Dtjytyk

How does one intimidate enemies without having the capacity for violence?

Minkowski space

Dragon forelimb placement

Has the BBC provided arguments for saying Brexit being cancelled is unlikely?

Why dont electromagnetic waves interact with each other?

Problem of parity - Can we draw a closed path made up of 20 line segments...

Can a Warlock become Neutral Good?

Why does Kotter return in Welcome Back Kotter?

How is it possible to have an ability score that is less than 3?

What does it mean to describe someone as a butt steak?

What does "Puller Prush Person" mean?

A newer friend of my brother's gave him a load of baseball cards that are supposedly extremely valuable. Is this a scam?

Theorems that impeded progress

How can bays and straits be determined in a procedurally generated map?

Email Account under attack (really) - anything I can do?

Can an x86 CPU running in real mode be considered to be basically an 8086 CPU?

TGV timetables / schedules?

Why "Having chlorophyll without photosynthesis is actually very dangerous" and "like living with a bomb"?

Approximately how much travel time was saved by the opening of the Suez Canal in 1869?

Is it possible to do 50 km distance without any previous training?

Do I have a twin with permutated remainders?

The use of multiple foreign keys on same column in SQL Server

Why Is Death Allowed In the Matrix?

Font hinting is lost in Chrome-like browsers (for some languages )

Why $int_2^infty{1over{sqrt x(sqrt x-1)^2}}dx style{text-decoration:line-through}{approx} sum_{n=2}^{infty}{1over{sqrt n(sqrt n-1)^2}}$?

Why does the Euler-Maclaurin summation formula not exist in $2$ or more dimensions?Estimating the sum $sum_{k=2}^{infty} frac{1}{k ln^2(k)}$How close can $sum_{k=1}^n sqrt{k}$ be to an integer?Proof that $sumlimits_{i=1}^n cos sqrt{i}$ is unbounded.Turning infinite sum into integralWhat is $lim_{xto infty} 2sqrt{x}- sum_{n=1}^x {1over sqrt{n}}$?Prove that: $zeta(3)=lim_{Nto infty}{1over N}sum_{k=1}^{N}{1over k^{phi^2}ln{left(1+{k^{phi^{-2}}over N}right)}}$What is $sum_{n=1}^{infty} frac{1}{sqrt{n^{3} + 1}}$?Is there a closed form for $sum_{n=0}^{infty}{2^{n+1}over {2n choose n}}cdotleft({2n-1over 2n+1}right)^2?$Euler-Maclaurin Formula Definition Confusion

0

$begingroup$

I am aware that there are certain cases where the infinite sum does not equal the infinite integral. However, I am not yet advanced enough to be able to understand the Euler–Maclaurin formula, especially in terms of coming up with the $k$th Bernoulli Number. That said, I am wondering if there is a way somebody could explain to me the following inequality that came up in my course. When I use a calculator to estimate the infinite sum $sum_{n=2}^{infty}{1over{sqrt n(sqrt n-1)^2}}$, I get $$sum_{n=2}^{9999999}{1over{sqrt n(sqrt n-1)^2}} approx 7.47356$$

When I try to estimate using an integral, I get a huge discrepancy
$$int_2^infty{1over{sqrt x(sqrt x-1)^2}}dx=2^{3over2}+2 approx 4.8284$$

Does this require the Euler–Maclaurin formula to explain, or is there an easier way to understand what is going on over here?

calculus integration sequences-and-series

share|cite|improve this question

asked Mar 19 at 23:40

agbltagblt

350114

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You may want to take a look at this: en.wikipedia.org/wiki/Abel%E2%80%93Plana_formula
  $endgroup$
  – Zachary
  Mar 19 at 23:56
  

  
  
  
  

add a comment | 

0

$begingroup$

I am aware that there are certain cases where the infinite sum does not equal the infinite integral. However, I am not yet advanced enough to be able to understand the Euler–Maclaurin formula, especially in terms of coming up with the $k$th Bernoulli Number. That said, I am wondering if there is a way somebody could explain to me the following inequality that came up in my course. When I use a calculator to estimate the infinite sum $sum_{n=2}^{infty}{1over{sqrt n(sqrt n-1)^2}}$, I get $$sum_{n=2}^{9999999}{1over{sqrt n(sqrt n-1)^2}} approx 7.47356$$

When I try to estimate using an integral, I get a huge discrepancy
$$int_2^infty{1over{sqrt x(sqrt x-1)^2}}dx=2^{3over2}+2 approx 4.8284$$

