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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1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
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