Vjylku

What is an explicit bijection in combinatorics?

What does "south of due west" mean?

Algebraic proof that two statements of the fundamental theorem of algebra are equivalent

Is the UK legally prevented from having another referendum on Brexit?

How do I fight with Heavy Armor as a Wizard with Tenser's Transformation?

How can guns be countered by melee combat without raw-ability or exceptional explanations?

Isn't a semicolon (';') needed after a function declaration in C++?

Can you say "leftside right"?

How can I give a Ranger advantage on a check due to Favored Enemy without spoiling the story for the player?

Taking an academic pseudonym?

Why don't programs completely uninstall (remove all their files) when I remove them?

When distributing a Linux kernel driver as source code, what's the difference between Proprietary and GPL license?

Cartoon in which kids compete by fighting as monsters or creatures with special powers in a virtual world

Missing a connection and don't have money to book next flight

Partial derivative with respect to three variables

Piping Multiple Numbers into Sed

What sort of grammatical construct is ‘Quod per sortem sternit fortem’?

Boss asked me to sign a resignation paper without a date on it along with my new contract

bash aliases do not expand even with shopt expand_aliases

Why is it that Bernie Sanders always called a "socialist"?

How many copper coins fit inside a cubic foot?

Short story about a man betting a group he could tell a story, and one of them would disappear and the others would not notice

What is formjacking?

Are one-line email responses considered disrespectful?

Show that sequence is a Cauchy sequence

Let $a_n=frac{a_{n-1}+a_{n-2}}{2}$ for each positive integer $ngeq 2$. Show that ${a_n}_{n=1}^{infty}$ is CauchyShow that $langle f_n rangle$ is a Cauchy sequence, where $f_n=1-frac12+frac13-frac14+dots+frac{(-1)^{n-1}}{n}$Show that $(3x_{n}+4y_{n})$ is also Cauchy sequence.Uniformly Cauchy sequence of functionsIs this sequence Cauchy in the space of polynomials under the infinity norm?The sequence $b_n=pa_n+qa_{n+1}$, where $|p|<q$, is convergent. Prove that $a_n$ converges. If $|p|geq q > 0$ show that $a_n$ need not converge.Prove that it is a cauchy sequenceShowing sequence is Cauchy by DefinitionSequence of function on $mathbb{R}$ Cauchy iff convergentProve that $(a_n) preccurlyeq_1 (b_n) iff (a_n) preccurlyeq_2 (b_n)$ or $(a_n) approx (b_n)$ for Cauchy sequences

2

$begingroup$

Prove that given sequence $$langle f_nrangle =1-frac{1}{2}+frac{1}{3}-frac{1}{4}+.....+frac{(-1)^{n-1}}{n}$$

is a Cauchy sequence

My attempt :
$|f_{n}-f_{m}|=Biggl|dfrac{(-1)^{m}}{m+1}+dfrac{(-1)^{m+1}}{m+2}cdotsdots+dfrac{(-1)^{n-1}}{n}Biggr|$

using $ m+1>m implies dfrac{1}{m+1}<dfrac{1}{m} $

$|f_{n}-f_{m}|le dfrac{1}{m}+dfrac{1}{m}+dfrac{1}{m}cdotscdotsdfrac{1}{m}$

$|f_{n}-f_{m}|ledfrac{n-m}{m}$

I don't know if I am proceeding correctly or if I am, how to proceed further, any hint would be really helpful .

sequences-and-series cauchy-sequences

share|cite|improve this question

edited 4 hours ago

Bernard

121k740116

asked 4 hours ago

kira0705kira0705

1167

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, the limit of the sequence because of Leibniz' criterion.
  $endgroup$
  – egreg
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: a convergent sequence is Cauchy.
  $endgroup$
  – Bernard
  4 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Prove that given sequence $$langle f_nrangle =1-frac{1}{2}+frac{1}{3}-frac{1}{4}+.....+frac{(-1)^{n-1}}{n}$$

is a Cauchy sequence

My attempt :
$|f_{n}-f_{m}|=Biggl|dfrac{(-1)^{m}}{m+1}+dfrac{(-1)^{m+1}}{m+2}cdotsdots+dfrac{(-1)^{n-1}}{n}Biggr|$

using $ m+1>m implies dfrac{1}{m+1}<dfrac{1}{m} $

$|f_{n}-f_{m}|le dfrac{1}{m}+dfrac{1}{m}+dfrac{1}{m}cdotscdotsdfrac{1}{m}$

$|f_{n}-f_{m}|ledfrac{n-m}{m}$

I don't know if I am proceeding correctly or if I am, how to proceed further, any hint would be really helpful .

