Determine whether the series $frac{e^frac{1}{n}}{n^2}$ converges or diverges












0














The problem states




Determine whether the series converges or diverges: $$sum_{n=1}^inftyfrac{e^{1/n}}{n^2}$$




Immediately I always like to try the test for divergence:



$$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{1/n}}{n^2} = 0$$ I judge this to be true by looking at the numerator and denominator of the series. The numerator will tend to 1 as $nto infty$ and the denominator will grow bigger and bigger, so eventually you will have $frac{1}{infty}$ which will equal to zero. This tells us absolutely nothing about the convergence or divergence of this series.



Next I'll try the integral test because i see in the denominator a term resulting from the derivative of the numerator. To apply the integral test the sequence $a_n$ of our series $sum_{n=1}^infty a_n$ should be continuous, positive and decreasing. The summation begins at one and not zero so we have no issues with continuity here. Both numerator and denominator will yield positive terms as $n to infty$ and sequential terms will get smaller as the numerator shrinks whilst the denominator grows as $nto infty$.



$$text{let } f(x) = frac{e^{x^{-1}}}{x^2}$$
$$int_1^infty f(x) dx = int_1^infty frac{e^{x^{-1}}}{x^2}dx$$
$$= -e^{x{^{-1}}}|_1^infty$$
$$= lim_{xto infty}- e^{frac{1}{x}} +e$$
$$=-1 + e$$



because $int_1^infty f(x)$ converges then by the integral test so must our series.



I have two questions:




  1. How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?

  2. Would it be possible to determine whether this series converges or diverges using the limit comparison test?






sequences-and-series convergence







share|cite|improve this question




















  • 3




    Compare, to say $e/n^2$.
    – Lord Shark the Unknown
    Oct 1 '17 at 16:19










  • This makes no sense: $$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{frac{1}{n}}}{n^2} = 0$$
    – Simply Beautiful Art
    Oct 1 '17 at 16:27










  • The test for divergence is applied to the sequence and not the series? I.e) if i had written $$ lim_{ntoinfty} frac{e^{frac{1}{n}}}{n^2} = 0$$ that would be fine?
    – Blargian
    Oct 1 '17 at 16:49










  • Yes, that is correct.
    – Simply Beautiful Art
    Oct 1 '17 at 16:53










  • Don’t forget the $dx$ on your integral
    – Chase Ryan Taylor
    Oct 1 '17 at 18:24
















0














The problem states




Determine whether the series converges or diverges: $$sum_{n=1}^inftyfrac{e^{1/n}}{n^2}$$




Immediately I always like to try the test for divergence:



$$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{1/n}}{n^2} = 0$$ I judge this to be true by looking at the numerator and denominator of the series. The numerator will tend to 1 as $nto infty$ and the denominator will grow bigger and bigger, so eventually you will have $frac{1}{infty}$ which will equal to zero. This tells us absolutely nothing about the convergence or divergence of this series.



Next I'll try the integral test because i see in the denominator a term resulting from the derivative of the numerator. To apply the integral test the sequence $a_n$ of our series $sum_{n=1}^infty a_n$ should be continuous, positive and decreasing. The summation begins at one and not zero so we have no issues with continuity here. Both numerator and denominator will yield positive terms as $n to infty$ and sequential terms will get smaller as the numerator shrinks whilst the denominator grows as $nto infty$.



$$text{let } f(x) = frac{e^{x^{-1}}}{x^2}$$
$$int_1^infty f(x) dx = int_1^infty frac{e^{x^{-1}}}{x^2}dx$$
$$= -e^{x{^{-1}}}|_1^infty$$
$$= lim_{xto infty}- e^{frac{1}{x}} +e$$
$$=-1 + e$$



because $int_1^infty f(x)$ converges then by the integral test so must our series.



I have two questions:




  1. How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?

  2. Would it be possible to determine whether this series converges or diverges using the limit comparison test?






sequences-and-series convergence







share|cite|improve this question




















  • 3




    Compare, to say $e/n^2$.
    – Lord Shark the Unknown
    Oct 1 '17 at 16:19










  • This makes no sense: $$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{frac{1}{n}}}{n^2} = 0$$
    – Simply Beautiful Art
    Oct 1 '17 at 16:27










  • The test for divergence is applied to the sequence and not the series? I.e) if i had written $$ lim_{ntoinfty} frac{e^{frac{1}{n}}}{n^2} = 0$$ that would be fine?
    – Blargian
    Oct 1 '17 at 16:49










  • Yes, that is correct.
    – Simply Beautiful Art
    Oct 1 '17 at 16:53










  • Don’t forget the $dx$ on your integral
    – Chase Ryan Taylor
    Oct 1 '17 at 18:24














0












0








0


1





The problem states




Determine whether the series converges or diverges: $$sum_{n=1}^inftyfrac{e^{1/n}}{n^2}$$




Immediately I always like to try the test for divergence:



$$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{1/n}}{n^2} = 0$$ I judge this to be true by looking at the numerator and denominator of the series. The numerator will tend to 1 as $nto infty$ and the denominator will grow bigger and bigger, so eventually you will have $frac{1}{infty}$ which will equal to zero. This tells us absolutely nothing about the convergence or divergence of this series.



