Limit, Riemann Sum, Integration, Natural logarithm












3












$begingroup$


For any natural number $m$, $lim_{nrightarrow infty }left ( frac{1}{n+1}+frac{1}{n+2}+frac{1}{n+3}+cdots +frac{1}{mn} right )=ln (m)$.



I tried to prove the statement in the following way.



Proof:



$$lim_{nrightarrow infty }left ( frac{1}{n+1}+frac{1}{n+2}+frac{1}{n+3}+cdots +frac{1}{mn} right )=lim_{nrightarrow infty }sum_{r=1}^{(m-1)n}frac{1}{n+r}$$



Dividing the numerator and the denominator of $frac{1}{n+r}$ by $n$, we get $frac{1/n}{1+r/n}$.



Therefore,



$$lim_{nrightarrow infty }sum_{r=1}^{(m-1)n}frac{1}{n+r}=lim_{nrightarrow infty }frac{1}{n}sum_{r=1}^{(m-1)n}frac{1}{1+r/n}$$



this is a Riemann sum, so replacing $frac{1}{n}$ with $dx$, $frac{r}{n}$ with $x$, and integrating between the limits $x=0$ and $x=m-1$



we get $



$$int_{0}^{m-1}frac{dx}{1+x}=ln(1+x)|_{0}^{m-1}=ln(m)-ln(1)=ln(m)blacksquare$$



Is this a valid way?






integration limits logarithms riemann-integration riemann-sum







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why $$lim_{nrightarrow infty }left ( frac{1}{n+1}+frac{1}{n+2}+frac{1}{n+3}+cdots +frac{1}{m+n} right )=lim_{nrightarrow infty }sum_{r=1}^{(m-1)n}frac{1}{n+r}$$ it should be $$=lim_{nrightarrow infty }sum_{r=1}^{m}frac{1}{n+r}$$
    $endgroup$
    – gimusi
    Dec 4 '18 at 8:38












  • $begingroup$
    @gimusi Oops! The denominator of the last term in the expression of the limit should be $mn$ instead of $m+n$.
    $endgroup$
    – Hussain-Alqatari
    Dec 4 '18 at 8:46












  • $begingroup$
    Ah ok! I revise my answer accordingly!
    $endgroup$
    – gimusi
    Dec 4 '18 at 8:47
















3












$begingroup$


For any natural number $m$, $lim_{nrightarrow infty }left ( frac{1}{n+1}+frac{1}{n+2}+frac{1}{n+3}+cdots +frac{1}{mn} right )=ln (m)$.



I tried to prove the statement in the following way.



Proof:



$$lim_{nrightarrow infty }left ( frac{1}{n+1}+frac{1}{n+2}+frac{1}{n+3}+cdots +frac{1}{mn} right )=lim_{nrightarrow infty }sum_{r=1}^{(m-1)n}frac{1}{n+r}$$



Dividing the numerator and the denominator of $frac{1}{n+r}$ by $n$, we get $frac{1/n}{1+r/n}$.



Therefore,



$$lim_{nrightarrow infty }sum_{r=1}^{(m-1)n}frac{1}{n+r}=lim_{nrightarrow infty }frac{1}{n}sum_{r=1}^{(m-1)n}frac{1}{1+r/n}$$



this is a Riemann sum, so replacing $frac{1}{n}$ with $dx$, $frac{r}{n}$ with $x$, and integrating between the limits $x=0$ and $x=m-1$



we get $



$$int_{0}^{m-1}frac{dx}{1+x}=ln(1+x)|_{0}^{m-1}=ln(m)-ln(1)=ln(m)blacksquare$$



Is this a valid way?






integration limits logarithms riemann-integration riemann-sum







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why $$lim_{nrightarrow infty }left ( frac{1}{n+1}+frac{1}{n+2}+frac{1}{n+3}+cdots +frac{1}{m+n} right )=lim_{nrightarrow infty }sum_{r=1}^{(m-1)n}frac{1}{n+r}$$ it should be $$=lim_{nrightarrow infty }sum_{r=1}^{m}frac{1}{n+r}$$
    $endgroup$
    – gimusi
    Dec 4 '18 at 8:38












  • $begingroup$
    @gimusi Oops! The denominator of the last term in the expression of the limit should be $mn$ instead of $m+n$.
    $endgroup$
    – Hussain-Alqatari
    Dec 4 '18 at 8:46












  • $begingroup$
    Ah ok! I revise my answer accordingly!
    $endgroup$
    – gimusi
    Dec 4 '18 at 8:47














3












3








3





$begingroup$


For any natural number $m$, $lim_{nrightarrow infty }left ( frac{1}{n+1}+frac{1}{n+2}+frac{1}{n+3}+cdots +frac{1}{mn} right )=ln (m)$.