sequences-and-series cauchy-sequences

share|cite|improve this question

edited 4 hours ago

Bernard

121k740116

asked 4 hours ago

kira0705kira0705

1167

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, the limit of the sequence because of Leibniz' criterion.
  $endgroup$
  – egreg
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: a convergent sequence is Cauchy.
  $endgroup$
  – Bernard
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Prove that given sequence $$langle f_nrangle =1-frac{1}{2}+frac{1}{3}-frac{1}{4}+.....+frac{(-1)^{n-1}}{n}$$

is a Cauchy sequence

My attempt :
$|f_{n}-f_{m}|=Biggl|dfrac{(-1)^{m}}{m+1}+dfrac{(-1)^{m+1}}{m+2}cdotsdots+dfrac{(-1)^{n-1}}{n}Biggr|$

using $ m+1>m implies dfrac{1}{m+1}<dfrac{1}{m} $

$|f_{n}-f_{m}|le dfrac{1}{m}+dfrac{1}{m}+dfrac{1}{m}cdotscdotsdfrac{1}{m}$

$|f_{n}-f_{m}|ledfrac{n-m}{m}$

I don't know if I am proceeding correctly or if I am, how to proceed further, any hint would be really helpful .

sequences-and-series cauchy-sequences

share|cite|improve this question

edited 4 hours ago

Bernard

121k740116

asked 4 hours ago

kira0705kira0705

1167

$endgroup$

share|cite|improve this question

edited 4 hours ago

Bernard

121k740116

asked 4 hours ago

kira0705kira0705

1167

share|cite|improve this question

edited 4 hours ago

Bernard

121k740116

asked 4 hours ago

kira0705kira0705

1167

edited 4 hours ago

Bernard

121k740116

edited 4 hours ago

Bernard

121k740116

asked 4 hours ago

kira0705kira0705

1167

asked 4 hours ago

kira0705kira0705

1167

3 Answers
3

active

oldest

votes

2

$begingroup$

If you ignore the signs of the terms,
the result diverges.
So you can't do that.

$f_n
=sum_{k=1}^n dfrac{(-1)^k}{k}
$
so,
if $n > m$,
$f_n-f_m
=sum_{k=m+1}^n dfrac{(-1)^k}{k}
=sum_{k=1}^{n-m} dfrac{(-1)^{k+m}}{k+m}
=(-1)^msum_{k=1}^{n-m} dfrac{(-1)^{k}}{k+m}
$.

If
$n-m$ is even,
so $n-m = 2j$,
then

$begin{array}\
f_n-f_m
&=(-1)^msum_{k=1}^{2j} dfrac{(-1)^{k}}{k+m}\
&=(-1)^msum_{k=1}^{j} left(dfrac{(-1)^{2k-1}}{2k-1+m}+dfrac{(-1)^{2k}}{2k+m}right)\
&=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{-1}{2k-1+m}+dfrac{1}{2k+m}right)\
&=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{(2k-1+m)-(2k+m)}{(2k-1+m)(2k+m)}right)\
&=(-1)^{m+1}sum_{k=1}^{j} left(dfrac{-1}{(2k-1+m)(2k+m)}right)\
&=(-1)^{m}sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
text{so}\
|f_n-f_m|
&=sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
&=sum_{k=1}^{j}dfrac14 left(dfrac{1}{(k-frac12+frac{m}{2})(k+frac{m}{2})}right)\
&lt dfrac14sum_{k=1}^{j} left(dfrac{1}{(k-1+frac{m}{2})(k+frac{m}{2})}right)
quadtext{this is the sneaky part}\
&lt dfrac14sum_{k=1}^{j} left(dfrac{1}{k-1+frac{m}{2}}-dfrac{1}{k+frac{m}{2}}right)\
&= dfrac14 left(dfrac{1}{frac{m}{2}}-dfrac{1}{j+frac{m}{2}}right)\
&= dfrac12 left(dfrac{1}{m}-dfrac{1}{2j+m}right)\
&= dfrac12 left(dfrac{1}{m}-dfrac{1}{n}right)\
&< dfrac{1}{2m}\
&to 0 text{ as } m to infty\
end{array}
$

