Otdfbt

Are modes in jazz primarily a melody thing?

Appropriate age to involve kids in life changing decisions

Why is the blank symbol not considered part of the input alphabet of a Turing machine?

Assuming a normal distribution: what is the sd for a given mean?

How do I minimise waste on a flight?

Can anyone identify this unknown 1988 PC card from The Palantir Corporation?

shebang or not shebang

Is it safe to keep the GPU on 100% utilization for a very long time?

Is there a reason why Turkey took the Balkan territories of the Ottoman Empire, instead of Greece or another of the Balkan states?

What is the Ancient One's mistake?

How to increase row height of a table and vertically "align middle"?

What is the meaning of "matter" in physics?

Convert Numbers To Emoji Math

Scaling rounded rectangles in Illustrator

Where do 5 or more U.S. counties meet in a single point?

The unknown and unexplained in science fiction

Do the Zhentarim fire members for killing fellow members?

Antivirus for Ubuntu 18.04

What chord could the notes 'F A♭ E♭' form?

How could a humanoid creature completely form within the span of 24 hours?

How to make a kid's bike easier to pedal

Gift for mentor after his thesis defense?

Extracting the parent, leaf, and extension from a valid path

Why were the rules for Proliferate changed?

Finding floor of reciprocal sum

Find floor of sum $sum_k=1^80 k^-1/2$How should I handle the limit with the floor function?Is it possible to find sum of this series?Sum of a decreasing geometric series of integersIf $ A=frac12sqrt1+frac13sqrt2+frac14sqrt3+…+frac1100sqrt99;,$ Then $lfloor A rfloor =$floor sum of first $11$ terms of Harmonic SeriesWhat is the integer part of the sum of the reciprocals of the first $2017$ positive integers?On a series involving harmonic numbersSum of given telescopic seriesSum of series $sum^10_i=1ibigg(frac1^21+i+frac2^22+i+cdots cdots +frac10^210+ibigg)$Floor value of sum whose general term satisfy recursive relation.

4

2

$begingroup$

Evaluation of

$$bigg lfloor frac1sqrt[3]1+frac1sqrt[3]2^2+frac1sqrt[3]3^2+cdots +frac1sqrt[3](1000)^2biggrfloor$$

Where $lfloor xrfloor $ is the floor of $x$

Try: It seems like we can solve it using Telescopic sums and that the sum lies between $2$ Telescopic sums, but could not figure out how to solve it.

Could someone help me to solve it? Thanks.

sequences-and-series

share|cite|improve this question

edited Apr 28 at 18:46

Stelios Sachpazis

1,085416

asked Apr 28 at 18:22

DXTDXT

6,0202733

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note the definition of the Generalized Harmonic numbers: $$H_N^(s)=sum_n=1^Nfrac1n^s$$ Your value is $$lfloor H_1000^(2/3) rfloor$$
  $endgroup$
  – clathratus
  Apr 28 at 18:25
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I don’t think there is something telescopic here. Comparing this sum with an integral, and bounding the error term might help.
  $endgroup$
  – HAMIDINE SOUMARE
  Apr 28 at 18:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In particular, you can check the answer of the user hypergeometric at math.stackexchange.com/questions/2570782/…. Follow his idea in your case. You can also work with Abel (partial) summation to estimate your sum.
  $endgroup$
  – Stelios Sachpazis
  Apr 28 at 18:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer is $27$. See here
  $endgroup$
  – clathratus
  Apr 28 at 18:30
  

  
  
  
  

add a comment | 

4

2

$begingroup$

Evaluation of

$$bigg lfloor frac1sqrt[3]1+frac1sqrt[3]2^2+frac1sqrt[3]3^2+cdots +frac1sqrt[3](1000)^2biggrfloor$$

Where $lfloor xrfloor $ is the floor of $x$

Try: It seems like we can solve it using Telescopic sums and that the sum lies between $2$ Telescopic sums, but could not figure out how to solve it.