I tried to prove the statement in the following way.



Proof:



$$lim_{nrightarrow infty }left ( frac{1}{n+1}+frac{1}{n+2}+frac{1}{n+3}+cdots +frac{1}{mn} right )=lim_{nrightarrow infty }sum_{r=1}^{(m-1)n}frac{1}{n+r}$$



Dividing the numerator and the denominator of $frac{1}{n+r}$ by $n$, we get $frac{1/n}{1+r/n}$.



Therefore,



$$lim_{nrightarrow infty }sum_{r=1}^{(m-1)n}frac{1}{n+r}=lim_{nrightarrow infty }frac{1}{n}sum_{r=1}^{(m-1)n}frac{1}{1+r/n}$$



this is a Riemann sum, so replacing $frac{1}{n}$ with $dx$, $frac{r}{n}$ with $x$, and integrating between the limits $x=0$ and $x=m-1$



we get $



$$int_{0}^{m-1}frac{dx}{1+x}=ln(1+x)|_{0}^{m-1}=ln(m)-ln(1)=ln(m)blacksquare$$



Is this a valid way?






integration limits logarithms riemann-integration riemann-sum







share|cite|improve this question











$endgroup$




For any natural number $m$, $lim_{nrightarrow infty }left ( frac{1}{n+1}+frac{1}{n+2}+frac{1}{n+3}+cdots +frac{1}{mn} right )=ln (m)$.



I tried to prove the statement in the following way.



Proof:



$$lim_{nrightarrow infty }left ( frac{1}{n+1}+frac{1}{n+2}+frac{1}{n+3}+cdots +frac{1}{mn} right )=lim_{nrightarrow infty }sum_{r=1}^{(m-1)n}frac{1}{n+r}$$



Dividing the numerator and the denominator of $frac{1}{n+r}$ by $n$, we get $frac{1/n}{1+r/n}$.



Therefore,



$$lim_{nrightarrow infty }sum_{r=1}^{(m-1)n}frac{1}{n+r}=lim_{nrightarrow infty }frac{1}{n}sum_{r=1}^{(m-1)n}frac{1}{1+r/n}$$



this is a Riemann sum, so replacing $frac{1}{n}$ with $dx$, $frac{r}{n}$ with $x$, and integrating between the limits $x=0$ and $x=m-1$



we get $



$$int_{0}^{m-1}frac{dx}{1+x}=ln(1+x)|_{0}^{m-1}=ln(m)-ln(1)=ln(m)blacksquare$$



Is this a valid way?






integration limits logarithms riemann-integration riemann-sum




integration limits logarithms riemann-integration riemann-sum






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 10:45









gimusi

92.8k84494




92.8k84494










asked Dec 4 '18 at 8:16









Hussain-AlqatariHussain-Alqatari

3187




3187












  • $begingroup$
    Why $$lim_{nrightarrow infty }left ( frac{1}{n+1}+frac{1}{n+2}+frac{1}{n+3}+cdots +frac{1}{m+n} right )=lim_{nrightarrow infty }sum_{r=1}^{(m-1)n}frac{1}{n+r}$$ it should be $$=lim_{nrightarrow infty }sum_{r=1}^{m}frac{1}{n+r}$$
    $endgroup$
    – gimusi
    Dec 4 '18 at 8:38












  • $begingroup$
    @gimusi Oops! The denominator of the last term in the expression of the limit should be $mn$ instead of $m+n$.
    $endgroup$
    – Hussain-Alqatari
    Dec 4 '18 at 8:46












  • $begingroup$
    Ah ok! I revise my answer accordingly!
    $endgroup$
    – gimusi
    Dec 4 '18 at 8:47


















  • $begingroup$
    Why $$lim_{nrightarrow infty }left ( frac{1}{n+1}+frac{1}{n+2}+frac{1}{n+3}+cdots +frac{1}{m+n} right )=lim_{nrightarrow infty }sum_{r=1}^{(m-1)n}frac{1}{n+r}$$ it should be $$=lim_{nrightarrow infty }sum_{r=1}^{m}frac{1}{n+r}$$
    $endgroup$
    – gimusi
    Dec 4 '18 at 8:38












  • $begingroup$
    @gimusi Oops! The denominator of the last term in the expression of the limit should be $mn$ instead of $m+n$.
    $endgroup$
    – Hussain-Alqatari
    Dec 4 '18 at 8:46












  • $begingroup$
    Ah ok! I revise my answer accordingly!
    $endgroup$
    – gimusi
    Dec 4 '18 at 8:47
















