Mixing
How to show that lim $(1+1/n^k) = 1$ for $ 0<k<1$
$begingroup$
We know that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = e$ for $k=1$.
But for $0< k< 1$ how do I show that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = 1$ for $ 0<k<1$?
real-analysis limits
edited Jan 24 at 2:15
Namaste
1
asked Jan 24 at 2:00
Hs PHs P
275
$endgroup$
add a comment |
$begingroup$
We know that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = e$ for $k=1$.
But for $0< k< 1$ how do I show that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = 1$ for $ 0<k<1$?
real-analysis limits
edited Jan 24 at 2:15
Namaste
1
asked Jan 24 at 2:00
Hs PHs P
275
$endgroup$
4
$begingroup$
You forgot a power outside of your parentheses.
$endgroup$
– lightxbulb
Jan 24 at 2:06
1
$begingroup$
Yes, so I modified it
$endgroup$
– Hs P
Jan 24 at 2:19
add a comment |
$begingroup$
We know that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = e$ for $k=1$.
But for $0< k< 1$ how do I show that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = 1$ for $ 0<k<1$?
real-analysis limits
edited Jan 24 at 2:15
Namaste
1
asked Jan 24 at 2:00
Hs PHs P
275
$endgroup$
We know that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = e$ for $k=1$.
But for $0< k< 1$ how do I show that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = 1$ for $ 0<k<1$?
real-analysis limits
real-analysis limits
edited Jan 24 at 2:15
Namaste
1
asked Jan 24 at 2:00
Hs PHs P
275
edited Jan 24 at 2:15
Namaste
1
asked Jan 24 at 2:00
Hs PHs P
275
edited Jan 24 at 2:15
Namaste
1
edited Jan 24 at 2:15
Namaste
1
edited Jan 24 at 2:15
Namaste
1
1
asked Jan 24 at 2:00
Hs PHs P
275
asked Jan 24 at 2:00
Hs PHs P
275
asked Jan 24 at 2:00
Hs PHs P
275
275
4
$begingroup$
You forgot a power outside of your parentheses.
$endgroup$
– lightxbulb
Jan 24 at 2:06
1
$begingroup$
Yes, so I modified it
$endgroup$
– Hs P
Jan 24 at 2:19
add a comment |
4
$begingroup$
You forgot a power outside of your parentheses.
$endgroup$
– lightxbulb
Jan 24 at 2:06
1
$begingroup$
Yes, so I modified it
$endgroup$
– Hs P
Jan 24 at 2:19
4
4
$begingroup$
You forgot a power outside of your parentheses.
$endgroup$
– lightxbulb
Jan 24 at 2:06
$begingroup$
You forgot a power outside of your parentheses.
$endgroup$
– lightxbulb
Jan 24 at 2:06
1
1
$begingroup$
Yes, so I modified it
$endgroup$
– Hs P
Jan 24 at 2:19
$begingroup$
Yes, so I modified it
$endgroup$
– Hs P
Jan 24 at 2:19
add a comment |
1 Answer
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$begingroup$
Well, $lim_{n rightarrow infty} frac{1}{n^k} = 0$ for fixed $k in (0,1)$, so $lim_{n rightarrow infty} (1-frac{1}{k})$ $ = lim_{n rightarrow infty} 1 + lim_{n rightarrow infty} frac{1}{n^k}$ $ = 1 +0 = 1.$
[Note that $lim_{n rightarrow infty} (1+frac{1}{n})^n = e$ i.e., there is the exponent $n$, and for $k<1$,
$lim_{n rightarrow infty} (1+frac{1}{n^k})^n =$ $lim_{n rightarrow infty} ((1+frac{1}{n^k})^{n^k})^{n^{1-k}} = e^{n^{1-k}} = infty $ ]
answered Jan 24 at 2:10
MikeMike
4,396412
$endgroup$
$begingroup$
Thank you! Maybe you assume that lim$(1+1/n)^{a_{n}}=$ lim$e^{a_{n}/n}$. Is this assumption true?
$endgroup$
– Hs P
Jan 24 at 2:27
add a comment |
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1 Answer
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$begingroup$
Well, $lim_{n rightarrow infty} frac{1}{n^k} = 0$ for fixed $k in (0,1)$, so $lim_{n rightarrow infty} (1-frac{1}{k})$ $ = lim_{n rightarrow infty} 1 + lim_{n rightarrow infty} frac{1}{n^k}$ $ = 1 +0 = 1.$
[Note that $lim_{n rightarrow infty} (1+frac{1}{n})^n = e$ i.e., there is the exponent $n$, and for $k<1$,
$lim_{n rightarrow infty} (1+frac{1}{n^k})^n =$ $lim_{n rightarrow infty} ((1+frac{1}{n^k})^{n^k})^{n^{1-k}} = e^{n^{1-k}} = infty $ ]
answered Jan 24 at 2:10
MikeMike
4,396412
$endgroup$
$begingroup$
Thank you! Maybe you assume that lim$(1+1/n)^{a_{n}}=$ lim$e^{a_{n}/n}$. Is this assumption true?
