Conversion of $int_0^1fraclog( 1+x)xdx$ to summation
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up vote
1
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We know,
$$lim_ntoinftyfrac1nsum_r=1^nfleft(frac rnright)=int_0^1f(x)dx$$
Now,we have $int_0^1fraclog( 1+x)xdx$
So,$f(x)=fraclog (1+x)x$.So we find $f(r/n)$ by replacing $x$ with $r/n$. Then putting in the summation form on LHS.
Converting to summation,
$$lim_ntoinftysum_r=1^nfraclogleft(1+fracrnright)r$$
Now this should evaluate to $1-frac12^2+frac13^3-cdots$.How do I carry on the simplification?
Thanks for any help!
calculus integration limits summation
edited Sep 3 at 11:53
Batominovski
25.7k22881
asked Sep 2 at 7:55
tatan
5,13542355
add a comment |Â
up vote
1
down vote
favorite
We know,
$$lim_ntoinftyfrac1nsum_r=1^nfleft(frac rnright)=int_0^1f(x)dx$$
Now,we have $int_0^1fraclog( 1+x)xdx$
So,$f(x)=fraclog (1+x)x$.So we find $f(r/n)$ by replacing $x$ with $r/n$. Then putting in the summation form on LHS.
Converting to summation,
$$lim_ntoinftysum_r=1^nfraclogleft(1+fracrnright)r$$
Now this should evaluate to $1-frac12^2+frac13^3-cdots$.How do I carry on the simplification?
Thanks for any help!
calculus integration limits summation
edited Sep 3 at 11:53
Batominovski
25.7k22881
asked Sep 2 at 7:55
tatan
5,13542355
Assuming $log(x)$ here is a natural log ( or it can be converted to one by appropriately scaling it). You can use the taylor series approximation of log(1+x) and further for cases where x tends to 0 ( because x = r/n and n tends to infinity) log(1+x) can be approximated to just x by ignoring the higher power terms in the taylor series.
– Aditya
Sep 2 at 8:06
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
We know,
$$lim_ntoinftyfrac1nsum_r=1^nfleft(frac rnright)=int_0^1f(x)dx$$
Now,we have $int_0^1fraclog( 1+x)xdx$
So,$f(x)=fraclog (1+x)x$.So we find $f(r/n)$ by replacing $x$ with $r/n$. Then putting in the summation form on LHS.
Converting to summation,
$$lim_ntoinftysum_r=1^nfraclogleft(1+fracrnright)r$$
Now this should evaluate to $1-frac12^2+frac13^3-cdots$.How do I carry on the simplification?
Thanks for any help!
calculus integration limits summation
edited Sep 3 at 11:53
Batominovski
25.7k22881
asked Sep 2 at 7:55
tatan
5,13542355
We know,
$$lim_ntoinftyfrac1nsum_r=1^nfleft(frac rnright)=int_0^1f(x)dx$$
Now,we have $int_0^1fraclog( 1+x)xdx$
So,$f(x)=fraclog (1+x)x$.So we find $f(r/n)$ by replacing $x$ with $r/n$. Then putting in the summation form on LHS.
Converting to summation,
$$lim_ntoinftysum_r=1^nfraclogleft(1+fracrnright)r$$
Now this should evaluate to $1-frac12^2+frac13^3-cdots$.How do I carry on the simplification?
Thanks for any help!
calculus integration limits summation
calculus integration limits summation
edited Sep 3 at 11:53
Batominovski
25.7k22881
asked Sep 2 at 7:55
tatan
5,13542355
edited Sep 3 at 11:53
Batominovski
25.7k22881
asked Sep 2 at 7:55
tatan
5,13542355
edited Sep 3 at 11:53
Batominovski
25.7k22881
edited Sep 3 at 11:53
Batominovski
25.7k22881
edited Sep 3 at 11:53
Batominovski
25.7k22881
25.7k22881
asked Sep 2 at 7:55
tatan
5,13542355
asked Sep 2 at 7:55
tatan
5,13542355
asked Sep 2 at 7:55
tatan
5,13542355
5,13542355
Assuming $log(x)$ here is a natural log ( or it can be converted to one by appropriately scaling it). You can use the taylor series approximation of log(1+x) and further for cases where x tends to 0 ( because x = r/n and n tends to infinity) log(1+x) can be approximated to just x by ignoring the higher power terms in the taylor series.
– Aditya
Sep 2 at 8:06
add a comment |Â
Assuming $log(x)$ here is a natural log ( or it can be converted to one by appropriately scaling it). You can use the taylor series approximation of log(1+x) and further for cases where x tends to 0 ( because x = r/n and n tends to infinity) log(1+x) can be approximated to just x by ignoring the higher power terms in the taylor series.
