convergence $ sum_{n=1}^{infty} 2^n cdot left(frac{n}{n+1}right)^{n^2} $












3












$begingroup$


How can I check convergence of $ sum_{n=1}^{infty} 2^n cdot left(frac{n}{n+1}right)^{n^2} $ ?
If I want check necessary condition $u_n rightarrow 0$ I need to do sth like that:
$$ u_n = 2^n cdot left(frac{n}{n+1}right)^{n^2} = 2^n cdot left(left(frac{1}{1+frac{1}{n}}right)^{n}right)^2 $$ but now I can't write just
$$ left(left(frac{1}{1+frac{1}{n}}right)^{n}right)^2 rightarrow frac{1}{e^2}$$ because I have $ 2^n $ part...
Thanks for your time.






sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hmmm... $$left(frac{n}{n+1}right)^{n^2} ne left(left(frac{1}{1+frac{1}{n}}right)^{n}right)^2=left(frac{n}{n+1}right)^{2n} $$
    $endgroup$
    – Did
    Dec 13 '18 at 16:27












  • $begingroup$
    Ahh.. My fail, your right, thanks
    $endgroup$
    – VirtualUser
    Dec 13 '18 at 16:30
















3












$begingroup$


How can I check convergence of $ sum_{n=1}^{infty} 2^n cdot left(frac{n}{n+1}right)^{n^2} $ ?
If I want check necessary condition $u_n rightarrow 0$ I need to do sth like that:
$$ u_n = 2^n cdot left(frac{n}{n+1}right)^{n^2} = 2^n cdot left(left(frac{1}{1+frac{1}{n}}right)^{n}right)^2 $$ but now I can't write just
$$ left(left(frac{1}{1+frac{1}{n}}right)^{n}right)^2 rightarrow frac{1}{e^2}$$ because I have $ 2^n $ part...
Thanks for your time.






sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hmmm... $$left(frac{n}{n+1}right)^{n^2} ne left(left(frac{1}{1+frac{1}{n}}right)^{n}right)^2=left(frac{n}{n+1}right)^{2n} $$
    $endgroup$
    – Did
    Dec 13 '18 at 16:27












  • $begingroup$
    Ahh.. My fail, your right, thanks
    $endgroup$
    – VirtualUser
    Dec 13 '18 at 16:30














3












3








3





$begingroup$


How can I check convergence of $ sum_{n=1}^{infty} 2^n cdot left(frac{n}{n+1}right)^{n^2} $ ?
If I want check necessary condition $u_n rightarrow 0$ I need to do sth like that:
$$ u_n = 2^n cdot left(frac{n}{n+1}right)^{n^2} = 2^n cdot left(left(frac{1}{1+frac{1}{n}}right)^{n}right)^2 $$ but now I can't write just
$$ left(left(frac{1}{1+frac{1}{n}}right)^{n}right)^2 rightarrow frac{1}{e^2}$$ because I have $ 2^n $ part...
Thanks for your time.






sequences-and-series







share|cite|improve this question











$endgroup$




How can I check convergence of $ sum_{n=1}^{infty} 2^n cdot left(frac{n}{n+1}right)^{n^2} $ ?
If I want check necessary condition $u_n rightarrow 0$ I need to do sth like that:
$$ u_n = 2^n cdot left(frac{n}{n+1}right)^{n^2} = 2^n cdot left(left(frac{1}{1+frac{1}{n}}right)^{n}right)^2 $$ but now I can't write just
$$ left(left(frac{1}{1+frac{1}{n}}right)^{n}right)^2 rightarrow frac{1}{e^2}$$ because I have $ 2^n $ part...
Thanks for your time.






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 16:30







VirtualUser

















asked Dec 13 '18 at 16:24









VirtualUserVirtualUser

1,293317




1,293317








  • 1




    $begingroup$
    Hmmm... $$left(frac{n}{n+1}right)^{n^2} ne left(left(frac{1}{1+frac{1}{n}}right)^{n}right)^2=left(frac{n}{n+1}right)^{2n} $$
    $endgroup$
    – Did
    Dec 13 '18 at 16:27












  • $begingroup$
    Ahh.. My fail, your right, thanks
    $endgroup$
    – VirtualUser
    Dec 13 '18 at 16:30














