Finding the limit of this functionFinding the limit of function - irrational functionset of subsequential...
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Finding the limit of this function
Finding the limit of function - irrational functionset of subsequential limitFinding the limit this recursive functionfinding an 'e' based limit for this sequenceShow this function can be defined as the limit functionHow does this inequality help me evaluate the limit?How do you solve for the limit of this series?Limit of zeta function in $x = 1$Finding the limit of a function .Show that the limit is divergent
$begingroup$
I know that
$$lim_{n rightarrow infty} frac{m^{n}}{e^{n}n^{frac12}} = 0$$ if $|m|lt 1$, otherwise the limit is $infty.$
How can I show that it converges if $|m|lt 1?$
real-analysis limits
asked Mar 14 at 11:20
rodger_kicksrodger_kicks
7910
$endgroup$
add a comment |
$begingroup$
I know that
$$lim_{n rightarrow infty} frac{m^{n}}{e^{n}n^{frac12}} = 0$$ if $|m|lt 1$, otherwise the limit is $infty.$
How can I show that it converges if $|m|lt 1?$
real-analysis limits
asked Mar 14 at 11:20
rodger_kicksrodger_kicks
7910
$endgroup$
$begingroup$
Not true, $|m|<e$ is sufficient for the convergence to $0$
$endgroup$
– Peter
Mar 14 at 11:24
add a comment |
$begingroup$
I know that
$$lim_{n rightarrow infty} frac{m^{n}}{e^{n}n^{frac12}} = 0$$ if $|m|lt 1$, otherwise the limit is $infty.$
How can I show that it converges if $|m|lt 1?$
real-analysis limits
asked Mar 14 at 11:20
rodger_kicksrodger_kicks
7910
$endgroup$
I know that
$$lim_{n rightarrow infty} frac{m^{n}}{e^{n}n^{frac12}} = 0$$ if $|m|lt 1$, otherwise the limit is $infty.$
How can I show that it converges if $|m|lt 1?$
real-analysis limits
real-analysis limits
asked Mar 14 at 11:20
rodger_kicksrodger_kicks
7910
asked Mar 14 at 11:20
rodger_kicksrodger_kicks
7910
asked Mar 14 at 11:20
rodger_kicksrodger_kicks
7910
asked Mar 14 at 11:20
rodger_kicksrodger_kicks
7910
asked Mar 14 at 11:20
rodger_kicksrodger_kicks
7910
7910
$begingroup$
Not true, $|m|<e$ is sufficient for the convergence to $0$
$endgroup$
– Peter
Mar 14 at 11:24
add a comment |
$begingroup$
Not true, $|m|<e$ is sufficient for the convergence to $0$
$endgroup$
– Peter
Mar 14 at 11:24
$begingroup$
Not true, $|m|<e$ is sufficient for the convergence to $0$
$endgroup$
– Peter
Mar 14 at 11:24
$begingroup$
Not true, $|m|<e$ is sufficient for the convergence to $0$
$endgroup$
– Peter
Mar 14 at 11:24
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $a_n=frac{m^{n}}{e^{n}n^{frac12}}.$ Then $|a_n|^{1/n} to frac{|m|}{e}.$
Hence, if $|m|<e$, the series $ sum a_n$ converges, thus $a_n to 0.$
answered Mar 14 at 11:30
FredFred
48.7k11849
$endgroup$
add a comment |
$begingroup$
Hint: For $|m| leq e$ you have $left|frac{m^n}{e^n}right| leq 1$.
answered Mar 14 at 11:31
KlausKlaus
2,732113
$endgroup$
add a comment |
$begingroup$
Clearly $lim_{nto infty}mid a/bmid^n$ converges if $mid a/bmid lt 1$. In your case $mid m/emid lt 1$ or $mid mmidlt e$.
