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Let $langle a_nrangle$ be a sequence where $a_n = sumlimits_j=1^n(-1)^j2/n$ then

Let $(a_n)$ be a sequence of real numbers. Suppose each of the subsequences $(a_2n),(a_2n+1)$ and $(a_3n)$ converges to $p,q$ and $r$If $prodlimits_n(1+a_n)$ converges, does $sumlimits_nfraca_n1+a_n$ converge?Let $a_n rightarrow 0$. Show $sumlimits_n=0^infty (a_n - a_n+1) = a_0$.If $langlea_nrangle$ is convergent with limit l, how would I show that $langler_nrangle$ also converges?Convergence of infinite series for $sum a_n $ and $sum b_n$Let $a_n$ be a sequence of real numbers. which one of the following is always true?If a sequence diverges to infinity then so do all its subsequences. Then why does the sum for $frac1n$ diverge but not $frac1n^2$?Series with $sum a_n$ converges but $sum n a_n^2$ diverges, and $a_n$ is decreasingConvergence of $sumlimits_n=1^inftya_n$ implies convergence of $sumlimits_n=1^inftya_n^sigma_n$ where $sigma_n=fracnn+1$?$a_n$ is an infinite sequence with $sumlimits_n=1^infty a_nleq1$ and $0leq a_n<1$. Prove that $sumlimits_n=1^infty a_n/(a_n-1)$ converges?

1

$begingroup$

Let $langle a_nrangle$ be a sequence where $a_n = sumlimits_j=1^n(-1)^j2/n$

1. is oscillatory




2. diverges to $+infty$




3. diverges to $-infty$




4. converges to $0$

I think by taking subsequences $langle a_2nrangle =(2/n) +(2/n) +ldots +(2/n)$ ($n$-times) $= 2$ and $langle a_2n-1rangle = -2/n -2/n -ldots -2/n$ ($n$-times) $=-2$ which is cunverging to $2$ and $-2$ respectively. So that the sequence is oscillatory. I have doubt taking this type of subsequence is right or wrong.

real-analysis sequences-and-series

share|cite|improve this question

edited Mar 13 at 10:18

rtybase

11.5k31534

asked Mar 13 at 8:37

800123800123

73

$endgroup$

add a comment | 

1

$begingroup$

Let $langle a_nrangle$ be a sequence where $a_n = sumlimits_j=1^n(-1)^j2/n$

1. is oscillatory




2. diverges to $+infty$




3. diverges to $-infty$




4. converges to $0$

I think by taking subsequences $langle a_2nrangle =(2/n) +(2/n) +ldots +(2/n)$ ($n$-times) $= 2$ and $langle a_2n-1rangle = -2/n -2/n -ldots -2/n$ ($n$-times) $=-2$ which is cunverging to $2$ and $-2$ respectively. So that the sequence is oscillatory. I have doubt taking this type of subsequence is right or wrong.

real-analysis sequences-and-series

share|cite|improve this question

edited Mar 13 at 10:18

rtybase

11.5k31534

asked Mar 13 at 8:37

800123800123

73

$endgroup$

add a comment | 

$begingroup$

Let $langle a_nrangle$ be a sequence where $a_n = sumlimits_j=1^n(-1)^j2/n$

1. is oscillatory




2. diverges to $+infty$




3. diverges to $-infty$




4. converges to $0$

I think by taking subsequences $langle a_2nrangle =(2/n) +(2/n) +ldots +(2/n)$ ($n$-times) $= 2$ and $langle a_2n-1rangle = -2/n -2/n -ldots -2/n$ ($n$-times) $=-2$ which is cunverging to $2$ and $-2$ respectively. So that the sequence is oscillatory. I have doubt taking this type of subsequence is right or wrong.

real-analysis sequences-and-series

share|cite|improve this question

edited Mar 13 at 10:18

rtybase

11.5k31534

asked Mar 13 at 8:37

800123800123

73

$endgroup$

share|cite|improve this question

edited Mar 13 at 10:18

rtybase

11.5k31534

asked Mar 13 at 8:37

800123800123

73

share|cite|improve this question

edited Mar 13 at 10:18

rtybase

11.5k31534

asked Mar 13 at 8:37

800123800123

73

edited Mar 13 at 10:18

rtybase

11.5k31534

edited Mar 13 at 10:18

rtybase

11.5k31534

asked Mar 13 at 8:37

800123800123

73

asked Mar 13 at 8:37

800123800123

73

4 Answers
4

active

oldest

votes

1

$begingroup$

The correct answer is 4). Write down the sums $sumlimits_k=1^n (-1)^k$ and convince yourself that these are bounded. So if you multiply these by $frac 2n $ and let $n to infty$ you get $0$.

