$lim_{ntoinfty}frac{sum_{i=1}^ni^{4lambda}}{Big(sum_{i=1}^ni^{2lambda}Big)^2}=0$
$begingroup$
Is it true that $forall lambda>0$ $$lim_{ntoinfty}frac{sum_{i=1}^ni^{4lambda}}{Big(sum_{i=1}^ni^{2lambda}Big)^2}=0$$
I cannot find a way to prove it, nor can I find a counterexample.
Any help is greatly appreciated!
limits sums-of-squares gauss-sums
asked Dec 15 '18 at 15:07
newbienewbie
313212
$endgroup$
add a comment |
$begingroup$
Is it true that $forall lambda>0$ $$lim_{ntoinfty}frac{sum_{i=1}^ni^{4lambda}}{Big(sum_{i=1}^ni^{2lambda}Big)^2}=0$$
I cannot find a way to prove it, nor can I find a counterexample.
Any help is greatly appreciated!
limits sums-of-squares gauss-sums
asked Dec 15 '18 at 15:07
newbienewbie
313212
$endgroup$
4
$begingroup$
For $k>0$, $$sum_{i=1}^n i^ksimfrac{n^{k+1}}{k+1}.$$
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 15:10
1
$begingroup$
To calculate, use Cesàro--Stolz theorem.
$endgroup$
– xbh
Dec 15 '18 at 15:30
add a comment |
$begingroup$
Is it true that $forall lambda>0$ $$lim_{ntoinfty}frac{sum_{i=1}^ni^{4lambda}}{Big(sum_{i=1}^ni^{2lambda}Big)^2}=0$$
I cannot find a way to prove it, nor can I find a counterexample.
Any help is greatly appreciated!
limits sums-of-squares gauss-sums
asked Dec 15 '18 at 15:07
newbienewbie
313212
$endgroup$
Is it true that $forall lambda>0$ $$lim_{ntoinfty}frac{sum_{i=1}^ni^{4lambda}}{Big(sum_{i=1}^ni^{2lambda}Big)^2}=0$$
I cannot find a way to prove it, nor can I find a counterexample.
Any help is greatly appreciated!
limits sums-of-squares gauss-sums
limits sums-of-squares gauss-sums
asked Dec 15 '18 at 15:07
newbienewbie
313212
asked Dec 15 '18 at 15:07
newbienewbie
313212
asked Dec 15 '18 at 15:07
newbienewbie
313212
asked Dec 15 '18 at 15:07
newbienewbie
313212
asked Dec 15 '18 at 15:07
newbienewbie
313212
313212
4
$begingroup$
For $k>0$, $$sum_{i=1}^n i^ksimfrac{n^{k+1}}{k+1}.$$
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 15:10
1
$begingroup$
To calculate, use Cesàro--Stolz theorem.
$endgroup$
– xbh
Dec 15 '18 at 15:30
add a comment |
4
$begingroup$
For $k>0$, $$sum_{i=1}^n i^ksimfrac{n^{k+1}}{k+1}.$$
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 15:10
1
$begingroup$
To calculate, use Cesàro--Stolz theorem.
$endgroup$
– xbh
Dec 15 '18 at 15:30
4
4
$begingroup$
For $k>0$, $$sum_{i=1}^n i^ksimfrac{n^{k+1}}{k+1}.$$
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 15:10
$begingroup$
For $k>0$, $$sum_{i=1}^n i^ksimfrac{n^{k+1}}{k+1}.$$
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 15:10
1
1
$begingroup$
To calculate, use Cesàro--Stolz theorem.
$endgroup$
– xbh
Dec 15 '18 at 15:30
$begingroup$
To calculate, use Cesàro--Stolz theorem.
$endgroup$
– xbh
Dec 15 '18 at 15:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The numerator is less than $ncdot n^{4lambda}.$ In the denominator there are at least $n/2$ terms that are at least $(n/2)^{2lambda}.$ Thus our expression is less than
$$frac{ncdot n^{4lambda}}{[(n/2)(n/2)^{2lambda}]^2}.$$
This is on the order of $dfrac{1}{n}$ as $nto infty,$ hence the limit is $0.$
answered Dec 15 '18 at 17:56
zhw.zhw.
72.9k43175
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add a comment |
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$begingroup$
The numerator is less than $ncdot n^{4lambda}.$ In the denominator there are at least $n/2$ terms that are at least $(n/2)^{2lambda}.$ Thus our expression is less than
$$frac{ncdot n^{4lambda}}{[(n/2)(n/2)^{2lambda}]^2}.$$
This is on the order of $dfrac{1}{n}$ as $nto infty,$ hence the limit is $0.$
answered Dec 15 '18 at 17:56
zhw.zhw.
72.9k43175
$endgroup$
add a comment |
$begingroup$
The numerator is less than $ncdot n^{4lambda}.$ In the denominator there are at least $n/2$ terms that are at least $(n/2)^{2lambda}.$ Thus our expression is less than
$$frac{ncdot n^{4lambda}}{[(n/2)(n/2)^{2lambda}]^2}.$$
This is on the order of $dfrac{1}{n}$ as $nto infty,$ hence the limit is $0.$
answered Dec 15 '18 at 17:56
zhw.zhw.
72.9k43175
$endgroup$
add a comment |
$begingroup$
The numerator is less than $ncdot n^{4lambda}.$ In the denominator there are at least $n/2$ terms that are at least $(n/2)^{2lambda}.$ Thus our expression is less than
$$frac{ncdot n^{4lambda}}{[(n/2)(n/2)^{2lambda}]^2}.$$
This is on the order of $dfrac{1}{n}$ as $nto infty,$ hence the limit is $0.$
answered Dec 15 '18 at 17:56
zhw.zhw.
72.9k43175
$endgroup$
The numerator is less than $ncdot n^{4lambda}.$ In the denominator there are at least $n/2$ terms that are at least $(n/2)^{2lambda}.$ Thus our expression is less than
$$frac{ncdot n^{4lambda}}{[(n/2)(n/2)^{2lambda}]^2}.$$
This is on the order of $dfrac{1}{n}$ as $nto infty,$ hence the limit is $0.$
answered Dec 15 '18 at 17:56
zhw.zhw.
72.9k43175
answered Dec 15 '18 at 17:56
zhw.zhw.
72.9k43175
answered Dec 15 '18 at 17:56
zhw.zhw.
72.9k43175
answered Dec 15 '18 at 17:56
zhw.zhw.
72.9k43175
72.9k43175
add a comment |
add a comment |
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4
$begingroup$
For $k>0$, $$sum_{i=1}^n i^ksimfrac{n^{k+1}}{k+1}.$$
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 15:10
1
$begingroup$
To calculate, use Cesàro--Stolz theorem.
$endgroup$
– xbh
Dec 15 '18 at 15:30