Next I'll try the integral test because i see in the denominator a term resulting from the derivative of the numerator. To apply the integral test the sequence $a_n$ of our series $sum_{n=1}^infty a_n$ should be continuous, positive and decreasing. The summation begins at one and not zero so we have no issues with continuity here. Both numerator and denominator will yield positive terms as $n to infty$ and sequential terms will get smaller as the numerator shrinks whilst the denominator grows as $nto infty$.



$$text{let } f(x) = frac{e^{x^{-1}}}{x^2}$$
$$int_1^infty f(x) dx = int_1^infty frac{e^{x^{-1}}}{x^2}dx$$
$$= -e^{x{^{-1}}}|_1^infty$$
$$= lim_{xto infty}- e^{frac{1}{x}} +e$$
$$=-1 + e$$



because $int_1^infty f(x)$ converges then by the integral test so must our series.



I have two questions:




  1. How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?

  2. Would it be possible to determine whether this series converges or diverges using the limit comparison test?






sequences-and-series convergence







share|cite|improve this question















The problem states




Determine whether the series converges or diverges: $$sum_{n=1}^inftyfrac{e^{1/n}}{n^2}$$




Immediately I always like to try the test for divergence:



$$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{1/n}}{n^2} = 0$$ I judge this to be true by looking at the numerator and denominator of the series. The numerator will tend to 1 as $nto infty$ and the denominator will grow bigger and bigger, so eventually you will have $frac{1}{infty}$ which will equal to zero. This tells us absolutely nothing about the convergence or divergence of this series.



Next I'll try the integral test because i see in the denominator a term resulting from the derivative of the numerator. To apply the integral test the sequence $a_n$ of our series $sum_{n=1}^infty a_n$ should be continuous, positive and decreasing. The summation begins at one and not zero so we have no issues with continuity here. Both numerator and denominator will yield positive terms as $n to infty$ and sequential terms will get smaller as the numerator shrinks whilst the denominator grows as $nto infty$.



$$text{let } f(x) = frac{e^{x^{-1}}}{x^2}$$
$$int_1^infty f(x) dx = int_1^infty frac{e^{x^{-1}}}{x^2}dx$$
$$= -e^{x{^{-1}}}|_1^infty$$
$$= lim_{xto infty}- e^{frac{1}{x}} +e$$
$$=-1 + e$$



because $int_1^infty f(x)$ converges then by the integral test so must our series.



I have two questions:




  1. How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?

  2. Would it be possible to determine whether this series converges or diverges using the limit comparison test?






sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 at 11:35









amWhy

191k28224439




191k28224439










asked Oct 1 '17 at 16:16









Blargian

161314




161314








  • 3




    Compare, to say $e/n^2$.
    – Lord Shark the Unknown
    Oct 1 '17 at 16:19










  • This makes no sense: $$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{frac{1}{n}}}{n^2} = 0$$
    – Simply Beautiful Art
    Oct 1 '17 at 16:27










  • The test for divergence is applied to the sequence and not the series? I.e) if i had written $$ lim_{ntoinfty} frac{e^{frac{1}{n}}}{n^2} = 0$$ that would be fine?
    – Blargian
    Oct 1 '17 at 16:49










  • Yes, that is correct.
    – Simply Beautiful Art
    Oct 1 '17 at 16:53










  • Don’t forget the $dx$ on your integral
    – Chase Ryan Taylor
    Oct 1 '17 at 18:24














  • 3




    Compare, to say $e/n^2$.
    – Lord Shark the Unknown
    Oct 1 '17 at 16:19










  • This makes no sense: $$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{frac{1}{n}}}{n^2} = 0$$
    – Simply Beautiful Art
    Oct 1 '17 at 16:27










  • The test for divergence is applied to the sequence and not the series? I.e) if i had written $$ lim_{ntoinfty} frac{e^{frac{1}{n}}}{n^2} = 0$$ that would be fine?
    – Blargian
    Oct 1 '17 at 16:49










  • Yes, that is correct.
    – Simply Beautiful Art
    Oct 1 '17 at 16:53










  • Don’t forget the $dx$ on your integral
    – Chase Ryan Taylor
    Oct 1 '17 at 18:24








3




3




Compare, to say $e/n^2$.
– Lord Shark the Unknown
Oct 1 '17 at 16:19




Compare, to say $e/n^2$.
– Lord Shark the Unknown
Oct 1 '17 at 16:19












This makes no sense: $$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{frac{1}{n}}}{n^2} = 0$$
– Simply Beautiful Art
Oct 1 '17 at 16:27