If $n-m$ is odd,
the sum changes
by at most $frac1{n}$
so it still goes to zero.

share|cite|improve this answer

answered 3 hours ago

marty cohenmarty cohen

73.8k549128

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for such an elaborate proof , not ignoring the signs was an important step indeed.
  $endgroup$
  – kira0705
  3 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint :

$$frac{1}{2n-1}-frac{1}{2n}=frac{1}{(2n-1)2n}leq frac{1}{(n-1)n}= frac{1}{n-1}-frac{1}{n} $$

share|cite|improve this answer

answered 3 hours ago

Clément GuérinClément Guérin

10k1736

$endgroup$

add a comment | 

0

$begingroup$

Fix $epsilon > 0$, then for $n > frac{2}{sqrt{epsilon}}$ (So since both sides of the inequality are positive, $n^2 > frac{4}{epsilon} implies frac{epsilon}{2} > frac{2}{n^2}$), observe that $$|f_n - f_{n+1}| = |frac{1}{n} + frac{1}{n+1}| = |frac{n+2}{n(n+1)}| = frac{n+2}{n(n+1)} < frac{n+2}{n^2} =frac{1}{n} +frac{2}{n^2} < frac{1}{frac{2}{sqrt{epsilon}}} + frac{epsilon}{2} = frac{sqrt{epsilon}}{2} + frac{epsilon}{2} < frac{epsilon}{2} + frac{epsilon}{2} = epsilon $$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

user516079user516079

318210

$endgroup$

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%2f3123120%2fshow-that-sequence-is-a-cauchy-sequence%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

If you ignore the signs of the terms,
the result diverges.
So you can't do that.

$f_n
=sum_{k=1}^n dfrac{(-1)^k}{k}
$
so,
if $n > m$,
$f_n-f_m
=sum_{k=m+1}^n dfrac{(-1)^k}{k}
=sum_{k=1}^{n-m} dfrac{(-1)^{k+m}}{k+m}
=(-1)^msum_{k=1}^{n-m} dfrac{(-1)^{k}}{k+m}
$.

If
$n-m$ is even,
so $n-m = 2j$,
then

$begin{array}\
f_n-f_m
&=(-1)^msum_{k=1}^{2j} dfrac{(-1)^{k}}{k+m}\
&=(-1)^msum_{k=1}^{j} left(dfrac{(-1)^{2k-1}}{2k-1+m}+dfrac{(-1)^{2k}}{2k+m}right)\
&=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{-1}{2k-1+m}+dfrac{1}{2k+m}right)\
&=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{(2k-1+m)-(2k+m)}{(2k-1+m)(2k+m)}right)\
&=(-1)^{m+1}sum_{k=1}^{j} left(dfrac{-1}{(2k-1+m)(2k+m)}right)\
&=(-1)^{m}sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
text{so}\
|f_n-f_m|
&=sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
&=sum_{k=1}^{j}dfrac14 left(dfrac{1}{(k-frac12+frac{m}{2})(k+frac{m}{2})}right)\
&lt dfrac14sum_{k=1}^{j} left(dfrac{1}{(k-1+frac{m}{2})(k+frac{m}{2})}right)
quadtext{this is the sneaky part}\
&lt dfrac14sum_{k=1}^{j} left(dfrac{1}{k-1+frac{m}{2}}-dfrac{1}{k+frac{m}{2}}right)\
&= dfrac14 left(dfrac{1}{frac{m}{2}}-dfrac{1}{j+frac{m}{2}}right)\
&= dfrac12 left(dfrac{1}{m}-dfrac{1}{2j+m}right)\
&= dfrac12 left(dfrac{1}{m}-dfrac{1}{n}right)\
&< dfrac{1}{2m}\
&to 0 text{ as } m to infty\
end{array}
$

If $n-m$ is odd,
the sum changes
by at most $frac1{n}$
so it still goes to zero.

share|cite|improve this answer

answered 3 hours ago

marty cohenmarty cohen

73.8k549128

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for such an elaborate proof , not ignoring the signs was an important step indeed.
  $endgroup$
  – kira0705
  3 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

If you ignore the signs of the terms,
the result diverges.
So you can't do that.