share|cite|improve this answer

answered Mar 13 at 8:42

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

$endgroup$

add a comment | 

1

$begingroup$

We have $a_n=frac2nsum_j=1^n(-1)^j=frac1n cdot (1-(-1)^n+1)$ and

$|a_n| le 2/n.$ for all $n$

Can you proceed ?

share|cite|improve this answer

edited Mar 13 at 9:11

answered Mar 13 at 8:43

FredFred

48.5k1849

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  this means $-2/nleq a_nleq2/n$ then by sandwitch theorem $<a_n> converges to 0
  $endgroup$
  – 800123
  Mar 13 at 11:04
  

  
  
  
  

add a comment | 

0

$begingroup$

1)$a_n= (-2/n)$, $n$ is odd.

2) $a_n=0$, $n$ is even.

1')$lim_ k rightarrow infty a_2k+1 =0$, $k=0,1,2,..;$

2')$ lim_l rightarrow infty a_2l=0, l=1,2,3,...;$

Lemma:

The $2$ subsequences $a_2l, a_2k+1$,

$l=1,2,..,$ $k=0,1,2,..,$

converge to $0$, hence $a_n$ converges to $0$.

Can you prove the Lemma?

share|cite|improve this answer

answered Mar 13 at 9:02

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

add a comment | 

0

$begingroup$

$$a_n = sumlimits_j=1^n(-1)^j2/n\to a_2k=0\,\a_2k+1=-2over 2k+1$$then this sequence oscillatorily converges to 0.

share|cite|improve this answer

answered Mar 13 at 11:58

Mostafa AyazMostafa Ayaz

16.5k3939

$endgroup$

add a comment | 

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1

$begingroup$

The correct answer is 4). Write down the sums $sumlimits_k=1^n (-1)^k$ and convince yourself that these are bounded. So if you multiply these by $frac 2n $ and let $n to infty$ you get $0$.

share|cite|improve this answer

answered Mar 13 at 8:42

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

$endgroup$

add a comment | 

1

$begingroup$

The correct answer is 4). Write down the sums $sumlimits_k=1^n (-1)^k$ and convince yourself that these are bounded. So if you multiply these by $frac 2n $ and let $n to infty$ you get $0$.

share|cite|improve this answer

answered Mar 13 at 8:42

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

$endgroup$

add a comment | 

$begingroup$

The correct answer is 4). Write down the sums $sumlimits_k=1^n (-1)^k$ and convince yourself that these are bounded. So if you multiply these by $frac 2n $ and let $n to infty$ you get $0$.

share|cite|improve this answer

answered Mar 13 at 8:42

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

$endgroup$

share|cite|improve this answer

answered Mar 13 at 8:42

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

answered Mar 13 at 8:42

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

answered Mar 13 at 8:42

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

1

$begingroup$

We have $a_n=frac2nsum_j=1^n(-1)^j=frac1n cdot (1-(-1)^n+1)$ and

$|a_n| le 2/n.$ for all $n$

Can you proceed ?

share|cite|improve this answer

edited Mar 13 at 9:11

answered Mar 13 at 8:43

FredFred

48.5k1849

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  this means $-2/nleq a_nleq2/n$ then by sandwitch theorem $<a_n> converges to 0
  $endgroup$
  – 800123
  Mar 13 at 11:04
  

  
  
  
  

add a comment | 

1

$begingroup$

We have $a_n=frac2nsum_j=1^n(-1)^j=frac1n cdot (1-(-1)^n+1)$ and

$|a_n| le 2/n.$ for all $n$

Can you proceed ?

share|cite|improve this answer

edited Mar 13 at 9:11

answered Mar 13 at 8:43

FredFred

48.5k1849

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  this means $-2/nleq a_nleq2/n$ then by sandwitch theorem $<a_n> converges to 0
  $endgroup$
  – 800123
  Mar 13 at 11:04
  