This makes no sense: $$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{frac{1}{n}}}{n^2} = 0$$
– Simply Beautiful Art
Oct 1 '17 at 16:27












The test for divergence is applied to the sequence and not the series? I.e) if i had written $$ lim_{ntoinfty} frac{e^{frac{1}{n}}}{n^2} = 0$$ that would be fine?
– Blargian
Oct 1 '17 at 16:49




The test for divergence is applied to the sequence and not the series? I.e) if i had written $$ lim_{ntoinfty} frac{e^{frac{1}{n}}}{n^2} = 0$$ that would be fine?
– Blargian
Oct 1 '17 at 16:49












Yes, that is correct.
– Simply Beautiful Art
Oct 1 '17 at 16:53




Yes, that is correct.
– Simply Beautiful Art
Oct 1 '17 at 16:53












Don’t forget the $dx$ on your integral
– Chase Ryan Taylor
Oct 1 '17 at 18:24




Don’t forget the $dx$ on your integral
– Chase Ryan Taylor
Oct 1 '17 at 18:24










1 Answer
1






active

oldest

votes


















1














As Lord Shark the Unknown comments, simply note that



$$e^{1/n}<e,quad n>1$$



And so,



$$0<sum_{n=1}^inftyfrac{e^{1/n}}{n^2}<sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6<infty$$




How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?




If $f(x)$ is continuous, $f(x)ge0$, and $f'(x)le0$, then...




Would it be possible to determine whether this series converges or diverges using the limit comparison test?




Sure. You could apply the limit comparison test against $1/n^2$, since $e^{1/n}to1$ as $ntoinfty$.






share|cite|improve this answer





















  • Could you please explain how $$sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6$$
    – Blargian
    Oct 1 '17 at 16:50






  • 1




    See the Basel problem.
    – Simply Beautiful Art
    Oct 1 '17 at 16:52






  • 1




    You probably don't want this, but another interpretation of the conditions for the integral test would be$$(fin C^1)landforall xge0[(f(x)ge0)land f'(x)le0)]impliesleft(sum_{n=0}^infty f(n)<inftyiffint_0^infty f(x)~mathrm dx<inftyright)$$
    – Simply Beautiful Art
    Oct 1 '17 at 16:56






  • 1




    If you don't want to restrict this to differentiable functions, then use $forall xge0forall yge0[x<yimplies f(x)>f(y)]$
    – Simply Beautiful Art
    Oct 1 '17 at 16:57











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














As Lord Shark the Unknown comments, simply note that



$$e^{1/n}<e,quad n>1$$



And so,



$$0<sum_{n=1}^inftyfrac{e^{1/n}}{n^2}<sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6<infty$$




How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?




If $f(x)$ is continuous, $f(x)ge0$, and $f'(x)le0$, then...




Would it be possible to determine whether this series converges or diverges using the limit comparison test?




Sure. You could apply the limit comparison test against $1/n^2$, since $e^{1/n}to1$ as $ntoinfty$.






share|cite|improve this answer





















  • Could you please explain how $$sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6$$
    – Blargian
    Oct 1 '17 at 16:50






  • 1




    See the Basel problem.
    – Simply Beautiful Art
    Oct 1 '17 at 16:52






  • 1




    You probably don't want this, but another interpretation of the conditions for the integral test would be$$(fin C^1)landforall xge0[(f(x)ge0)land f'(x)le0)]impliesleft(sum_{n=0}^infty f(n)<inftyiffint_0^infty f(x)~mathrm dx<inftyright)$$
    – Simply Beautiful Art
    Oct 1 '17 at 16:56






  • 1




    If you don't want to restrict this to differentiable functions, then use $forall xge0forall yge0[x<yimplies f(x)>f(y)]$
    – Simply Beautiful Art
    Oct 1 '17 at 16:57
















1














As Lord Shark the Unknown comments, simply note that



$$e^{1/n}<e,quad n>1$$



And so,



$$0<sum_{n=1}^inftyfrac{e^{1/n}}{n^2}<sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6<infty$$




How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?




If $f(x)$ is continuous, $f(x)ge0$, and $f'(x)le0$, then...




Would it be possible to determine whether this series converges or diverges using the limit comparison test?