  • 1




    $begingroup$
    Hmmm... $$left(frac{n}{n+1}right)^{n^2} ne left(left(frac{1}{1+frac{1}{n}}right)^{n}right)^2=left(frac{n}{n+1}right)^{2n} $$
    $endgroup$
    – Did
    Dec 13 '18 at 16:27












  • $begingroup$
    Ahh.. My fail, your right, thanks
    $endgroup$
    – VirtualUser
    Dec 13 '18 at 16:30








1




1




$begingroup$
Hmmm... $$left(frac{n}{n+1}right)^{n^2} ne left(left(frac{1}{1+frac{1}{n}}right)^{n}right)^2=left(frac{n}{n+1}right)^{2n} $$
$endgroup$
– Did
Dec 13 '18 at 16:27






$begingroup$
Hmmm... $$left(frac{n}{n+1}right)^{n^2} ne left(left(frac{1}{1+frac{1}{n}}right)^{n}right)^2=left(frac{n}{n+1}right)^{2n} $$
$endgroup$
– Did
Dec 13 '18 at 16:27














$begingroup$
Ahh.. My fail, your right, thanks
$endgroup$
– VirtualUser
Dec 13 '18 at 16:30




$begingroup$
Ahh.. My fail, your right, thanks
$endgroup$
– VirtualUser
Dec 13 '18 at 16:30










1 Answer
1






active

oldest

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4












$begingroup$

Using Root test,



$$lim_{ntoinfty}sqrt[n]{u_n}=dfrac2{lim_{ntoinfty}left(1+dfrac1nright)^n}=dfrac2e<1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It is great idea! $frac{1}{e} in left(frac{1}{2};frac{1}{3} right) $ so it is clearly under $1$. But how can I solve necessary condition? I can't use there root test
    $endgroup$
    – VirtualUser
    Dec 13 '18 at 16:32














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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Using Root test,



$$lim_{ntoinfty}sqrt[n]{u_n}=dfrac2{lim_{ntoinfty}left(1+dfrac1nright)^n}=dfrac2e<1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It is great idea! $frac{1}{e} in left(frac{1}{2};frac{1}{3} right) $ so it is clearly under $1$. But how can I solve necessary condition? I can't use there root test
    $endgroup$
    – VirtualUser
    Dec 13 '18 at 16:32


















4












$begingroup$

Using Root test,



$$lim_{ntoinfty}sqrt[n]{u_n}=dfrac2{lim_{ntoinfty}left(1+dfrac1nright)^n}=dfrac2e<1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It is great idea! $frac{1}{e} in left(frac{1}{2};frac{1}{3} right) $ so it is clearly under $1$. But how can I solve necessary condition? I can't use there root test
    $endgroup$
    – VirtualUser
    Dec 13 '18 at 16:32
















4












4








4





$begingroup$

Using Root test,



$$lim_{ntoinfty}sqrt[n]{u_n}=dfrac2{lim_{ntoinfty}left(1+dfrac1nright)^n}=dfrac2e<1$$






share|cite|improve this answer









$endgroup$



Using Root test,



$$lim_{ntoinfty}sqrt[n]{u_n}=dfrac2{lim_{ntoinfty}left(1+dfrac1nright)^n}=dfrac2e<1$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 16:26









lab bhattacharjeelab bhattacharjee

228k15158279




228k15158279












  • $begingroup$
    It is great idea! $frac{1}{e} in left(frac{1}{2};frac{1}{3} right) $ so it is clearly under $1$. But how can I solve necessary condition? I can't use there root test
    $endgroup$
    – VirtualUser
    Dec 13 '18 at 16:32




















  • $begingroup$
    It is great idea! $frac{1}{e} in left(frac{1}{2};frac{1}{3} right) $ so it is clearly under $1$. But how can I solve necessary condition? I can't use there root test
    $endgroup$
    – VirtualUser
    Dec 13 '18 at 16:32


















$begingroup$
It is great idea! $frac{1}{e} in left(frac{1}{2};frac{1}{3} right) $ so it is clearly under $1$. But how can I solve necessary condition? I can't use there root test
$endgroup$
– VirtualUser
Dec 13 '18 at 16:32






$begingroup$
It is great idea! $frac{1}{e} in left(frac{1}{2};frac{1}{3} right) $ so it is clearly under $1$. But how can I solve necessary condition? I can't use there root test
$endgroup$
– VirtualUser
Dec 13 '18 at 16:32




















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