Proving ${frac{n+2}{2n+3}} $ converges to $frac{1}{2}$
I'd just like to verify whether I've done it correctly.
Proof strategy. First we determine what we need to set $N$ equal to;
$$left|frac{n+2}{2n+3}-frac12right|=left|frac{2n+4-(2n+3)}{4n+6}right|= frac1{4n+6}<epsilon$$
Some rewriting get us,
$$n>frac{1}{4epsilon}-frac{6}4{}$$
When $epsilon$ is $frac{1}{6}$ $N$ will be 0 but this is not allowed since N should be a positive integer. We notice:
$$n>frac1{4epsilon}>frac1{4epsilon} -frac64$$
Hence choose $N=lceilfrac1{4epsilon}rceil$
Proof: Let $epsilon>0$ and choose $N=lceil1/4epsilonrceil$. Let $n>N$ where $n in mathbb{Z}$. Then $n>frac1{4epsilon}>frac1{4epsilon}-frac64$ .
Thus, rewrite to get, $ frac{1}{4n+6} < epsilon$
Therefore, $$left|frac{n+2}{2n+3}-frac12right|=left|frac{2n+4-(2n+3)}{4n+6}right|=frac1{4n+6}<epsilon$$
Hence the sequence converges to $frac{1}{2}$. $$space blacksquare$$
real-analysis proof-verification
asked Oct 8 '18 at 16:37
Cro-MagnonCro-Magnon
501212
add a comment |
I'd just like to verify whether I've done it correctly.
Proof strategy. First we determine what we need to set $N$ equal to;
$$left|frac{n+2}{2n+3}-frac12right|=left|frac{2n+4-(2n+3)}{4n+6}right|= frac1{4n+6}<epsilon$$
Some rewriting get us,
$$n>frac{1}{4epsilon}-frac{6}4{}$$
When $epsilon$ is $frac{1}{6}$ $N$ will be 0 but this is not allowed since N should be a positive integer. We notice:
$$n>frac1{4epsilon}>frac1{4epsilon} -frac64$$
Hence choose $N=lceilfrac1{4epsilon}rceil$
Proof: Let $epsilon>0$ and choose $N=lceil1/4epsilonrceil$. Let $n>N$ where $n in mathbb{Z}$. Then $n>frac1{4epsilon}>frac1{4epsilon}-frac64$ .
Thus, rewrite to get, $ frac{1}{4n+6} < epsilon$
Therefore, $$left|frac{n+2}{2n+3}-frac12right|=left|frac{2n+4-(2n+3)}{4n+6}right|=frac1{4n+6}<epsilon$$
Hence the sequence converges to $frac{1}{2}$. $$space blacksquare$$
real-analysis proof-verification
asked Oct 8 '18 at 16:37
Cro-MagnonCro-Magnon
501212
3
$frac{1}{4epsilon}$ isn't always an integer, though. You can fix that by, for instance, using the ceiling function, or being less specific and saying that you pick an arbitrary $N$ greater than or equal to $frac{1}{4epsilon}$.
– Arthur
Oct 8 '18 at 16:39
@Arthur I did so in my proof but forgot to include it in the proof strategy.
– Cro-Magnon
Oct 8 '18 at 16:42
add a comment |
I'd just like to verify whether I've done it correctly.
Proof strategy. First we determine what we need to set $N$ equal to;
$$left|frac{n+2}{2n+3}-frac12right|=left|frac{2n+4-(2n+3)}{4n+6}right|= frac1{4n+6}<epsilon$$
Some rewriting get us,
$$n>frac{1}{4epsilon}-frac{6}4{}$$
When $epsilon$ is $frac{1}{6}$ $N$ will be 0 but this is not allowed since N should be a positive integer. We notice:
$$n>frac1{4epsilon}>frac1{4epsilon} -frac64$$
Hence choose $N=lceilfrac1{4epsilon}rceil$
Proof: Let $epsilon>0$ and choose $N=lceil1/4epsilonrceil$. Let $n>N$ where $n in mathbb{Z}$. Then $n>frac1{4epsilon}>frac1{4epsilon}-frac64$ .
Thus, rewrite to get, $ frac{1}{4n+6} < epsilon$
Therefore, $$left|frac{n+2}{2n+3}-frac12right|=left|frac{2n+4-(2n+3)}{4n+6}right|=frac1{4n+6}<epsilon$$
Hence the sequence converges to $frac{1}{2}$. $$space blacksquare$$
real-analysis proof-verification
asked Oct 8 '18 at 16:37
Cro-MagnonCro-Magnon
501212
I'd just like to verify whether I've done it correctly.
Proof strategy. First we determine what we need to set $N$ equal to;
$$left|frac{n+2}{2n+3}-frac12right|=left|frac{2n+4-(2n+3)}{4n+6}right|= frac1{4n+6}<epsilon$$
Some rewriting get us,
$$n>frac{1}{4epsilon}-frac{6}4{}$$
When $epsilon$ is $frac{1}{6}$ $N$ will be 0 but this is not allowed since N should be a positive integer. We notice:
$$n>frac1{4epsilon}>frac1{4epsilon} -frac64$$
Hence choose $N=lceilfrac1{4epsilon}rceil$
Proof: Let $epsilon>0$ and choose $N=lceil1/4epsilonrceil$. Let $n>N$ where $n in mathbb{Z}$. Then $n>frac1{4epsilon}>frac1{4epsilon}-frac64$ .
