Prove: $sum _{n=1}^{infty} left(frac{n^{2} + 1}{n^{2} +n +1}right)^{n^2} < infty$ without...












3












$begingroup$


Given $sum _{n=1}^{infty} left(frac{n^{2} + 1}{n^{2} +n +1}right)^{n^2}$, prove that it converges.



I tried to use the Ratio test.
I got a terrible algebraic expression: $$lim_{n to infty} frac{a_{n+1}}{a_n} = lim_{n to infty} left(frac{n^4 +2n^3 +3n^2 + 2n +2}{n^4 + 4n^3 + 7n^2 + 6n +3}right)cdot left(frac{n^2 + 2n + 2}{n^2 + 3n + 3}right)^{2n +1}$$



Now problem is I'm not allowed to use L'Hoptial and I don't really know what to do with this disaster.






real-analysis calculus sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Tip: Note that the summands are equal to $(1-frac{n}{n^2+n+1})^{n^2}$.
    $endgroup$
    – Serg
    Dec 8 '18 at 15:44
















3












$begingroup$


Given $sum _{n=1}^{infty} left(frac{n^{2} + 1}{n^{2} +n +1}right)^{n^2}$, prove that it converges.



I tried to use the Ratio test.
I got a terrible algebraic expression: $$lim_{n to infty} frac{a_{n+1}}{a_n} = lim_{n to infty} left(frac{n^4 +2n^3 +3n^2 + 2n +2}{n^4 + 4n^3 + 7n^2 + 6n +3}right)cdot left(frac{n^2 + 2n + 2}{n^2 + 3n + 3}right)^{2n +1}$$



Now problem is I'm not allowed to use L'Hoptial and I don't really know what to do with this disaster.






real-analysis calculus sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Tip: Note that the summands are equal to $(1-frac{n}{n^2+n+1})^{n^2}$.
    $endgroup$
    – Serg
    Dec 8 '18 at 15:44














3












3








3


1



$begingroup$


Given $sum _{n=1}^{infty} left(frac{n^{2} + 1}{n^{2} +n +1}right)^{n^2}$, prove that it converges.



I tried to use the Ratio test.
I got a terrible algebraic expression: $$lim_{n to infty} frac{a_{n+1}}{a_n} = lim_{n to infty} left(frac{n^4 +2n^3 +3n^2 + 2n +2}{n^4 + 4n^3 + 7n^2 + 6n +3}right)cdot left(frac{n^2 + 2n + 2}{n^2 + 3n + 3}right)^{2n +1}$$



Now problem is I'm not allowed to use L'Hoptial and I don't really know what to do with this disaster.






real-analysis calculus sequences-and-series







share|cite|improve this question











$endgroup$




Given $sum _{n=1}^{infty} left(frac{n^{2} + 1}{n^{2} +n +1}right)^{n^2}$, prove that it converges.



I tried to use the Ratio test.
I got a terrible algebraic expression: $$lim_{n to infty} frac{a_{n+1}}{a_n} = lim_{n to infty} left(frac{n^4 +2n^3 +3n^2 + 2n +2}{n^4 + 4n^3 + 7n^2 + 6n +3}right)cdot left(frac{n^2 + 2n + 2}{n^2 + 3n + 3}right)^{2n +1}$$



Now problem is I'm not allowed to use L'Hoptial and I don't really know what to do with this disaster.






real-analysis calculus sequences-and-series




real-analysis calculus 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 8 '18 at 15:41









Dando18

4,73241235




4,73241235










asked Dec 8 '18 at 15:40









JnevenJneven

943322




943322












  • $begingroup$
    Tip: Note that the summands are equal to $(1-frac{n}{n^2+n+1})^{n^2}$.
    $endgroup$
    – Serg
    Dec 8 '18 at 15:44


















  • $begingroup$
    Tip: Note that the summands are equal to $(1-frac{n}{n^2+n+1})^{n^2}$.
    $endgroup$
    – Serg
    Dec 8 '18 at 15:44
















$begingroup$
Tip: Note that the summands are equal to $(1-frac{n}{n^2+n+1})^{n^2}$.
$endgroup$
– Serg
Dec 8 '18 at 15:44




$begingroup$
Tip: Note that the summands are equal to $(1-frac{n}{n^2+n+1})^{n^2}$.
$endgroup$
– Serg
Dec 8 '18 at 15:44










3 Answers
3






active

oldest

votes


















2












$begingroup$

Try the root criterion.