$endgroup$
– Hs P
Jan 24 at 2:27
add a comment |
$begingroup$
Well, $lim_{n rightarrow infty} frac{1}{n^k} = 0$ for fixed $k in (0,1)$, so $lim_{n rightarrow infty} (1-frac{1}{k})$ $ = lim_{n rightarrow infty} 1 + lim_{n rightarrow infty} frac{1}{n^k}$ $ = 1 +0 = 1.$
[Note that $lim_{n rightarrow infty} (1+frac{1}{n})^n = e$ i.e., there is the exponent $n$, and for $k<1$,
$lim_{n rightarrow infty} (1+frac{1}{n^k})^n =$ $lim_{n rightarrow infty} ((1+frac{1}{n^k})^{n^k})^{n^{1-k}} = e^{n^{1-k}} = infty $ ]
answered Jan 24 at 2:10
MikeMike
4,396412
$endgroup$
$begingroup$
Thank you! Maybe you assume that lim$(1+1/n)^{a_{n}}=$ lim$e^{a_{n}/n}$. Is this assumption true?
$endgroup$
– Hs P
Jan 24 at 2:27
add a comment |
$begingroup$
Well, $lim_{n rightarrow infty} frac{1}{n^k} = 0$ for fixed $k in (0,1)$, so $lim_{n rightarrow infty} (1-frac{1}{k})$ $ = lim_{n rightarrow infty} 1 + lim_{n rightarrow infty} frac{1}{n^k}$ $ = 1 +0 = 1.$
[Note that $lim_{n rightarrow infty} (1+frac{1}{n})^n = e$ i.e., there is the exponent $n$, and for $k<1$,
$lim_{n rightarrow infty} (1+frac{1}{n^k})^n =$ $lim_{n rightarrow infty} ((1+frac{1}{n^k})^{n^k})^{n^{1-k}} = e^{n^{1-k}} = infty $ ]
answered Jan 24 at 2:10
MikeMike
4,396412
$endgroup$
Well, $lim_{n rightarrow infty} frac{1}{n^k} = 0$ for fixed $k in (0,1)$, so $lim_{n rightarrow infty} (1-frac{1}{k})$ $ = lim_{n rightarrow infty} 1 + lim_{n rightarrow infty} frac{1}{n^k}$ $ = 1 +0 = 1.$
[Note that $lim_{n rightarrow infty} (1+frac{1}{n})^n = e$ i.e., there is the exponent $n$, and for $k<1$,
$lim_{n rightarrow infty} (1+frac{1}{n^k})^n =$ $lim_{n rightarrow infty} ((1+frac{1}{n^k})^{n^k})^{n^{1-k}} = e^{n^{1-k}} = infty $ ]
answered Jan 24 at 2:10
MikeMike
4,396412
answered Jan 24 at 2:10
MikeMike
4,396412
answered Jan 24 at 2:10
MikeMike
4,396412
answered Jan 24 at 2:10
MikeMike
4,396412
4,396412
$begingroup$
Thank you! Maybe you assume that lim$(1+1/n)^{a_{n}}=$ lim$e^{a_{n}/n}$. Is this assumption true?
$endgroup$
– Hs P
Jan 24 at 2:27
add a comment |
$begingroup$
Thank you! Maybe you assume that lim$(1+1/n)^{a_{n}}=$ lim$e^{a_{n}/n}$. Is this assumption true?
$endgroup$
– Hs P
Jan 24 at 2:27
$begingroup$
Thank you! Maybe you assume that lim$(1+1/n)^{a_{n}}=$ lim$e^{a_{n}/n}$. Is this assumption true?
$endgroup$
– Hs P
Jan 24 at 2:27
$begingroup$
Thank you! Maybe you assume that lim$(1+1/n)^{a_{n}}=$ lim$e^{a_{n}/n}$. Is this assumption true?
$endgroup$
– Hs P
Jan 24 at 2:27
add a comment |
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4
$begingroup$
You forgot a power outside of your parentheses.
$endgroup$
– lightxbulb
Jan 24 at 2:06
1
$begingroup$
Yes, so I modified it
$endgroup$
– Hs P
Jan 24 at 2:19