Thus, rewrite to get, $ frac{1}{4n+6} < epsilon$
Therefore, $$left|frac{n+2}{2n+3}-frac12right|=left|frac{2n+4-(2n+3)}{4n+6}right|=frac1{4n+6}<epsilon$$
Hence the sequence converges to $frac{1}{2}$. $$space blacksquare$$
real-analysis proof-verification
real-analysis proof-verification
asked Oct 8 '18 at 16:37
Cro-MagnonCro-Magnon
501212
asked Oct 8 '18 at 16:37
Cro-MagnonCro-Magnon
501212
edited Oct 8 '18 at 16:41
Cro-Magnon
asked Oct 8 '18 at 16:37
Cro-MagnonCro-Magnon
501212
asked Oct 8 '18 at 16:37
Cro-MagnonCro-Magnon
501212
asked Oct 8 '18 at 16:37
Cro-MagnonCro-Magnon
501212
501212
3
$frac{1}{4epsilon}$ isn't always an integer, though. You can fix that by, for instance, using the ceiling function, or being less specific and saying that you pick an arbitrary $N$ greater than or equal to $frac{1}{4epsilon}$.
– Arthur
Oct 8 '18 at 16:39
@Arthur I did so in my proof but forgot to include it in the proof strategy.
– Cro-Magnon
Oct 8 '18 at 16:42
add a comment |
3
$frac{1}{4epsilon}$ isn't always an integer, though. You can fix that by, for instance, using the ceiling function, or being less specific and saying that you pick an arbitrary $N$ greater than or equal to $frac{1}{4epsilon}$.
– Arthur
Oct 8 '18 at 16:39
@Arthur I did so in my proof but forgot to include it in the proof strategy.
– Cro-Magnon
Oct 8 '18 at 16:42
3
3
$frac{1}{4epsilon}$ isn't always an integer, though. You can fix that by, for instance, using the ceiling function, or being less specific and saying that you pick an arbitrary $N$ greater than or equal to $frac{1}{4epsilon}$.
– Arthur
Oct 8 '18 at 16:39
$frac{1}{4epsilon}$ isn't always an integer, though. You can fix that by, for instance, using the ceiling function, or being less specific and saying that you pick an arbitrary $N$ greater than or equal to $frac{1}{4epsilon}$.
– Arthur
Oct 8 '18 at 16:39
@Arthur I did so in my proof but forgot to include it in the proof strategy.
– Cro-Magnon
Oct 8 '18 at 16:42
@Arthur I did so in my proof but forgot to include it in the proof strategy.
– Cro-Magnon
Oct 8 '18 at 16:42
add a comment |
1 Answer
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Yes, it is correct, apart from the part where you write "We notice n > ...". You should omit the part "n > " there and simply write "We notice $frac1{4varepsilon} > frac1{4varepsilon} - frac64.$"
answered Dec 5 '18 at 12:08
StefanieStefanie
506
add a comment |
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Yes, it is correct, apart from the part where you write "We notice n > ...". You should omit the part "n > " there and simply write "We notice $frac1{4varepsilon} > frac1{4varepsilon} - frac64.$"
answered Dec 5 '18 at 12:08
StefanieStefanie
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add a comment |
Yes, it is correct, apart from the part where you write "We notice n > ...". You should omit the part "n > " there and simply write "We notice $frac1{4varepsilon} > frac1{4varepsilon} - frac64.$"
answered Dec 5 '18 at 12:08
StefanieStefanie
506
add a comment |
Yes, it is correct, apart from the part where you write "We notice n > ...". You should omit the part "n > " there and simply write "We notice $frac1{4varepsilon} > frac1{4varepsilon} - frac64.$"
answered Dec 5 '18 at 12:08
StefanieStefanie
506
Yes, it is correct, apart from the part where you write "We notice n > ...". You should omit the part "n > " there and simply write "We notice $frac1{4varepsilon} > frac1{4varepsilon} - frac64.$"
answered Dec 5 '18 at 12:08
StefanieStefanie
506
answered Dec 5 '18 at 12:08
StefanieStefanie
506
answered Dec 5 '18 at 12:08
StefanieStefanie
506
answered Dec 5 '18 at 12:08
StefanieStefanie
506
506
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$frac{1}{4epsilon}$ isn't always an integer, though. You can fix that by, for instance, using the ceiling function, or being less specific and saying that you pick an arbitrary $N$ greater than or equal to $frac{1}{4epsilon}$.
– Arthur
Oct 8 '18 at 16:39
@Arthur I did so in my proof but forgot to include it in the proof strategy.
– Cro-Magnon
Oct 8 '18 at 16:42