Prove that
$$sqrt[n]{left(frac{n^2+1}{n^2+n+1}right)^{n^2}}=left(1-frac{n}{n^2+n+1}right)^n$$
and that this tends to $L<1$ as $nto infty$. Then, the criterion implies that the sum converges.



To find the limit, remember that if $a_nto 0$ then
$$(1+a_n)^{frac1{a_n}}to e.$$



So
$$left(1-frac{n}{n^2+n+1}right)^n=left(1-frac{n}{n^2+n+1}right)^{left(-frac{n^2+n+1}{n}right)cdot left(-frac{n}{n^2+n+1}right)cdot n}=$$
$$=left[left(1-frac{n}{n^2+n+1}right)^{-frac{n^2+n+1}{n}}right]^{ left(-frac{n}{n^2+n+1}right)cdot n}.$$



But the bracketed expression is exactly of the form
$$(1+a_n)^{frac1{a_n}},quad a_nto 0,$$
so it tends to $e$. And the expression in the exponent clearly goes to $-1$, so the limit is $e^{-1}<1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    yes, I tried this version too. couldn't prove that this goes to 0, or even just smaller then 1.
    $endgroup$
    – Jneven
    Dec 8 '18 at 16:25










  • $begingroup$
    Check the update.
    $endgroup$
    – Alejandro Nasif Salum
    Dec 8 '18 at 19:42



















2












$begingroup$

Another approach using the Root test.



Let $L = lim_{ntoinfty} sqrt[n]{left(frac{n^2+1}{n^2+n+1}right)^{n^2}} $



$$ begin{align}
L &= lim_{ntoinfty} exp log sqrt[n]{left(frac{n^2+1}{n^2+n+1}right)^{n^2}} \
&= exp lim_{ntoinfty} frac{log left(left( frac{n^2+1}{n^2+n+1} right)^{n^2}right)}{n} \
&= exp lim_{ntoinfty} frac{n^2 log frac{n^2+1}{n^2+n+1}}{n} \
&= exp lim_{ntoinfty} n log frac{n^2+1}{n^2+n+1} tag{1} \
&= exp(-1) \
end{align}$$



and since $L = exp(-1) < 1$ the series is absolutely convergent by the root test.



* (1) can be shown by the fact that the Laurent expansion for $nlogfrac{n^2+1}{n^2+n+1}$ at $n=infty$ is $-1 + textrm{O}(frac{1}{n})$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    how did you come up to the the term exp(-1) from $n log {frac{n^2 +1}{n^2 + n +1}}$? that's exactly the problem
    $endgroup$
    – Jneven
    Dec 8 '18 at 16:35












  • $begingroup$
    @Jneven see my edit
    $endgroup$
    – Dando18
    Dec 8 '18 at 16:50










  • $begingroup$
    Laurent expansion is much higher level of mathematics then what i'm famiilair with, at this moment. this is of course a solid solution - but as I don't get it, I can't solve my problem using it...
    $endgroup$
    – Jneven
    Dec 9 '18 at 11:08



















1












$begingroup$

By Taylor's expansion we have



$$left(frac{n^{2} + 1}{n^{2} +n +1}right)^{n^2}=left(1-frac{n}{n^{2} +n +1}right)^{n^2}=e^{n^2 logleft(1-frac{n}{n^{2} +n +1}right)}=e^{n^2 left(frac{-n}{n^{2} +n +1}+O(1/n^2)right)}sim frac c{e^n}$$



and then refer to limit comparison test.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    that's perfect! incredibly aesthetic!
    $endgroup$
    – Jneven
    Dec 8 '18 at 16:27






  • 1




    $begingroup$
    @Jneven I'm glad you appreciate that so much! Thanks :)
    $endgroup$
    – gimusi
    Dec 8 '18 at 16:28











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Try the root criterion.



Prove that
$$sqrt[n]{left(frac{n^2+1}{n^2+n+1}right)^{n^2}}=left(1-frac{n}{n^2+n+1}right)^n$$
and that this tends to $L<1$ as $nto infty$. Then, the criterion implies that the sum converges.



