for large $n$, both limits $s$ and $-s$ must be equal to $0$. Since $liminf=limsup=0$ then $s=0$.
real-analysis sequences-and-series proof-verification
real-analysis sequences-and-series proof-verification
asked Nov 20 '18 at 14:43
DMH16
578217
asked Nov 20 '18 at 14:43
DMH16
578217
edited Nov 20 '18 at 14:47
asked Nov 20 '18 at 14:43
DMH16
578217
asked Nov 20 '18 at 14:43
DMH16
578217
asked Nov 20 '18 at 14:43
DMH16
578217
578217
"For these to be equal" -> the sequences are not equal. Their limits (respectively $s$ and $-s$) have to be.
– Clement C.
Nov 20 '18 at 14:45
Sorry, that is what I intended to write
– DMH16
Nov 20 '18 at 14:46
add a comment |
"For these to be equal" -> the sequences are not equal. Their limits (respectively $s$ and $-s$) have to be.
– Clement C.
Nov 20 '18 at 14:45
Sorry, that is what I intended to write
– DMH16
Nov 20 '18 at 14:46
"For these to be equal" -> the sequences are not equal. Their limits (respectively $s$ and $-s$) have to be.
– Clement C.
Nov 20 '18 at 14:45
"For these to be equal" -> the sequences are not equal. Their limits (respectively $s$ and $-s$) have to be.
– Clement C.
Nov 20 '18 at 14:45
Sorry, that is what I intended to write
– DMH16
Nov 20 '18 at 14:46
Sorry, that is what I intended to write
– DMH16
Nov 20 '18 at 14:46
add a comment |
2 Answers
2
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oldest
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Your proof is fine. Here a proof without $ lim inf$ and $lim sup$:
Suppose that $s ne 0$. WLOG we assume thatb $s>0$. Since $s_{n_k} to s$ as $ k to infty$, there is $K$ such that $s_{n_k}>0$ for all $k>K$. If $k>K$ and $k$ is odd, we get $(-1)^ks_{n_k}<0$, a contradiction.
answered Nov 20 '18 at 14:50
Fred
44.2k1845
add a comment |
$lim_{n rightarrow infty} s_n=s.$
Then every subsequence converges to $s$.
Given:
$s_{n_k}$ converges to $s$, and $(-1)^k s_{n_k} ge 0$.
1) $k =2m$ then $s_{n_{2m}} ge 0$, and this subsequence of $s_{n_k}$ converges to $s$:
$lim_{m rightarrow infty} s_{n_{2m}} =s ge 0.$
Similarly;
2)$ k=(2m+1)$ then $s_{n_{2m+1}} le 0$, and this subsequence of $s_{n_k}$ converges to $s$.
$lim_{ m rightarrow infty}s_{n_{2m+1}}=s le 0.$
Hence?
Used : If every term of a convergent sequence is $ge 0,$ then the limit is $ge 0$.
answered Nov 20 '18 at 16:18
Peter Szilas
10.7k2720
I am not sure how your answer differs from mine
– DMH16
Nov 21 '18 at 4:11
DMH16.Same idea, without limsup,liminf.
– Peter Szilas
Nov 21 '18 at 8:36
add a comment |
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2 Answers
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oldest
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2 Answers
2
active
oldest
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active
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active
oldest
votes
Your proof is fine. Here a proof without $ lim inf$ and $lim sup$:
Suppose that $s ne 0$. WLOG we assume thatb $s>0$. Since $s_{n_k} to s$ as $ k to infty$, there is $K$ such that $s_{n_k}>0$ for all $k>K$. If $k>K$ and $k$ is odd, we get $(-1)^ks_{n_k}<0$, a contradiction.
answered Nov 20 '18 at 14:50
Fred
44.2k1845
add a comment |
Your proof is fine. Here a proof without $ lim inf$ and $lim sup$:
Suppose that $s ne 0$. WLOG we assume thatb $s>0$. Since $s_{n_k} to s$ as $ k to infty$, there is $K$ such that $s_{n_k}>0$ for all $k>K$. If $k>K$ and $k$ is odd, we get $(-1)^ks_{n_k}<0$, a contradiction.
answered Nov 20 '18 at 14:50
Fred
44.2k1845
add a comment |
Your proof is fine. Here a proof without $ lim inf$ and $lim sup$:
Suppose that $s ne 0$. WLOG we assume thatb $s>0$. Since $s_{n_k} to s$ as $ k to infty$, there is $K$ such that $s_{n_k}>0$ for all $k>K$. If $k>K$ and $k$ is odd, we get $(-1)^ks_{n_k}<0$, a contradiction.
answered Nov 20 '18 at 14:50
Fred
44.2k1845
Your proof is fine. Here a proof without $ lim inf$ and $lim sup$:
Suppose that $s ne 0$. WLOG we assume thatb $s>0$. Since $s_{n_k} to s$ as $ k to infty$, there is $K$ such that $s_{n_k}>0$ for all $k>K$. If $k>K$ and $k$ is odd, we get $(-1)^ks_{n_k}<0$, a contradiction.
answered Nov 20 '18 at 14:50
Fred
44.2k1845
answered Nov 20 '18 at 14:50
Fred
44.2k1845
answered Nov 20 '18 at 14:50
Fred
44.2k1845
answered Nov 20 '18 at 14:50
Fred
44.2k1845
44.2k1845
add a comment |
add a comment |
$lim_{n rightarrow infty} s_n=s.$
Then every subsequence converges to $s$.