$begingroup$
Why $$lim_{nrightarrow infty }left ( frac{1}{n+1}+frac{1}{n+2}+frac{1}{n+3}+cdots +frac{1}{m+n} right )=lim_{nrightarrow infty }sum_{r=1}^{(m-1)n}frac{1}{n+r}$$ it should be $$=lim_{nrightarrow infty }sum_{r=1}^{m}frac{1}{n+r}$$
$endgroup$
– gimusi
Dec 4 '18 at 8:38






$begingroup$
Why $$lim_{nrightarrow infty }left ( frac{1}{n+1}+frac{1}{n+2}+frac{1}{n+3}+cdots +frac{1}{m+n} right )=lim_{nrightarrow infty }sum_{r=1}^{(m-1)n}frac{1}{n+r}$$ it should be $$=lim_{nrightarrow infty }sum_{r=1}^{m}frac{1}{n+r}$$
$endgroup$
– gimusi
Dec 4 '18 at 8:38














$begingroup$
@gimusi Oops! The denominator of the last term in the expression of the limit should be $mn$ instead of $m+n$.
$endgroup$
– Hussain-Alqatari
Dec 4 '18 at 8:46






$begingroup$
@gimusi Oops! The denominator of the last term in the expression of the limit should be $mn$ instead of $m+n$.
$endgroup$
– Hussain-Alqatari
Dec 4 '18 at 8:46














$begingroup$
Ah ok! I revise my answer accordingly!
$endgroup$
– gimusi
Dec 4 '18 at 8:47




$begingroup$
Ah ok! I revise my answer accordingly!
$endgroup$
– gimusi
Dec 4 '18 at 8:47










2 Answers
2






active

oldest

votes


















1












$begingroup$

Yes your method is correct.



As an alternative by harmonic series



$$sum_{r=1}^{(m-1)n}frac{1}{n+r}=sum_{r=1}^{mn}frac{1}{r}-sum_{r=1}^{n}frac{1}{r}sim ln(mn)-ln(n)=ln m$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you please check again?
    $endgroup$
    – Hussain-Alqatari
    Dec 4 '18 at 8:50










  • $begingroup$
    Thanks dear gimusi
    $endgroup$
    – Hussain-Alqatari
    Dec 5 '18 at 15:25










  • $begingroup$
    @Hussain-Alqatari You are welcome and you did a great work there! Bye
    $endgroup$
    – gimusi
    Dec 5 '18 at 15:27





















2












$begingroup$

A simpler and more direct choice is to write
$$lim_{n to infty} left( frac{1}{n+1} + frac{1}{n+2} + cdots + frac{1}{mn} right) = lim_{n to infty} sum_{r=n+1}^{mn} frac{1}{r} = lim_{n to infty} frac{1}{n} sum_{r=n+1}^{mn} frac{1}{r/n} = int_{x=1}^m frac{1}{x} , dx = log m,$$ but your solution is valid.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Many Thanks heropup!
    $endgroup$
    – Hussain-Alqatari
    Dec 5 '18 at 15:22











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Yes your method is correct.



As an alternative by harmonic series



$$sum_{r=1}^{(m-1)n}frac{1}{n+r}=sum_{r=1}^{mn}frac{1}{r}-sum_{r=1}^{n}frac{1}{r}sim ln(mn)-ln(n)=ln m$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you please check again?
    $endgroup$
    – Hussain-Alqatari
    Dec 4 '18 at 8:50










  • $begingroup$
    Thanks dear gimusi
    $endgroup$
    – Hussain-Alqatari
    Dec 5 '18 at 15:25










  • $begingroup$
    @Hussain-Alqatari You are welcome and you did a great work there! Bye
    $endgroup$
    – gimusi
    Dec 5 '18 at 15:27


















1












$begingroup$

Yes your method is correct.



As an alternative by harmonic series



$$sum_{r=1}^{(m-1)n}frac{1}{n+r}=sum_{r=1}^{mn}frac{1}{r}-sum_{r=1}^{n}frac{1}{r}sim ln(mn)-ln(n)=ln m$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you please check again?
    $endgroup$
    – Hussain-Alqatari
    Dec 4 '18 at 8:50










  • $begingroup$
    Thanks dear gimusi
    $endgroup$
    – Hussain-Alqatari
    Dec 5 '18 at 15:25










  • $begingroup$
    @Hussain-Alqatari You are welcome and you did a great work there! Bye
    $endgroup$
    – gimusi
    Dec 5 '18 at 15:27
















1












1








1





$begingroup$

Yes your method is correct.



As an alternative by harmonic series



$$sum_{r=1}^{(m-1)n}frac{1}{n+r}=sum_{r=1}^{mn}frac{1}{r}-sum_{r=1}^{n}frac{1}{r}sim ln(mn)-ln(n)=ln m$$






share|cite|improve this answer











$endgroup$



Yes your method is correct.