To find the limit, remember that if $a_nto 0$ then
$$(1+a_n)^{frac1{a_n}}to e.$$



So
$$left(1-frac{n}{n^2+n+1}right)^n=left(1-frac{n}{n^2+n+1}right)^{left(-frac{n^2+n+1}{n}right)cdot left(-frac{n}{n^2+n+1}right)cdot n}=$$
$$=left[left(1-frac{n}{n^2+n+1}right)^{-frac{n^2+n+1}{n}}right]^{ left(-frac{n}{n^2+n+1}right)cdot n}.$$



But the bracketed expression is exactly of the form
$$(1+a_n)^{frac1{a_n}},quad a_nto 0,$$
so it tends to $e$. And the expression in the exponent clearly goes to $-1$, so the limit is $e^{-1}<1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    yes, I tried this version too. couldn't prove that this goes to 0, or even just smaller then 1.
    $endgroup$
    – Jneven
    Dec 8 '18 at 16:25










  • $begingroup$
    Check the update.
    $endgroup$
    – Alejandro Nasif Salum
    Dec 8 '18 at 19:42
















2












$begingroup$

Try the root criterion.



Prove that
$$sqrt[n]{left(frac{n^2+1}{n^2+n+1}right)^{n^2}}=left(1-frac{n}{n^2+n+1}right)^n$$
and that this tends to $L<1$ as $nto infty$. Then, the criterion implies that the sum converges.



To find the limit, remember that if $a_nto 0$ then
$$(1+a_n)^{frac1{a_n}}to e.$$



So
$$left(1-frac{n}{n^2+n+1}right)^n=left(1-frac{n}{n^2+n+1}right)^{left(-frac{n^2+n+1}{n}right)cdot left(-frac{n}{n^2+n+1}right)cdot n}=$$
$$=left[left(1-frac{n}{n^2+n+1}right)^{-frac{n^2+n+1}{n}}right]^{ left(-frac{n}{n^2+n+1}right)cdot n}.$$



But the bracketed expression is exactly of the form
$$(1+a_n)^{frac1{a_n}},quad a_nto 0,$$
so it tends to $e$. And the expression in the exponent clearly goes to $-1$, so the limit is $e^{-1}<1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    yes, I tried this version too. couldn't prove that this goes to 0, or even just smaller then 1.
    $endgroup$
    – Jneven
    Dec 8 '18 at 16:25










  • $begingroup$
    Check the update.
    $endgroup$
    – Alejandro Nasif Salum
    Dec 8 '18 at 19:42














2












2








2





$begingroup$

Try the root criterion.



Prove that
$$sqrt[n]{left(frac{n^2+1}{n^2+n+1}right)^{n^2}}=left(1-frac{n}{n^2+n+1}right)^n$$
and that this tends to $L<1$ as $nto infty$. Then, the criterion implies that the sum converges.



To find the limit, remember that if $a_nto 0$ then
$$(1+a_n)^{frac1{a_n}}to e.$$



So
$$left(1-frac{n}{n^2+n+1}right)^n=left(1-frac{n}{n^2+n+1}right)^{left(-frac{n^2+n+1}{n}right)cdot left(-frac{n}{n^2+n+1}right)cdot n}=$$
$$=left[left(1-frac{n}{n^2+n+1}right)^{-frac{n^2+n+1}{n}}right]^{ left(-frac{n}{n^2+n+1}right)cdot n}.$$



But the bracketed expression is exactly of the form
$$(1+a_n)^{frac1{a_n}},quad a_nto 0,$$
so it tends to $e$. And the expression in the exponent clearly goes to $-1$, so the limit is $e^{-1}<1$.






share|cite|improve this answer











$endgroup$



Try the root criterion.



Prove that
$$sqrt[n]{left(frac{n^2+1}{n^2+n+1}right)^{n^2}}=left(1-frac{n}{n^2+n+1}right)^n$$
and that this tends to $L<1$ as $nto infty$. Then, the criterion implies that the sum converges.



To find the limit, remember that if $a_nto 0$ then
$$(1+a_n)^{frac1{a_n}}to e.$$



So
$$left(1-frac{n}{n^2+n+1}right)^n=left(1-frac{n}{n^2+n+1}right)^{left(-frac{n^2+n+1}{n}right)cdot left(-frac{n}{n^2+n+1}right)cdot n}=$$
$$=left[left(1-frac{n}{n^2+n+1}right)^{-frac{n^2+n+1}{n}}right]^{ left(-frac{n}{n^2+n+1}right)cdot n}.$$