$f_n
=sum_{k=1}^n dfrac{(-1)^k}{k}
$
so,
if $n > m$,
$f_n-f_m
=sum_{k=m+1}^n dfrac{(-1)^k}{k}
=sum_{k=1}^{n-m} dfrac{(-1)^{k+m}}{k+m}
=(-1)^msum_{k=1}^{n-m} dfrac{(-1)^{k}}{k+m}
$.

If
$n-m$ is even,
so $n-m = 2j$,
then

$begin{array}\
f_n-f_m
&=(-1)^msum_{k=1}^{2j} dfrac{(-1)^{k}}{k+m}\
&=(-1)^msum_{k=1}^{j} left(dfrac{(-1)^{2k-1}}{2k-1+m}+dfrac{(-1)^{2k}}{2k+m}right)\
&=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{-1}{2k-1+m}+dfrac{1}{2k+m}right)\
&=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{(2k-1+m)-(2k+m)}{(2k-1+m)(2k+m)}right)\
&=(-1)^{m+1}sum_{k=1}^{j} left(dfrac{-1}{(2k-1+m)(2k+m)}right)\
&=(-1)^{m}sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
text{so}\
|f_n-f_m|
&=sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
&=sum_{k=1}^{j}dfrac14 left(dfrac{1}{(k-frac12+frac{m}{2})(k+frac{m}{2})}right)\
&lt dfrac14sum_{k=1}^{j} left(dfrac{1}{(k-1+frac{m}{2})(k+frac{m}{2})}right)
quadtext{this is the sneaky part}\
&lt dfrac14sum_{k=1}^{j} left(dfrac{1}{k-1+frac{m}{2}}-dfrac{1}{k+frac{m}{2}}right)\
&= dfrac14 left(dfrac{1}{frac{m}{2}}-dfrac{1}{j+frac{m}{2}}right)\
&= dfrac12 left(dfrac{1}{m}-dfrac{1}{2j+m}right)\
&= dfrac12 left(dfrac{1}{m}-dfrac{1}{n}right)\
&< dfrac{1}{2m}\
&to 0 text{ as } m to infty\
end{array}
$

If $n-m$ is odd,
the sum changes
by at most $frac1{n}$
so it still goes to zero.

share|cite|improve this answer

answered 3 hours ago

marty cohenmarty cohen

73.8k549128

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for such an elaborate proof , not ignoring the signs was an important step indeed.
  $endgroup$
  – kira0705
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

If you ignore the signs of the terms,
the result diverges.
So you can't do that.

$f_n
=sum_{k=1}^n dfrac{(-1)^k}{k}
$
so,
if $n > m$,
$f_n-f_m
=sum_{k=m+1}^n dfrac{(-1)^k}{k}
=sum_{k=1}^{n-m} dfrac{(-1)^{k+m}}{k+m}
=(-1)^msum_{k=1}^{n-m} dfrac{(-1)^{k}}{k+m}
$.

If
$n-m$ is even,
so $n-m = 2j$,
then

$begin{array}\
f_n-f_m
&=(-1)^msum_{k=1}^{2j} dfrac{(-1)^{k}}{k+m}\
&=(-1)^msum_{k=1}^{j} left(dfrac{(-1)^{2k-1}}{2k-1+m}+dfrac{(-1)^{2k}}{2k+m}right)\
&=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{-1}{2k-1+m}+dfrac{1}{2k+m}right)\
&=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{(2k-1+m)-(2k+m)}{(2k-1+m)(2k+m)}right)\
&=(-1)^{m+1}sum_{k=1}^{j} left(dfrac{-1}{(2k-1+m)(2k+m)}right)\
&=(-1)^{m}sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
text{so}\
|f_n-f_m|
&=sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
&=sum_{k=1}^{j}dfrac14 left(dfrac{1}{(k-frac12+frac{m}{2})(k+frac{m}{2})}right)\
&lt dfrac14sum_{k=1}^{j} left(dfrac{1}{(k-1+frac{m}{2})(k+frac{m}{2})}right)
quadtext{this is the sneaky part}\
&lt dfrac14sum_{k=1}^{j} left(dfrac{1}{k-1+frac{m}{2}}-dfrac{1}{k+frac{m}{2}}right)\
&= dfrac14 left(dfrac{1}{frac{m}{2}}-dfrac{1}{j+frac{m}{2}}right)\
&= dfrac12 left(dfrac{1}{m}-dfrac{1}{2j+m}right)\
&= dfrac12 left(dfrac{1}{m}-dfrac{1}{n}right)\
&< dfrac{1}{2m}\
&to 0 text{ as } m to infty\
end{array}
$