Given:
$s_{n_k}$ converges to $s$, and $(-1)^k s_{n_k} ge 0$.
1) $k =2m$ then $s_{n_{2m}} ge 0$, and this subsequence of $s_{n_k}$ converges to $s$:
$lim_{m rightarrow infty} s_{n_{2m}} =s ge 0.$
Similarly;
2)$ k=(2m+1)$ then $s_{n_{2m+1}} le 0$, and this subsequence of $s_{n_k}$ converges to $s$.
$lim_{ m rightarrow infty}s_{n_{2m+1}}=s le 0.$
Hence?
Used : If every term of a convergent sequence is $ge 0,$ then the limit is $ge 0$.
answered Nov 20 '18 at 16:18
Peter Szilas
10.7k2720
I am not sure how your answer differs from mine
– DMH16
Nov 21 '18 at 4:11
DMH16.Same idea, without limsup,liminf.
– Peter Szilas
Nov 21 '18 at 8:36
add a comment |
$lim_{n rightarrow infty} s_n=s.$
Then every subsequence converges to $s$.
Given:
$s_{n_k}$ converges to $s$, and $(-1)^k s_{n_k} ge 0$.
1) $k =2m$ then $s_{n_{2m}} ge 0$, and this subsequence of $s_{n_k}$ converges to $s$:
$lim_{m rightarrow infty} s_{n_{2m}} =s ge 0.$
Similarly;
2)$ k=(2m+1)$ then $s_{n_{2m+1}} le 0$, and this subsequence of $s_{n_k}$ converges to $s$.
$lim_{ m rightarrow infty}s_{n_{2m+1}}=s le 0.$
Hence?
Used : If every term of a convergent sequence is $ge 0,$ then the limit is $ge 0$.
answered Nov 20 '18 at 16:18
Peter Szilas
10.7k2720
I am not sure how your answer differs from mine
– DMH16
Nov 21 '18 at 4:11
DMH16.Same idea, without limsup,liminf.
– Peter Szilas
Nov 21 '18 at 8:36
add a comment |
$lim_{n rightarrow infty} s_n=s.$
Then every subsequence converges to $s$.
Given:
$s_{n_k}$ converges to $s$, and $(-1)^k s_{n_k} ge 0$.
1) $k =2m$ then $s_{n_{2m}} ge 0$, and this subsequence of $s_{n_k}$ converges to $s$:
$lim_{m rightarrow infty} s_{n_{2m}} =s ge 0.$
Similarly;
2)$ k=(2m+1)$ then $s_{n_{2m+1}} le 0$, and this subsequence of $s_{n_k}$ converges to $s$.
$lim_{ m rightarrow infty}s_{n_{2m+1}}=s le 0.$
Hence?
Used : If every term of a convergent sequence is $ge 0,$ then the limit is $ge 0$.
answered Nov 20 '18 at 16:18
Peter Szilas
10.7k2720
$lim_{n rightarrow infty} s_n=s.$
Then every subsequence converges to $s$.
Given:
$s_{n_k}$ converges to $s$, and $(-1)^k s_{n_k} ge 0$.
1) $k =2m$ then $s_{n_{2m}} ge 0$, and this subsequence of $s_{n_k}$ converges to $s$:
$lim_{m rightarrow infty} s_{n_{2m}} =s ge 0.$
Similarly;
2)$ k=(2m+1)$ then $s_{n_{2m+1}} le 0$, and this subsequence of $s_{n_k}$ converges to $s$.
$lim_{ m rightarrow infty}s_{n_{2m+1}}=s le 0.$
Hence?
Used : If every term of a convergent sequence is $ge 0,$ then the limit is $ge 0$.
answered Nov 20 '18 at 16:18
Peter Szilas
10.7k2720
answered Nov 20 '18 at 16:18
Peter Szilas
10.7k2720
answered Nov 20 '18 at 16:18
Peter Szilas
10.7k2720
answered Nov 20 '18 at 16:18
Peter Szilas
10.7k2720
10.7k2720
I am not sure how your answer differs from mine
– DMH16
Nov 21 '18 at 4:11
DMH16.Same idea, without limsup,liminf.
– Peter Szilas
Nov 21 '18 at 8:36
add a comment |
I am not sure how your answer differs from mine
– DMH16
Nov 21 '18 at 4:11
DMH16.Same idea, without limsup,liminf.
– Peter Szilas
Nov 21 '18 at 8:36
I am not sure how your answer differs from mine
– DMH16
Nov 21 '18 at 4:11
I am not sure how your answer differs from mine
– DMH16
Nov 21 '18 at 4:11
DMH16.Same idea, without limsup,liminf.
– Peter Szilas
Nov 21 '18 at 8:36
DMH16.Same idea, without limsup,liminf.
– Peter Szilas
Nov 21 '18 at 8:36
add a comment |
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"For these to be equal" -> the sequences are not equal. Their limits (respectively $s$ and $-s$) have to be.
– Clement C.
Nov 20 '18 at 14:45
Sorry, that is what I intended to write
– DMH16
Nov 20 '18 at 14:46