Sure. You could apply the limit comparison test against $1/n^2$, since $e^{1/n}to1$ as $ntoinfty$.






share|cite|improve this answer





















  • Could you please explain how $$sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6$$
    – Blargian
    Oct 1 '17 at 16:50






  • 1




    See the Basel problem.
    – Simply Beautiful Art
    Oct 1 '17 at 16:52






  • 1




    You probably don't want this, but another interpretation of the conditions for the integral test would be$$(fin C^1)landforall xge0[(f(x)ge0)land f'(x)le0)]impliesleft(sum_{n=0}^infty f(n)<inftyiffint_0^infty f(x)~mathrm dx<inftyright)$$
    – Simply Beautiful Art
    Oct 1 '17 at 16:56






  • 1




    If you don't want to restrict this to differentiable functions, then use $forall xge0forall yge0[x<yimplies f(x)>f(y)]$
    – Simply Beautiful Art
    Oct 1 '17 at 16:57














1












1








1






As Lord Shark the Unknown comments, simply note that



$$e^{1/n}<e,quad n>1$$



And so,



$$0<sum_{n=1}^inftyfrac{e^{1/n}}{n^2}<sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6<infty$$




How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?




If $f(x)$ is continuous, $f(x)ge0$, and $f'(x)le0$, then...




Would it be possible to determine whether this series converges or diverges using the limit comparison test?




Sure. You could apply the limit comparison test against $1/n^2$, since $e^{1/n}to1$ as $ntoinfty$.






share|cite|improve this answer












As Lord Shark the Unknown comments, simply note that



$$e^{1/n}<e,quad n>1$$



And so,



$$0<sum_{n=1}^inftyfrac{e^{1/n}}{n^2}<sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6<infty$$




How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?




If $f(x)$ is continuous, $f(x)ge0$, and $f'(x)le0$, then...




Would it be possible to determine whether this series converges or diverges using the limit comparison test?




Sure. You could apply the limit comparison test against $1/n^2$, since $e^{1/n}to1$ as $ntoinfty$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 1 '17 at 16:31









Simply Beautiful Art

50.2k578181




50.2k578181












  • Could you please explain how $$sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6$$
    – Blargian
    Oct 1 '17 at 16:50






  • 1




    See the Basel problem.
    – Simply Beautiful Art
    Oct 1 '17 at 16:52






  • 1




    You probably don't want this, but another interpretation of the conditions for the integral test would be$$(fin C^1)landforall xge0[(f(x)ge0)land f'(x)le0)]impliesleft(sum_{n=0}^infty f(n)<inftyiffint_0^infty f(x)~mathrm dx<inftyright)$$
    – Simply Beautiful Art
    Oct 1 '17 at 16:56






  • 1




    If you don't want to restrict this to differentiable functions, then use $forall xge0forall yge0[x<yimplies f(x)>f(y)]$
    – Simply Beautiful Art
    Oct 1 '17 at 16:57


















  • Could you please explain how $$sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6$$
    – Blargian
    Oct 1 '17 at 16:50






  • 1




    See the Basel problem.
    – Simply Beautiful Art
    Oct 1 '17 at 16:52






  • 1




    You probably don't want this, but another interpretation of the conditions for the integral test would be$$(fin C^1)landforall xge0[(f(x)ge0)land f'(x)le0)]impliesleft(sum_{n=0}^infty f(n)<inftyiffint_0^infty f(x)~mathrm dx<inftyright)$$
    – Simply Beautiful Art
    Oct 1 '17 at 16:56






  • 1




    If you don't want to restrict this to differentiable functions, then use $forall xge0forall yge0[x<yimplies f(x)>f(y)]$
    – Simply Beautiful Art
    Oct 1 '17 at 16:57
















Could you please explain how $$sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6$$
– Blargian
Oct 1 '17 at 16:50




Could you please explain how $$sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6$$
– Blargian
Oct 1 '17 at 16:50




1




1




See the Basel problem.
– Simply Beautiful Art
Oct 1 '17 at 16:52




See the Basel problem.
– Simply Beautiful Art
Oct 1 '17 at 16:52




1




1




You probably don't want this, but another interpretation of the conditions for the integral test would be$$(fin C^1)landforall xge0[(f(x)ge0)land f'(x)le0)]impliesleft(sum_{n=0}^infty f(n)<inftyiffint_0^infty f(x)~mathrm dx<inftyright)$$
– Simply Beautiful Art
Oct 1 '17 at 16:56




You probably don't want this, but another interpretation of the conditions for the integral test would be$$(fin C^1)landforall xge0[(f(x)ge0)land f'(x)le0)]impliesleft(sum_{n=0}^infty f(n)<inftyiffint_0^infty f(x)~mathrm dx<inftyright)$$
– Simply Beautiful Art
Oct 1 '17 at 16:56




1




1




If you don't want to restrict this to differentiable functions, then use $forall xge0forall yge0[x<yimplies f(x)>f(y)]$
– Simply Beautiful Art
Oct 1 '17 at 16:57




If you don't want to restrict this to differentiable functions, then use $forall xge0forall yge0[x<yimplies f(x)>f(y)]$
– Simply Beautiful Art
Oct 1 '17 at 16:57


















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Verónica Boquete

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Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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