If $n-m$ is odd,
the sum changes
by at most $frac1{n}$
so it still goes to zero.

share|cite|improve this answer

answered 3 hours ago

marty cohenmarty cohen

73.8k549128

$endgroup$

share|cite|improve this answer

answered 3 hours ago

marty cohenmarty cohen

73.8k549128

answered 3 hours ago

marty cohenmarty cohen

73.8k549128

answered 3 hours ago

marty cohenmarty cohen

73.8k549128

2

$begingroup$

Hint :

$$frac{1}{2n-1}-frac{1}{2n}=frac{1}{(2n-1)2n}leq frac{1}{(n-1)n}= frac{1}{n-1}-frac{1}{n} $$

share|cite|improve this answer

answered 3 hours ago

Clément GuérinClément Guérin

10k1736

$endgroup$

add a comment | 

2

$begingroup$

Hint :

$$frac{1}{2n-1}-frac{1}{2n}=frac{1}{(2n-1)2n}leq frac{1}{(n-1)n}= frac{1}{n-1}-frac{1}{n} $$

share|cite|improve this answer

answered 3 hours ago

Clément GuérinClément Guérin

10k1736

$endgroup$

add a comment | 

$begingroup$

Hint :

$$frac{1}{2n-1}-frac{1}{2n}=frac{1}{(2n-1)2n}leq frac{1}{(n-1)n}= frac{1}{n-1}-frac{1}{n} $$

share|cite|improve this answer

answered 3 hours ago

Clément GuérinClément Guérin

10k1736

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Clément GuérinClément Guérin

10k1736

answered 3 hours ago

Clément GuérinClément Guérin

10k1736

answered 3 hours ago

Clément GuérinClément Guérin

10k1736

0

$begingroup$

Fix $epsilon > 0$, then for $n > frac{2}{sqrt{epsilon}}$ (So since both sides of the inequality are positive, $n^2 > frac{4}{epsilon} implies frac{epsilon}{2} > frac{2}{n^2}$), observe that $$|f_n - f_{n+1}| = |frac{1}{n} + frac{1}{n+1}| = |frac{n+2}{n(n+1)}| = frac{n+2}{n(n+1)} < frac{n+2}{n^2} =frac{1}{n} +frac{2}{n^2} < frac{1}{frac{2}{sqrt{epsilon}}} + frac{epsilon}{2} = frac{sqrt{epsilon}}{2} + frac{epsilon}{2} < frac{epsilon}{2} + frac{epsilon}{2} = epsilon $$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

user516079user516079

318210

$endgroup$

add a comment | 

0

$begingroup$

Fix $epsilon > 0$, then for $n > frac{2}{sqrt{epsilon}}$ (So since both sides of the inequality are positive, $n^2 > frac{4}{epsilon} implies frac{epsilon}{2} > frac{2}{n^2}$), observe that $$|f_n - f_{n+1}| = |frac{1}{n} + frac{1}{n+1}| = |frac{n+2}{n(n+1)}| = frac{n+2}{n(n+1)} < frac{n+2}{n^2} =frac{1}{n} +frac{2}{n^2} < frac{1}{frac{2}{sqrt{epsilon}}} + frac{epsilon}{2} = frac{sqrt{epsilon}}{2} + frac{epsilon}{2} < frac{epsilon}{2} + frac{epsilon}{2} = epsilon $$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

user516079user516079

318210

$endgroup$

add a comment | 

$begingroup$

Fix $epsilon > 0$, then for $n > frac{2}{sqrt{epsilon}}$ (So since both sides of the inequality are positive, $n^2 > frac{4}{epsilon} implies frac{epsilon}{2} > frac{2}{n^2}$), observe that $$|f_n - f_{n+1}| = |frac{1}{n} + frac{1}{n+1}| = |frac{n+2}{n(n+1)}| = frac{n+2}{n(n+1)} < frac{n+2}{n^2} =frac{1}{n} +frac{2}{n^2} < frac{1}{frac{2}{sqrt{epsilon}}} + frac{epsilon}{2} = frac{sqrt{epsilon}}{2} + frac{epsilon}{2} < frac{epsilon}{2} + frac{epsilon}{2} = epsilon $$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

user516079user516079

318210

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

user516079user516079

318210

answered 3 hours ago

user516079user516079

318210

answered 3 hours ago

user516079user516079

318210