Solution verification: evaluate $limlimits_{n to infty}frac{1^{lambda n}+2^{lambda n}+cdots+n^{lambda...












6












$begingroup$


Problem



Evaluate
$$lim_{n to infty}frac{1^{lambda n}+2^{lambda n}+cdots+n^{lambda n}}{n^{lambda n}}$$
where $lambda>0.$



Solution



Denote
$$S_n:=sum_{k=1}^{n}left(frac{k}{n}right)^{lambda n}=sum_{k=0}^{n-1}left(1-frac{k}{n}right)^{lambda n}.tag1$$



On one hand, we choose some $pin mathbb{N+}$, fix it, and let $n>p+1$. Then
$$S_n geq sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}.tag2$$
Let $n to infty$ within $(2)$. We obtain
$$liminf_{n to infty}S_ngeq liminf_{n to infty} sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}liminf_{n to infty}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}e^{-lambda k}.tag3$$
Notice that $(3)$ holds for all $p$. We may let $p to infty$ and obtain
$$liminf_{n to infty}S_ngeq sum_{k=0}^{infty}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag4$$



On the other hand, from $1+xleq e^x$ we may derive
$left(1-dfrac{k}{n}right)^{lambda n}leq e^{-lambda k}.$
Thus
$$S_n leq sum_{k=0}^{n-1}e^{-lambda k}.tag5$$
Let $n to infty$ within $(5)$. We obtain
$$limsup_{n to infty}S_n leq limsup_{n to infty}sum_{k=0}^{n-1}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag6$$
See $(4)$ and $(6)$. We may conclude
$$frac{e^{lambda}}{e^{lambda}-1}leq liminf_{n to infty}S_nleq limsup_{n to infty}S_nleq frac{e^{lambda}}{e^{lambda}-1},tag7$$
which implies
$$lim_{n to infty}S_n=frac{e^{lambda}}{e^{lambda}-1}.tag8$$



Please correct me if I'm wrong. Thanks.






limits proof-verification limsup-and-liminf







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  • $begingroup$
    Yes, it's correct.
    $endgroup$
    – zoidberg
    Dec 7 '18 at 4:35
















6












$begingroup$


Problem



Evaluate
$$lim_{n to infty}frac{1^{lambda n}+2^{lambda n}+cdots+n^{lambda n}}{n^{lambda n}}$$
where $lambda>0.$



Solution



Denote
$$S_n:=sum_{k=1}^{n}left(frac{k}{n}right)^{lambda n}=sum_{k=0}^{n-1}left(1-frac{k}{n}right)^{lambda n}.tag1$$



On one hand, we choose some $pin mathbb{N+}$, fix it, and let $n>p+1$. Then
$$S_n geq sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}.tag2$$
Let $n to infty$ within $(2)$. We obtain
$$liminf_{n to infty}S_ngeq liminf_{n to infty} sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}liminf_{n to infty}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}e^{-lambda k}.tag3$$
Notice that $(3)$ holds for all $p$. We may let $p to infty$ and obtain
$$liminf_{n to infty}S_ngeq sum_{k=0}^{infty}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag4$$



On the other hand, from $1+xleq e^x$ we may derive
$left(1-dfrac{k}{n}right)^{lambda n}leq e^{-lambda k}.$
Thus
$$S_n leq sum_{k=0}^{n-1}e^{-lambda k}.tag5$$
Let $n to infty$ within $(5)$. We obtain
$$limsup_{n to infty}S_n leq limsup_{n to infty}sum_{k=0}^{n-1}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag6$$
See $(4)$ and $(6)$. We may conclude
$$frac{e^{lambda}}{e^{lambda}-1}leq liminf_{n to infty}S_nleq limsup_{n to infty}S_nleq frac{e^{lambda}}{e^{lambda}-1},tag7$$
which implies
$$lim_{n to infty}S_n=frac{e^{lambda}}{e^{lambda}-1}.tag8$$



Please correct me if I'm wrong. Thanks.






limits proof-verification limsup-and-liminf







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, it's correct.
    $endgroup$
    – zoidberg
    Dec 7 '18 at 4:35














6












6








6


1



$begingroup$


Problem



Evaluate
$$lim_{n to infty}frac{1^{lambda n}+2^{lambda n}+cdots+n^{lambda n}}{n^{lambda n}}$$
where $lambda>0.$