answered Mar 14 at 11:31
Paras KhoslaParas Khosla
2,620323
$endgroup$
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
$begingroup$
Let $a_n=frac{m^{n}}{e^{n}n^{frac12}}.$ Then $|a_n|^{1/n} to frac{|m|}{e}.$
Hence, if $|m|<e$, the series $ sum a_n$ converges, thus $a_n to 0.$
answered Mar 14 at 11:30
FredFred
48.7k11849
$endgroup$
add a comment |
$begingroup$
Let $a_n=frac{m^{n}}{e^{n}n^{frac12}}.$ Then $|a_n|^{1/n} to frac{|m|}{e}.$
Hence, if $|m|<e$, the series $ sum a_n$ converges, thus $a_n to 0.$
answered Mar 14 at 11:30
FredFred
48.7k11849
$endgroup$
add a comment |
$begingroup$
Let $a_n=frac{m^{n}}{e^{n}n^{frac12}}.$ Then $|a_n|^{1/n} to frac{|m|}{e}.$
Hence, if $|m|<e$, the series $ sum a_n$ converges, thus $a_n to 0.$
answered Mar 14 at 11:30
FredFred
48.7k11849
$endgroup$
Let $a_n=frac{m^{n}}{e^{n}n^{frac12}}.$ Then $|a_n|^{1/n} to frac{|m|}{e}.$
Hence, if $|m|<e$, the series $ sum a_n$ converges, thus $a_n to 0.$
answered Mar 14 at 11:30
FredFred
48.7k11849
answered Mar 14 at 11:30
FredFred
48.7k11849
answered Mar 14 at 11:30
FredFred
48.7k11849
answered Mar 14 at 11:30
FredFred
48.7k11849
48.7k11849
add a comment |
add a comment |
$begingroup$
Hint: For $|m| leq e$ you have $left|frac{m^n}{e^n}right| leq 1$.
answered Mar 14 at 11:31
KlausKlaus
2,732113
$endgroup$
add a comment |
$begingroup$
Hint: For $|m| leq e$ you have $left|frac{m^n}{e^n}right| leq 1$.
answered Mar 14 at 11:31
KlausKlaus
2,732113
$endgroup$
add a comment |
$begingroup$
Hint: For $|m| leq e$ you have $left|frac{m^n}{e^n}right| leq 1$.
answered Mar 14 at 11:31
KlausKlaus
2,732113
$endgroup$
Hint: For $|m| leq e$ you have $left|frac{m^n}{e^n}right| leq 1$.
answered Mar 14 at 11:31
KlausKlaus
2,732113
answered Mar 14 at 11:31
KlausKlaus
2,732113
answered Mar 14 at 11:31
KlausKlaus
2,732113
answered Mar 14 at 11:31
KlausKlaus
2,732113
2,732113
add a comment |
add a comment |
$begingroup$
Clearly $lim_{nto infty}mid a/bmid^n$ converges if $mid a/bmid lt 1$. In your case $mid m/emid lt 1$ or $mid mmidlt e$.
answered Mar 14 at 11:31
Paras KhoslaParas Khosla
2,620323
$endgroup$
add a comment |
$begingroup$
Clearly $lim_{nto infty}mid a/bmid^n$ converges if $mid a/bmid lt 1$. In your case $mid m/emid lt 1$ or $mid mmidlt e$.
answered Mar 14 at 11:31
Paras KhoslaParas Khosla
2,620323
$endgroup$
add a comment |
$begingroup$
Clearly $lim_{nto infty}mid a/bmid^n$ converges if $mid a/bmid lt 1$. In your case $mid m/emid lt 1$ or $mid mmidlt e$.
answered Mar 14 at 11:31
Paras KhoslaParas Khosla
2,620323
$endgroup$
Clearly $lim_{nto infty}mid a/bmid^n$ converges if $mid a/bmid lt 1$. In your case $mid m/emid lt 1$ or $mid mmidlt e$.
answered Mar 14 at 11:31
Paras KhoslaParas Khosla
2,620323
answered Mar 14 at 11:31
Paras KhoslaParas Khosla
2,620323
answered Mar 14 at 11:31
Paras KhoslaParas Khosla
2,620323
answered Mar 14 at 11:31
Paras KhoslaParas Khosla
2,620323
2,620323
add a comment |
add a comment |
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$begingroup$
Not true, $|m|<e$ is sufficient for the convergence to $0$
$endgroup$
– Peter
Mar 14 at 11:24