  
  
  
  

add a comment | 

$begingroup$

We have $a_n=frac2nsum_j=1^n(-1)^j=frac1n cdot (1-(-1)^n+1)$ and

$|a_n| le 2/n.$ for all $n$

Can you proceed ?

share|cite|improve this answer

edited Mar 13 at 9:11

answered Mar 13 at 8:43

FredFred

48.5k1849

$endgroup$

share|cite|improve this answer

edited Mar 13 at 9:11

answered Mar 13 at 8:43

FredFred

48.5k1849

answered Mar 13 at 8:43

FredFred

48.5k1849

answered Mar 13 at 8:43

FredFred

48.5k1849

0

$begingroup$

1)$a_n= (-2/n)$, $n$ is odd.

2) $a_n=0$, $n$ is even.

1')$lim_ k rightarrow infty a_2k+1 =0$, $k=0,1,2,..;$

2')$ lim_l rightarrow infty a_2l=0, l=1,2,3,...;$

Lemma:

The $2$ subsequences $a_2l, a_2k+1$,

$l=1,2,..,$ $k=0,1,2,..,$

converge to $0$, hence $a_n$ converges to $0$.

Can you prove the Lemma?

share|cite|improve this answer

answered Mar 13 at 9:02

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

add a comment | 

0

$begingroup$

1)$a_n= (-2/n)$, $n$ is odd.

2) $a_n=0$, $n$ is even.

1')$lim_ k rightarrow infty a_2k+1 =0$, $k=0,1,2,..;$

2')$ lim_l rightarrow infty a_2l=0, l=1,2,3,...;$

Lemma:

The $2$ subsequences $a_2l, a_2k+1$,

$l=1,2,..,$ $k=0,1,2,..,$

converge to $0$, hence $a_n$ converges to $0$.

Can you prove the Lemma?

share|cite|improve this answer

answered Mar 13 at 9:02

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

add a comment | 

$begingroup$

1)$a_n= (-2/n)$, $n$ is odd.

2) $a_n=0$, $n$ is even.

1')$lim_ k rightarrow infty a_2k+1 =0$, $k=0,1,2,..;$

2')$ lim_l rightarrow infty a_2l=0, l=1,2,3,...;$

Lemma:

The $2$ subsequences $a_2l, a_2k+1$,

$l=1,2,..,$ $k=0,1,2,..,$

converge to $0$, hence $a_n$ converges to $0$.

Can you prove the Lemma?

share|cite|improve this answer

answered Mar 13 at 9:02

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

share|cite|improve this answer

answered Mar 13 at 9:02

Peter SzilasPeter Szilas

11.6k2822

answered Mar 13 at 9:02

Peter SzilasPeter Szilas

11.6k2822

answered Mar 13 at 9:02

Peter SzilasPeter Szilas

11.6k2822

0

$begingroup$

$$a_n = sumlimits_j=1^n(-1)^j2/n\to a_2k=0\,\a_2k+1=-2over 2k+1$$then this sequence oscillatorily converges to 0.

share|cite|improve this answer

answered Mar 13 at 11:58

Mostafa AyazMostafa Ayaz

16.5k3939

$endgroup$

add a comment | 

0

$begingroup$

$$a_n = sumlimits_j=1^n(-1)^j2/n\to a_2k=0\,\a_2k+1=-2over 2k+1$$then this sequence oscillatorily converges to 0.

share|cite|improve this answer

answered Mar 13 at 11:58

Mostafa AyazMostafa Ayaz

16.5k3939

$endgroup$

add a comment | 

$begingroup$

$$a_n = sumlimits_j=1^n(-1)^j2/n\to a_2k=0\,\a_2k+1=-2over 2k+1$$then this sequence oscillatorily converges to 0.

share|cite|improve this answer

answered Mar 13 at 11:58

Mostafa AyazMostafa Ayaz

16.5k3939

$endgroup$

share|cite|improve this answer

answered Mar 13 at 11:58

Mostafa AyazMostafa Ayaz

16.5k3939

answered Mar 13 at 11:58

Mostafa AyazMostafa Ayaz

16.5k3939

answered Mar 13 at 11:58

Mostafa AyazMostafa Ayaz

16.5k3939