As an alternative by harmonic series



$$sum_{r=1}^{(m-1)n}frac{1}{n+r}=sum_{r=1}^{mn}frac{1}{r}-sum_{r=1}^{n}frac{1}{r}sim ln(mn)-ln(n)=ln m$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 4 '18 at 8:51

























answered Dec 4 '18 at 8:21









gimusigimusi

92.8k84494




92.8k84494












  • $begingroup$
    Can you please check again?
    $endgroup$
    – Hussain-Alqatari
    Dec 4 '18 at 8:50










  • $begingroup$
    Thanks dear gimusi
    $endgroup$
    – Hussain-Alqatari
    Dec 5 '18 at 15:25










  • $begingroup$
    @Hussain-Alqatari You are welcome and you did a great work there! Bye
    $endgroup$
    – gimusi
    Dec 5 '18 at 15:27




















  • $begingroup$
    Can you please check again?
    $endgroup$
    – Hussain-Alqatari
    Dec 4 '18 at 8:50










  • $begingroup$
    Thanks dear gimusi
    $endgroup$
    – Hussain-Alqatari
    Dec 5 '18 at 15:25










  • $begingroup$
    @Hussain-Alqatari You are welcome and you did a great work there! Bye
    $endgroup$
    – gimusi
    Dec 5 '18 at 15:27


















$begingroup$
Can you please check again?
$endgroup$
– Hussain-Alqatari
Dec 4 '18 at 8:50




$begingroup$
Can you please check again?
$endgroup$
– Hussain-Alqatari
Dec 4 '18 at 8:50












$begingroup$
Thanks dear gimusi
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 15:25




$begingroup$
Thanks dear gimusi
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 15:25












$begingroup$
@Hussain-Alqatari You are welcome and you did a great work there! Bye
$endgroup$
– gimusi
Dec 5 '18 at 15:27






$begingroup$
@Hussain-Alqatari You are welcome and you did a great work there! Bye
$endgroup$
– gimusi
Dec 5 '18 at 15:27













2












$begingroup$

A simpler and more direct choice is to write
$$lim_{n to infty} left( frac{1}{n+1} + frac{1}{n+2} + cdots + frac{1}{mn} right) = lim_{n to infty} sum_{r=n+1}^{mn} frac{1}{r} = lim_{n to infty} frac{1}{n} sum_{r=n+1}^{mn} frac{1}{r/n} = int_{x=1}^m frac{1}{x} , dx = log m,$$ but your solution is valid.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Many Thanks heropup!
    $endgroup$
    – Hussain-Alqatari
    Dec 5 '18 at 15:22
















2












$begingroup$

A simpler and more direct choice is to write
$$lim_{n to infty} left( frac{1}{n+1} + frac{1}{n+2} + cdots + frac{1}{mn} right) = lim_{n to infty} sum_{r=n+1}^{mn} frac{1}{r} = lim_{n to infty} frac{1}{n} sum_{r=n+1}^{mn} frac{1}{r/n} = int_{x=1}^m frac{1}{x} , dx = log m,$$ but your solution is valid.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Many Thanks heropup!
    $endgroup$
    – Hussain-Alqatari
    Dec 5 '18 at 15:22














2












2








2





$begingroup$

A simpler and more direct choice is to write
$$lim_{n to infty} left( frac{1}{n+1} + frac{1}{n+2} + cdots + frac{1}{mn} right) = lim_{n to infty} sum_{r=n+1}^{mn} frac{1}{r} = lim_{n to infty} frac{1}{n} sum_{r=n+1}^{mn} frac{1}{r/n} = int_{x=1}^m frac{1}{x} , dx = log m,$$ but your solution is valid.






share|cite|improve this answer









$endgroup$



A simpler and more direct choice is to write
$$lim_{n to infty} left( frac{1}{n+1} + frac{1}{n+2} + cdots + frac{1}{mn} right) = lim_{n to infty} sum_{r=n+1}^{mn} frac{1}{r} = lim_{n to infty} frac{1}{n} sum_{r=n+1}^{mn} frac{1}{r/n} = int_{x=1}^m frac{1}{x} , dx = log m,$$ but your solution is valid.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 10:17









heropupheropup

63k66199




63k66199












  • $begingroup$
    Many Thanks heropup!
    $endgroup$
    – Hussain-Alqatari
    Dec 5 '18 at 15:22


















  • $begingroup$
    Many Thanks heropup!
    $endgroup$
    – Hussain-Alqatari
    Dec 5 '18 at 15:22
















$begingroup$
Many Thanks heropup!
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 15:22




$begingroup$
Many Thanks heropup!
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 15:22


















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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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