Solution



Denote
$$S_n:=sum_{k=1}^{n}left(frac{k}{n}right)^{lambda n}=sum_{k=0}^{n-1}left(1-frac{k}{n}right)^{lambda n}.tag1$$



On one hand, we choose some $pin mathbb{N+}$, fix it, and let $n>p+1$. Then
$$S_n geq sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}.tag2$$
Let $n to infty$ within $(2)$. We obtain
$$liminf_{n to infty}S_ngeq liminf_{n to infty} sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}liminf_{n to infty}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}e^{-lambda k}.tag3$$
Notice that $(3)$ holds for all $p$. We may let $p to infty$ and obtain
$$liminf_{n to infty}S_ngeq sum_{k=0}^{infty}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag4$$



On the other hand, from $1+xleq e^x$ we may derive
$left(1-dfrac{k}{n}right)^{lambda n}leq e^{-lambda k}.$
Thus
$$S_n leq sum_{k=0}^{n-1}e^{-lambda k}.tag5$$
Let $n to infty$ within $(5)$. We obtain
$$limsup_{n to infty}S_n leq limsup_{n to infty}sum_{k=0}^{n-1}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag6$$
See $(4)$ and $(6)$. We may conclude
$$frac{e^{lambda}}{e^{lambda}-1}leq liminf_{n to infty}S_nleq limsup_{n to infty}S_nleq frac{e^{lambda}}{e^{lambda}-1},tag7$$
which implies
$$lim_{n to infty}S_n=frac{e^{lambda}}{e^{lambda}-1}.tag8$$



Please correct me if I'm wrong. Thanks.






limits proof-verification limsup-and-liminf







share|cite|improve this question











$endgroup$




Problem



Evaluate
$$lim_{n to infty}frac{1^{lambda n}+2^{lambda n}+cdots+n^{lambda n}}{n^{lambda n}}$$
where $lambda>0.$



Solution



Denote
$$S_n:=sum_{k=1}^{n}left(frac{k}{n}right)^{lambda n}=sum_{k=0}^{n-1}left(1-frac{k}{n}right)^{lambda n}.tag1$$



On one hand, we choose some $pin mathbb{N+}$, fix it, and let $n>p+1$. Then
$$S_n geq sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}.tag2$$
Let $n to infty$ within $(2)$. We obtain
$$liminf_{n to infty}S_ngeq liminf_{n to infty} sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}liminf_{n to infty}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}e^{-lambda k}.tag3$$
Notice that $(3)$ holds for all $p$. We may let $p to infty$ and obtain
$$liminf_{n to infty}S_ngeq sum_{k=0}^{infty}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag4$$



On the other hand, from $1+xleq e^x$ we may derive
$left(1-dfrac{k}{n}right)^{lambda n}leq e^{-lambda k}.$
Thus
$$S_n leq sum_{k=0}^{n-1}e^{-lambda k}.tag5$$
Let $n to infty$ within $(5)$. We obtain
$$limsup_{n to infty}S_n leq limsup_{n to infty}sum_{k=0}^{n-1}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag6$$
See $(4)$ and $(6)$. We may conclude
$$frac{e^{lambda}}{e^{lambda}-1}leq liminf_{n to infty}S_nleq limsup_{n to infty}S_nleq frac{e^{lambda}}{e^{lambda}-1},tag7$$
which implies
$$lim_{n to infty}S_n=frac{e^{lambda}}{e^{lambda}-1}.tag8$$



Please correct me if I'm wrong. Thanks.






limits proof-verification limsup-and-liminf




limits proof-verification limsup-and-liminf






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edited Dec 7 '18 at 13:22







mengdie1982

















asked Dec 7 '18 at 4:22









mengdie1982mengdie1982

4,912618




4,912618












  • $begingroup$
    Yes, it's correct.
    $endgroup$
    – zoidberg
    Dec 7 '18 at 4:35


















  • $begingroup$
    Yes, it's correct.
    $endgroup$
    – zoidberg
    Dec 7 '18 at 4:35
















$begingroup$
Yes, it's correct.
$endgroup$
– zoidberg
Dec 7 '18 at 4:35




$begingroup$
Yes, it's correct.
$endgroup$
– zoidberg
Dec 7 '18 at 4:35










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