Could someone help me to solve it? Thanks.

sequences-and-series

share|cite|improve this question

edited Apr 28 at 18:46

Stelios Sachpazis

1,085416

asked Apr 28 at 18:22

DXTDXT

6,0202733

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note the definition of the Generalized Harmonic numbers: $$H_N^(s)=sum_n=1^Nfrac1n^s$$ Your value is $$lfloor H_1000^(2/3) rfloor$$
  $endgroup$
  – clathratus
  Apr 28 at 18:25
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I don’t think there is something telescopic here. Comparing this sum with an integral, and bounding the error term might help.
  $endgroup$
  – HAMIDINE SOUMARE
  Apr 28 at 18:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In particular, you can check the answer of the user hypergeometric at math.stackexchange.com/questions/2570782/…. Follow his idea in your case. You can also work with Abel (partial) summation to estimate your sum.
  $endgroup$
  – Stelios Sachpazis
  Apr 28 at 18:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer is $27$. See here
  $endgroup$
  – clathratus
  Apr 28 at 18:30
  

  
  
  
  

add a comment | 

$begingroup$

Evaluation of

$$bigg lfloor frac1sqrt[3]1+frac1sqrt[3]2^2+frac1sqrt[3]3^2+cdots +frac1sqrt[3](1000)^2biggrfloor$$

Where $lfloor xrfloor $ is the floor of $x$

Try: It seems like we can solve it using Telescopic sums and that the sum lies between $2$ Telescopic sums, but could not figure out how to solve it.

Could someone help me to solve it? Thanks.

sequences-and-series

share|cite|improve this question

edited Apr 28 at 18:46

Stelios Sachpazis

1,085416

asked Apr 28 at 18:22

DXTDXT

6,0202733

$endgroup$

share|cite|improve this question

edited Apr 28 at 18:46

Stelios Sachpazis

1,085416

asked Apr 28 at 18:22

DXTDXT

6,0202733

share|cite|improve this question

edited Apr 28 at 18:46

Stelios Sachpazis

1,085416

asked Apr 28 at 18:22

DXTDXT

6,0202733

edited Apr 28 at 18:46

Stelios Sachpazis

1,085416

edited Apr 28 at 18:46

Stelios Sachpazis

1,085416

asked Apr 28 at 18:22

DXTDXT

6,0202733

asked Apr 28 at 18:22

DXTDXT

6,0202733

3 Answers
3

active

oldest

votes

1

$begingroup$

Note that $displaystyle sum_k=1^1000frac3sqrt[3](k+1)^2+sqrt[3]k(k+1)+sqrt[3]k^2<sum_k=1^1000frac1sqrt[3]k^2<1+sum_k=2^1000frac3sqrt[3](k-1)^2+sqrt[3]k(k-1)+sqrt[3]k^2$.

$displaystyle sum_k=1^1000frac3sqrt[3](k+1)^2+sqrt[3]k(k+1)+sqrt[3]k^2=sum_k=1^1000frac3(sqrt[3]k+1-sqrt[3]k)(k+1)-k=3(sqrt[3]1001-1)>27$

$displaystyle 1+sum_k=2^1000frac3sqrt[3](k-1)^2+sqrt[3]k(k-1)+sqrt[3]k^2=1+sum_k=2^1000frac3(sqrt[3]k-sqrt[3]k-1)(k-1)-k=1+3(sqrt[3]1000-1)=28$

So, $displaystyle leftlfloor sum_k=1^1000frac1sqrt[3]k^2rightrfloor=27$.

share|cite|improve this answer

edited Apr 29 at 3:47

answered Apr 29 at 3:36

CY AriesCY Aries

17.8k11743

$endgroup$

add a comment | 

4

$begingroup$

One approach is to approximate the sum with an integral, and show that the error is bounded by a number less than 1.