Does this require the Euler–Maclaurin formula to explain, or is there an easier way to understand what is going on over here?

calculus integration sequences-and-series

share|cite|improve this question

asked Mar 19 at 23:40

agbltagblt

350114

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You may want to take a look at this: en.wikipedia.org/wiki/Abel%E2%80%93Plana_formula
  $endgroup$
  – Zachary
  Mar 19 at 23:56
  

  
  
  
  

add a comment | 

$begingroup$

I am aware that there are certain cases where the infinite sum does not equal the infinite integral. However, I am not yet advanced enough to be able to understand the Euler–Maclaurin formula, especially in terms of coming up with the $k$th Bernoulli Number. That said, I am wondering if there is a way somebody could explain to me the following inequality that came up in my course. When I use a calculator to estimate the infinite sum $sum_{n=2}^{infty}{1over{sqrt n(sqrt n-1)^2}}$, I get $$sum_{n=2}^{9999999}{1over{sqrt n(sqrt n-1)^2}} approx 7.47356$$

When I try to estimate using an integral, I get a huge discrepancy
$$int_2^infty{1over{sqrt x(sqrt x-1)^2}}dx=2^{3over2}+2 approx 4.8284$$

Does this require the Euler–Maclaurin formula to explain, or is there an easier way to understand what is going on over here?

calculus integration sequences-and-series

share|cite|improve this question

asked Mar 19 at 23:40

agbltagblt

350114

$endgroup$

share|cite|improve this question

asked Mar 19 at 23:40

agbltagblt

350114

share|cite|improve this question

asked Mar 19 at 23:40

agbltagblt

350114

asked Mar 19 at 23:40

agbltagblt

350114

asked Mar 19 at 23:40

agbltagblt

350114

1 Answer
1

active

oldest

votes

2

$begingroup$

The discrepancy comes mostly from the first two terms: the integral from $2$ to $3$ is about $2.1$ while the sum term for $n=2$ is about $4.2$ Same from $3$ to $4$ integral is about $.73$, the $n=3$ sum term is about $1.1$ so if you start from $4$ on and noting that because the function is decreasing, the integral is always smaller than the sum starting from same bound, you get the integral to be about $2$ and the sum to be about $2.17$ so the discrepancy gets much smaller.

share|cite|improve this answer

answered Mar 20 at 0:53

ConradConrad

1,32745

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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%2f3154792%2fwhy-int-2-infty1-over-sqrt-x-sqrt-x-12dx-styletext-decorationline-t%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

The discrepancy comes mostly from the first two terms: the integral from $2$ to $3$ is about $2.1$ while the sum term for $n=2$ is about $4.2$ Same from $3$ to $4$ integral is about $.73$, the $n=3$ sum term is about $1.1$ so if you start from $4$ on and noting that because the function is decreasing, the integral is always smaller than the sum starting from same bound, you get the integral to be about $2$ and the sum to be about $2.17$ so the discrepancy gets much smaller.

share|cite|improve this answer

answered Mar 20 at 0:53

ConradConrad

1,32745

$endgroup$

add a comment | 

2

$begingroup$

The discrepancy comes mostly from the first two terms: the integral from $2$ to $3$ is about $2.1$ while the sum term for $n=2$ is about $4.2$ Same from $3$ to $4$ integral is about $.73$, the $n=3$ sum term is about $1.1$ so if you start from $4$ on and noting that because the function is decreasing, the integral is always smaller than the sum starting from same bound, you get the integral to be about $2$ and the sum to be about $2.17$ so the discrepancy gets much smaller.

share|cite|improve this answer

answered Mar 20 at 0:53

ConradConrad

1,32745

$endgroup$

add a comment | 

$begingroup$

The discrepancy comes mostly from the first two terms: the integral from $2$ to $3$ is about $2.1$ while the sum term for $n=2$ is about $4.2$ Same from $3$ to $4$ integral is about $.73$, the $n=3$ sum term is about $1.1$ so if you start from $4$ on and noting that because the function is decreasing, the integral is always smaller than the sum starting from same bound, you get the integral to be about $2$ and the sum to be about $2.17$ so the discrepancy gets much smaller.

share|cite|improve this answer

answered Mar 20 at 0:53

ConradConrad

1,32745

$endgroup$

share|cite|improve this answer

answered Mar 20 at 0:53

ConradConrad

1,32745

answered Mar 20 at 0:53

ConradConrad

1,32745

answered Mar 20 at 0:53

ConradConrad

1,32745