But the bracketed expression is exactly of the form
$$(1+a_n)^{frac1{a_n}},quad a_nto 0,$$
so it tends to $e$. And the expression in the exponent clearly goes to $-1$, so the limit is $e^{-1}<1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 16:44

























answered Dec 8 '18 at 15:47









Alejandro Nasif SalumAlejandro Nasif Salum

4,765118




4,765118












  • $begingroup$
    yes, I tried this version too. couldn't prove that this goes to 0, or even just smaller then 1.
    $endgroup$
    – Jneven
    Dec 8 '18 at 16:25










  • $begingroup$
    Check the update.
    $endgroup$
    – Alejandro Nasif Salum
    Dec 8 '18 at 19:42


















  • $begingroup$
    yes, I tried this version too. couldn't prove that this goes to 0, or even just smaller then 1.
    $endgroup$
    – Jneven
    Dec 8 '18 at 16:25










  • $begingroup$
    Check the update.
    $endgroup$
    – Alejandro Nasif Salum
    Dec 8 '18 at 19:42
















$begingroup$
yes, I tried this version too. couldn't prove that this goes to 0, or even just smaller then 1.
$endgroup$
– Jneven
Dec 8 '18 at 16:25




$begingroup$
yes, I tried this version too. couldn't prove that this goes to 0, or even just smaller then 1.
$endgroup$
– Jneven
Dec 8 '18 at 16:25












$begingroup$
Check the update.
$endgroup$
– Alejandro Nasif Salum
Dec 8 '18 at 19:42




$begingroup$
Check the update.
$endgroup$
– Alejandro Nasif Salum
Dec 8 '18 at 19:42











2












$begingroup$

Another approach using the Root test.



Let $L = lim_{ntoinfty} sqrt[n]{left(frac{n^2+1}{n^2+n+1}right)^{n^2}} $



$$ begin{align}
L &= lim_{ntoinfty} exp log sqrt[n]{left(frac{n^2+1}{n^2+n+1}right)^{n^2}} \
&= exp lim_{ntoinfty} frac{log left(left( frac{n^2+1}{n^2+n+1} right)^{n^2}right)}{n} \
&= exp lim_{ntoinfty} frac{n^2 log frac{n^2+1}{n^2+n+1}}{n} \
&= exp lim_{ntoinfty} n log frac{n^2+1}{n^2+n+1} tag{1} \
&= exp(-1) \
end{align}$$



and since $L = exp(-1) < 1$ the series is absolutely convergent by the root test.



* (1) can be shown by the fact that the Laurent expansion for $nlogfrac{n^2+1}{n^2+n+1}$ at $n=infty$ is $-1 + textrm{O}(frac{1}{n})$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    how did you come up to the the term exp(-1) from $n log {frac{n^2 +1}{n^2 + n +1}}$? that's exactly the problem
    $endgroup$
    – Jneven
    Dec 8 '18 at 16:35












  • $begingroup$
    @Jneven see my edit
    $endgroup$
    – Dando18
    Dec 8 '18 at 16:50










  • $begingroup$
    Laurent expansion is much higher level of mathematics then what i'm famiilair with, at this moment. this is of course a solid solution - but as I don't get it, I can't solve my problem using it...
    $endgroup$
    – Jneven
    Dec 9 '18 at 11:08
















2












$begingroup$

Another approach using the Root test.



Let $L = lim_{ntoinfty} sqrt[n]{left(frac{n^2+1}{n^2+n+1}right)^{n^2}} $



$$ begin{align}
L &= lim_{ntoinfty} exp log sqrt[n]{left(frac{n^2+1}{n^2+n+1}right)^{n^2}} \
&= exp lim_{ntoinfty} frac{log left(left( frac{n^2+1}{n^2+n+1} right)^{n^2}right)}{n} \
&= exp lim_{ntoinfty} frac{n^2 log frac{n^2+1}{n^2+n+1}}{n} \
&= exp lim_{ntoinfty} n log frac{n^2+1}{n^2+n+1} tag{1} \
&= exp(-1) \
end{align}$$



and since $L = exp(-1) < 1$ the series is absolutely convergent by the root test.



* (1) can be shown by the fact that the Laurent expansion for $nlogfrac{n^2+1}{n^2+n+1}$ at $n=infty$ is $-1 + textrm{O}(frac{1}{n})$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    how did you come up to the the term exp(-1) from $n log {frac{n^2 +1}{n^2 + n +1}}$? that's exactly the problem
    $endgroup$
    – Jneven
    Dec 8 '18 at 16:35












  • $begingroup$
    @Jneven see my edit
    $endgroup$
    – Dando18
    Dec 8 '18 at 16:50










  • $begingroup$
    Laurent expansion is much higher level of mathematics then what i'm famiilair with, at this moment. this is of course a solid solution - but as I don't get it, I can't solve my problem using it...
    $endgroup$
    – Jneven
    Dec 9 '18 at 11:08














2












2








2





$begingroup$

Another approach using the Root test.



