Test the convergence of series $sum frac{a^n}{a^n+x^n}$ when $xneq0$
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Let $ t_n=frac{a^n}{a^n+x^n}$ and using root test
$$ displaystyle lim_{n to infty} {t_n}^{frac{1}{n}}= lim_{n to infty} left(frac{a^n}{a^n+x^n}right)^{frac{1}{n}}= lim_{n to infty} frac{1}{big(1+(frac{x}{a})^nbig)^{frac{1}{n}}}$$
Now I am stuck, I don't know how to evaluate this limit
real-analysis calculus sequences-and-series
asked Dec 20 '18 at 17:35
jirenjiren
766
$endgroup$
add a comment |
$begingroup$
Let $ t_n=frac{a^n}{a^n+x^n}$ and using root test
$$ displaystyle lim_{n to infty} {t_n}^{frac{1}{n}}= lim_{n to infty} left(frac{a^n}{a^n+x^n}right)^{frac{1}{n}}= lim_{n to infty} frac{1}{big(1+(frac{x}{a})^nbig)^{frac{1}{n}}}$$
Now I am stuck, I don't know how to evaluate this limit
real-analysis calculus sequences-and-series
asked Dec 20 '18 at 17:35
jirenjiren
766
$endgroup$
6
$begingroup$
What’s a definition of $e^x$?
$endgroup$
– R.Jackson
Dec 20 '18 at 17:38
$begingroup$
@R.Jackson Thanks for replying. But I don't get it. The exponent $frac{1}{n}$ in the denominator doesn't tends to infinity so how can we use the definition of $e^x$.
$endgroup$
– jiren
Dec 20 '18 at 17:58
add a comment |
$begingroup$
Let $ t_n=frac{a^n}{a^n+x^n}$ and using root test
$$ displaystyle lim_{n to infty} {t_n}^{frac{1}{n}}= lim_{n to infty} left(frac{a^n}{a^n+x^n}right)^{frac{1}{n}}= lim_{n to infty} frac{1}{big(1+(frac{x}{a})^nbig)^{frac{1}{n}}}$$
Now I am stuck, I don't know how to evaluate this limit
real-analysis calculus sequences-and-series
asked Dec 20 '18 at 17:35
jirenjiren
766
$endgroup$
Let $ t_n=frac{a^n}{a^n+x^n}$ and using root test
$$ displaystyle lim_{n to infty} {t_n}^{frac{1}{n}}= lim_{n to infty} left(frac{a^n}{a^n+x^n}right)^{frac{1}{n}}= lim_{n to infty} frac{1}{big(1+(frac{x}{a})^nbig)^{frac{1}{n}}}$$
Now I am stuck, I don't know how to evaluate this limit
real-analysis calculus sequences-and-series
real-analysis calculus sequences-and-series
asked Dec 20 '18 at 17:35
jirenjiren
766
asked Dec 20 '18 at 17:35
jirenjiren
766
asked Dec 20 '18 at 17:35
jirenjiren
766
asked Dec 20 '18 at 17:35
jirenjiren
766
asked Dec 20 '18 at 17:35
jirenjiren
766
766
6
$begingroup$
What’s a definition of $e^x$?
$endgroup$
– R.Jackson
Dec 20 '18 at 17:38
$begingroup$
@R.Jackson Thanks for replying. But I don't get it. The exponent $frac{1}{n}$ in the denominator doesn't tends to infinity so how can we use the definition of $e^x$.
$endgroup$
– jiren
Dec 20 '18 at 17:58
add a comment |
6
$begingroup$
What’s a definition of $e^x$?
$endgroup$
– R.Jackson
Dec 20 '18 at 17:38
$begingroup$
@R.Jackson Thanks for replying. But I don't get it. The exponent $frac{1}{n}$ in the denominator doesn't tends to infinity so how can we use the definition of $e^x$.
$endgroup$
– jiren
Dec 20 '18 at 17:58
6
6
$begingroup$
What’s a definition of $e^x$?
$endgroup$
– R.Jackson
Dec 20 '18 at 17:38
$begingroup$
What’s a definition of $e^x$?
$endgroup$
– R.Jackson
Dec 20 '18 at 17:38
$begingroup$
@R.Jackson Thanks for replying. But I don't get it. The exponent $frac{1}{n}$ in the denominator doesn't tends to infinity so how can we use the definition of $e^x$.
$endgroup$
– jiren
Dec 20 '18 at 17:58
$begingroup$
@R.Jackson Thanks for replying. But I don't get it. The exponent $frac{1}{n}$ in the denominator doesn't tends to infinity so how can we use the definition of $e^x$.
$endgroup$
– jiren
Dec 20 '18 at 17:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$x=a implies t_n=frac 12 $$
$$implies sum t_n text{ diverges}$$
$$|x|<|a|implies lim_{nto+infty}t_n=1$$
$$implies sum t_ntext{ diverges}$$
$$|x|>|a|implies t_nsim (frac ax)^n$$
$$implies sum t_n text{ converges}$$
answered Dec 20 '18 at 18:41
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$$x=a implies t_n=frac 12 $$
$$implies sum t_n text{ diverges}$$
$$|x|<|a|implies lim_{nto+infty}t_n=1$$
$$implies sum t_ntext{ diverges}$$
$$|x|>|a|implies t_nsim (frac ax)^n$$
$$implies sum t_n text{ converges}$$
answered Dec 20 '18 at 18:41
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
$begingroup$
$$x=a implies t_n=frac 12 $$
$$implies sum t_n text{ diverges}$$
$$|x|<|a|implies lim_{nto+infty}t_n=1$$
$$implies sum t_ntext{ diverges}$$
$$|x|>|a|implies t_nsim (frac ax)^n$$
$$implies sum t_n text{ converges}$$
answered Dec 20 '18 at 18:41
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
$begingroup$
$$x=a implies t_n=frac 12 $$
$$implies sum t_n text{ diverges}$$
$$|x|<|a|implies lim_{nto+infty}t_n=1$$
$$implies sum t_ntext{ diverges}$$
$$|x|>|a|implies t_nsim (frac ax)^n$$
$$implies sum t_n text{ converges}$$
answered Dec 20 '18 at 18:41
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$$x=a implies t_n=frac 12 $$
$$implies sum t_n text{ diverges}$$
$$|x|<|a|implies lim_{nto+infty}t_n=1$$
$$implies sum t_ntext{ diverges}$$
$$|x|>|a|implies t_nsim (frac ax)^n$$
$$implies sum t_n text{ converges}$$
answered Dec 20 '18 at 18:41
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 20 '18 at 18:41
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 20 '18 at 18:41
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 20 '18 at 18:41
hamam_Abdallahhamam_Abdallah
38.1k21634
38.1k21634
add a comment |
add a comment |
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6
$begingroup$
What’s a definition of $e^x$?
$endgroup$
– R.Jackson
Dec 20 '18 at 17:38
$begingroup$
@R.Jackson Thanks for replying. But I don't get it. The exponent $frac{1}{n}$ in the denominator doesn't tends to infinity so how can we use the definition of $e^x$.
$endgroup$
– jiren
Dec 20 '18 at 17:58