In particular, let $$H_1000^(2/3) = sum_k=1^1000 k^-2/3, quad I = int_x=1^1000 x^-2/3 , dx.$$ Then we know $$I le H_1000^(2/3) < I+1.$$ But $I = 27$, and we are done.

share|cite|improve this answer

answered Apr 28 at 18:40

heropupheropup

66.2k866104

$endgroup$

add a comment | 

4

$begingroup$

You can bound this summation by two integrals;
$$int_1^1001 x^-2/3mathrmdxltsum_k=1^1000k^-2/3lt1+int_1^1000x^-2/3mathrmdx$$
Hence we have
$$27lt3sqrt[3]1001-3ltsum_k=1^1000k^-2/3lt28$$
So as the value of the sum is strictly between $27$ and $28$, the floor of the sum is $27$.

share|cite|improve this answer

answered Apr 28 at 18:41

Peter ForemanPeter Foreman

9,5241321

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3205926%2ffinding-floor-of-reciprocal-sum%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

Note that $displaystyle sum_k=1^1000frac3sqrt[3](k+1)^2+sqrt[3]k(k+1)+sqrt[3]k^2<sum_k=1^1000frac1sqrt[3]k^2<1+sum_k=2^1000frac3sqrt[3](k-1)^2+sqrt[3]k(k-1)+sqrt[3]k^2$.

$displaystyle sum_k=1^1000frac3sqrt[3](k+1)^2+sqrt[3]k(k+1)+sqrt[3]k^2=sum_k=1^1000frac3(sqrt[3]k+1-sqrt[3]k)(k+1)-k=3(sqrt[3]1001-1)>27$

$displaystyle 1+sum_k=2^1000frac3sqrt[3](k-1)^2+sqrt[3]k(k-1)+sqrt[3]k^2=1+sum_k=2^1000frac3(sqrt[3]k-sqrt[3]k-1)(k-1)-k=1+3(sqrt[3]1000-1)=28$

So, $displaystyle leftlfloor sum_k=1^1000frac1sqrt[3]k^2rightrfloor=27$.

share|cite|improve this answer

edited Apr 29 at 3:47

answered Apr 29 at 3:36

CY AriesCY Aries

17.8k11743

$endgroup$

add a comment | 

1

$begingroup$

Note that $displaystyle sum_k=1^1000frac3sqrt[3](k+1)^2+sqrt[3]k(k+1)+sqrt[3]k^2<sum_k=1^1000frac1sqrt[3]k^2<1+sum_k=2^1000frac3sqrt[3](k-1)^2+sqrt[3]k(k-1)+sqrt[3]k^2$.

$displaystyle sum_k=1^1000frac3sqrt[3](k+1)^2+sqrt[3]k(k+1)+sqrt[3]k^2=sum_k=1^1000frac3(sqrt[3]k+1-sqrt[3]k)(k+1)-k=3(sqrt[3]1001-1)>27$

$displaystyle 1+sum_k=2^1000frac3sqrt[3](k-1)^2+sqrt[3]k(k-1)+sqrt[3]k^2=1+sum_k=2^1000frac3(sqrt[3]k-sqrt[3]k-1)(k-1)-k=1+3(sqrt[3]1000-1)=28$

So, $displaystyle leftlfloor sum_k=1^1000frac1sqrt[3]k^2rightrfloor=27$.

share|cite|improve this answer

edited Apr 29 at 3:47

answered Apr 29 at 3:36

CY AriesCY Aries

17.8k11743

$endgroup$

add a comment | 

$begingroup$

Note that $displaystyle sum_k=1^1000frac3sqrt[3](k+1)^2+sqrt[3]k(k+1)+sqrt[3]k^2<sum_k=1^1000frac1sqrt[3]k^2<1+sum_k=2^1000frac3sqrt[3](k-1)^2+sqrt[3]k(k-1)+sqrt[3]k^2$.

$displaystyle sum_k=1^1000frac3sqrt[3](k+1)^2+sqrt[3]k(k+1)+sqrt[3]k^2=sum_k=1^1000frac3(sqrt[3]k+1-sqrt[3]k)(k+1)-k=3(sqrt[3]1001-1)>27$

$displaystyle 1+sum_k=2^1000frac3sqrt[3](k-1)^2+sqrt[3]k(k-1)+sqrt[3]k^2=1+sum_k=2^1000frac3(sqrt[3]k-sqrt[3]k-1)(k-1)-k=1+3(sqrt[3]1000-1)=28$

So, $displaystyle leftlfloor sum_k=1^1000frac1sqrt[3]k^2rightrfloor=27$.

share|cite|improve this answer

edited Apr 29 at 3:47

answered Apr 29 at 3:36

CY AriesCY Aries

17.8k11743

$endgroup$

share|cite|improve this answer

edited Apr 29 at 3:47

answered Apr 29 at 3:36

CY AriesCY Aries

17.8k11743

answered Apr 29 at 3:36

CY AriesCY Aries

17.8k11743

answered Apr 29 at 3:36

CY AriesCY Aries

17.8k11743

4

$begingroup$

One approach is to approximate the sum with an integral, and show that the error is bounded by a number less than 1.