Let $L = lim_{ntoinfty} sqrt[n]{left(frac{n^2+1}{n^2+n+1}right)^{n^2}} $



$$ begin{align}
L &= lim_{ntoinfty} exp log sqrt[n]{left(frac{n^2+1}{n^2+n+1}right)^{n^2}} \
&= exp lim_{ntoinfty} frac{log left(left( frac{n^2+1}{n^2+n+1} right)^{n^2}right)}{n} \
&= exp lim_{ntoinfty} frac{n^2 log frac{n^2+1}{n^2+n+1}}{n} \
&= exp lim_{ntoinfty} n log frac{n^2+1}{n^2+n+1} tag{1} \
&= exp(-1) \
end{align}$$



and since $L = exp(-1) < 1$ the series is absolutely convergent by the root test.



* (1) can be shown by the fact that the Laurent expansion for $nlogfrac{n^2+1}{n^2+n+1}$ at $n=infty$ is $-1 + textrm{O}(frac{1}{n})$.






share|cite|improve this answer











$endgroup$



Another approach using the Root test.



Let $L = lim_{ntoinfty} sqrt[n]{left(frac{n^2+1}{n^2+n+1}right)^{n^2}} $



$$ begin{align}
L &= lim_{ntoinfty} exp log sqrt[n]{left(frac{n^2+1}{n^2+n+1}right)^{n^2}} \
&= exp lim_{ntoinfty} frac{log left(left( frac{n^2+1}{n^2+n+1} right)^{n^2}right)}{n} \
&= exp lim_{ntoinfty} frac{n^2 log frac{n^2+1}{n^2+n+1}}{n} \
&= exp lim_{ntoinfty} n log frac{n^2+1}{n^2+n+1} tag{1} \
&= exp(-1) \
end{align}$$



and since $L = exp(-1) < 1$ the series is absolutely convergent by the root test.



* (1) can be shown by the fact that the Laurent expansion for $nlogfrac{n^2+1}{n^2+n+1}$ at $n=infty$ is $-1 + textrm{O}(frac{1}{n})$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 16:49

























answered Dec 8 '18 at 15:55









Dando18Dando18

4,73241235




4,73241235












  • $begingroup$
    how did you come up to the the term exp(-1) from $n log {frac{n^2 +1}{n^2 + n +1}}$? that's exactly the problem
    $endgroup$
    – Jneven
    Dec 8 '18 at 16:35












  • $begingroup$
    @Jneven see my edit
    $endgroup$
    – Dando18
    Dec 8 '18 at 16:50










  • $begingroup$
    Laurent expansion is much higher level of mathematics then what i'm famiilair with, at this moment. this is of course a solid solution - but as I don't get it, I can't solve my problem using it...
    $endgroup$
    – Jneven
    Dec 9 '18 at 11:08


















  • $begingroup$
    how did you come up to the the term exp(-1) from $n log {frac{n^2 +1}{n^2 + n +1}}$? that's exactly the problem
    $endgroup$
    – Jneven
    Dec 8 '18 at 16:35












  • $begingroup$
    @Jneven see my edit
    $endgroup$
    – Dando18
    Dec 8 '18 at 16:50










  • $begingroup$
    Laurent expansion is much higher level of mathematics then what i'm famiilair with, at this moment. this is of course a solid solution - but as I don't get it, I can't solve my problem using it...
    $endgroup$
    – Jneven
    Dec 9 '18 at 11:08
















$begingroup$
how did you come up to the the term exp(-1) from $n log {frac{n^2 +1}{n^2 + n +1}}$? that's exactly the problem
$endgroup$
– Jneven
Dec 8 '18 at 16:35






$begingroup$
how did you come up to the the term exp(-1) from $n log {frac{n^2 +1}{n^2 + n +1}}$? that's exactly the problem
$endgroup$
– Jneven
Dec 8 '18 at 16:35














$begingroup$
@Jneven see my edit
$endgroup$
– Dando18
Dec 8 '18 at 16:50




$begingroup$
@Jneven see my edit
$endgroup$
– Dando18
Dec 8 '18 at 16:50












$begingroup$
Laurent expansion is much higher level of mathematics then what i'm famiilair with, at this moment. this is of course a solid solution - but as I don't get it, I can't solve my problem using it...
$endgroup$
– Jneven
Dec 9 '18 at 11:08




$begingroup$
Laurent expansion is much higher level of mathematics then what i'm famiilair with, at this moment. this is of course a solid solution - but as I don't get it, I can't solve my problem using it...
$endgroup$
– Jneven
Dec 9 '18 at 11:08











1












$begingroup$

By Taylor's expansion we have



$$left(frac{n^{2} + 1}{n^{2} +n +1}right)^{n^2}=left(1-frac{n}{n^{2} +n +1}right)^{n^2}=e^{n^2 logleft(1-frac{n}{n^{2} +n +1}right)}=e^{n^2 left(frac{-n}{n^{2} +n +1}+O(1/n^2)right)}sim frac c{e^n}$$



and then refer to limit comparison test.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    that's perfect! incredibly aesthetic!
    $endgroup$
    – Jneven
    Dec 8 '18 at 16:27






  • 1




    $begingroup$
    @Jneven I'm glad you appreciate that so much! Thanks :)
    $endgroup$
    – gimusi
    Dec 8 '18 at 16:28
















1












$begingroup$

By Taylor's expansion we have



$$left(frac{n^{2} + 1}{n^{2} +n +1}right)^{n^2}=left(1-frac{n}{n^{2} +n +1}right)^{n^2}=e^{n^2 logleft(1-frac{n}{n^{2} +n +1}right)}=e^{n^2 left(frac{-n}{n^{2} +n +1}+O(1/n^2)right)}sim frac c{e^n}$$



and then refer to limit comparison test.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    that's perfect! incredibly aesthetic!
    $endgroup$
    – Jneven
    Dec 8 '18 at 16:27






  • 1




    $begingroup$
    @Jneven I'm glad you appreciate that so much! Thanks :)
    $endgroup$
    – gimusi
    Dec 8 '18 at 16:28














1












1








1





$begingroup$

By Taylor's expansion we have



$$left(frac{n^{2} + 1}{n^{2} +n +1}right)^{n^2}=left(1-frac{n}{n^{2} +n +1}right)^{n^2}=e^{n^2 logleft(1-frac{n}{n^{2} +n +1}right)}=e^{n^2 left(frac{-n}{n^{2} +n +1}+O(1/n^2)right)}sim frac c{e^n}$$



and then refer to limit comparison test.






share|cite|improve this answer









$endgroup$



By Taylor's expansion we have



$$left(frac{n^{2} + 1}{n^{2} +n +1}right)^{n^2}=left(1-frac{n}{n^{2} +n +1}right)^{n^2}=e^{n^2 logleft(1-frac{n}{n^{2} +n +1}right)}=e^{n^2 left(frac{-n}{n^{2} +n +1}+O(1/n^2)right)}sim frac c{e^n}$$



and then refer to limit comparison test.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 '18 at 16:01









gimusigimusi

93k84594




93k84594








  • 1




    $begingroup$
    that's perfect! incredibly aesthetic!
    $endgroup$
    – Jneven
    Dec 8 '18 at 16:27






  • 1




    $begingroup$
    @Jneven I'm glad you appreciate that so much! Thanks :)
    $endgroup$
    – gimusi
    Dec 8 '18 at 16:28














  • 1




    $begingroup$
    that's perfect! incredibly aesthetic!
    $endgroup$
    – Jneven
    Dec 8 '18 at 16:27






  • 1




    $begingroup$
    @Jneven I'm glad you appreciate that so much! Thanks :)
    $endgroup$
    – gimusi
    Dec 8 '18 at 16:28








1




1




$begingroup$
that's perfect! incredibly aesthetic!
$endgroup$
– Jneven
Dec 8 '18 at 16:27




$begingroup$
that's perfect! incredibly aesthetic!
$endgroup$
– Jneven
Dec 8 '18 at 16:27




1




1




$begingroup$
@Jneven I'm glad you appreciate that so much! Thanks :)
$endgroup$
– gimusi
Dec 8 '18 at 16:28




$begingroup$
@Jneven I'm glad you appreciate that so much! Thanks :)
$endgroup$
– gimusi
Dec 8 '18 at 16:28


















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