In particular, let $$H_1000^(2/3) = sum_k=1^1000 k^-2/3, quad I = int_x=1^1000 x^-2/3 , dx.$$ Then we know $$I le H_1000^(2/3) < I+1.$$ But $I = 27$, and we are done.

share|cite|improve this answer

answered Apr 28 at 18:40

heropupheropup

66.2k866104

$endgroup$

add a comment | 

4

$begingroup$

One approach is to approximate the sum with an integral, and show that the error is bounded by a number less than 1.

In particular, let $$H_1000^(2/3) = sum_k=1^1000 k^-2/3, quad I = int_x=1^1000 x^-2/3 , dx.$$ Then we know $$I le H_1000^(2/3) < I+1.$$ But $I = 27$, and we are done.

share|cite|improve this answer

answered Apr 28 at 18:40

heropupheropup

66.2k866104

$endgroup$

add a comment | 

$begingroup$

One approach is to approximate the sum with an integral, and show that the error is bounded by a number less than 1.

In particular, let $$H_1000^(2/3) = sum_k=1^1000 k^-2/3, quad I = int_x=1^1000 x^-2/3 , dx.$$ Then we know $$I le H_1000^(2/3) < I+1.$$ But $I = 27$, and we are done.

share|cite|improve this answer

answered Apr 28 at 18:40

heropupheropup

66.2k866104

$endgroup$

share|cite|improve this answer

answered Apr 28 at 18:40

heropupheropup

66.2k866104

answered Apr 28 at 18:40

heropupheropup

66.2k866104

answered Apr 28 at 18:40

heropupheropup

66.2k866104

4

$begingroup$

You can bound this summation by two integrals;
$$int_1^1001 x^-2/3mathrmdxltsum_k=1^1000k^-2/3lt1+int_1^1000x^-2/3mathrmdx$$
Hence we have
$$27lt3sqrt[3]1001-3ltsum_k=1^1000k^-2/3lt28$$
So as the value of the sum is strictly between $27$ and $28$, the floor of the sum is $27$.

share|cite|improve this answer

answered Apr 28 at 18:41

Peter ForemanPeter Foreman

9,5241321

$endgroup$

add a comment | 

4

$begingroup$

You can bound this summation by two integrals;
$$int_1^1001 x^-2/3mathrmdxltsum_k=1^1000k^-2/3lt1+int_1^1000x^-2/3mathrmdx$$
Hence we have
$$27lt3sqrt[3]1001-3ltsum_k=1^1000k^-2/3lt28$$
So as the value of the sum is strictly between $27$ and $28$, the floor of the sum is $27$.

share|cite|improve this answer

answered Apr 28 at 18:41

Peter ForemanPeter Foreman

9,5241321

$endgroup$

add a comment | 

$begingroup$

You can bound this summation by two integrals;
$$int_1^1001 x^-2/3mathrmdxltsum_k=1^1000k^-2/3lt1+int_1^1000x^-2/3mathrmdx$$
Hence we have
$$27lt3sqrt[3]1001-3ltsum_k=1^1000k^-2/3lt28$$
So as the value of the sum is strictly between $27$ and $28$, the floor of the sum is $27$.

share|cite|improve this answer

answered Apr 28 at 18:41

Peter ForemanPeter Foreman

9,5241321

$endgroup$

share|cite|improve this answer

answered Apr 28 at 18:41

Peter ForemanPeter Foreman

9,5241321

answered Apr 28 at 18:41

Peter ForemanPeter Foreman

9,5241321

answered Apr 28 at 18:41

Peter ForemanPeter Foreman

9,5241321