– Aditya
Sep 2 at 8:06
Assuming $log(x)$ here is a natural log ( or it can be converted to one by appropriately scaling it). You can use the taylor series approximation of log(1+x) and further for cases where x tends to 0 ( because x = r/n and n tends to infinity) log(1+x) can be approximated to just x by ignoring the higher power terms in the taylor series.
– Aditya
Sep 2 at 8:06
Assuming $log(x)$ here is a natural log ( or it can be converted to one by appropriately scaling it). You can use the taylor series approximation of log(1+x) and further for cases where x tends to 0 ( because x = r/n and n tends to infinity) log(1+x) can be approximated to just x by ignoring the higher power terms in the taylor series.
– Aditya
Sep 2 at 8:06
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
HINT:
Write beginalignfrac1rlogleft(1+frac rnright)&=frac1rleft[frac rn-frac12left(frac rnright)^2+frac13left(frac rnright)^3-cdotsright]\&=frac1n-frac r2n^2+fracr^23n^3-cdotsendalign so beginalignsum_r=1^nfraclogleft(1+fracrnright)r&=1-fracsum r2n^2+fracsum r^23n^3-cdots\&=1-frac12int_0^1r,dr+frac13int_0^1r^2,dr-cdotsendalign as each term is essentially a Riemann sum.
answered Sep 2 at 8:02
TheSimpliFire
10.8k62054
1
Or treat each term $$frac(-1)^k(k+1),n^k+1,sum_r=1^n,r^k$$ as the Riemann sum for $$frac(-1)^k(k+1),int_0^1,x^k,textdx,.$$
– Batominovski
Sep 2 at 8:09
@Batominovski Nice. I'll add this in.
– TheSimpliFire
Sep 2 at 8:10
Can't this be done without using $log(1+x)$ expansion? I was trying $lim_ntoinftysum_r=1^nfraclog(1+frac rn)r=lim_xto0sum_r=1^nfraclog(1+x)xcdotfrac1n$ where $x=frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here?
– tatan
Sep 2 at 8:13
@tatan You can't take limits whilst the term is being multiplied by another in a summation.
– TheSimpliFire
Sep 2 at 8:15
@TheSimpliFire Okay thanks. But why can you please explain?
– tatan
Sep 2 at 8:15
 |Â
show 4 more comments
up vote
1
down vote
HINT
Following the previous suggestion by TheSimpliFire
$$sum_r=1^nfraclogleft(1+fracrnright)r=1-fracsum r2n^2+fracsum r^23n^3-cdotsto1-frac12cdot 2+frac13cdot 3ldots=$$$$=sum_k=0^infty frac(-1)^k(k+1)^2=sum_k=1^infty frac(-1)^k+1k^2=fracpi^212$$
indeed from the well known results
- Different methods to compute $sumlimits_k=1^infty frac1k^2$ (Basel problem)
- Sum of Reciprocals of Squares of Odd Integers
we have
$$sum_k=1^infty frac(-1)^k+1k^2=sum_k=1^infty frac1(2k-1)^2-sum_k=1^infty frac1(2k)^2=sum_k=1^infty frac1(2k-1)^2-frac14sum_k=1^infty frac1k^2=fracpi^28-fracpi^224=fracpi^212$$
answered Sep 2 at 8:13
gimusi
72.2k73888
Can't this be done without using $log(1+x)$ expansion? I was trying $lim_ntoinftysum_r=1^nfraclog(1+frac rn)r=lim_xto0sum_r=1^nfraclog(1+x)xcdotfrac1n$ where $x=frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here?
– tatan
Sep 2 at 8:15
@tatan No we can't since we need all the terms of the expansion and take 1 corresponds to take only one terms which leads to a wrong result.
– gimusi
Sep 2 at 8:31
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
HINT:
Write beginalignfrac1rlogleft(1+frac rnright)&=frac1rleft[frac rn-frac12left(frac rnright)^2+frac13left(frac rnright)^3-cdotsright]\&=frac1n-frac r2n^2+fracr^23n^3-cdotsendalign so beginalignsum_r=1^nfraclogleft(1+fracrnright)r&=1-fracsum r2n^2+fracsum r^23n^3-cdots\&=1-frac12int_0^1r,dr+frac13int_0^1r^2,dr-cdotsendalign as each term is essentially a Riemann sum.
answered Sep 2 at 8:02
TheSimpliFire
10.8k62054
1
Or treat each term $$frac(-1)^k(k+1),n^k+1,sum_r=1^n,r^k$$ as the Riemann sum for $$frac(-1)^k(k+1),int_0^1,x^k,textdx,.$$
– Batominovski
Sep 2 at 8:09
@Batominovski Nice. I'll add this in.
– TheSimpliFire
Sep 2 at 8:10
Can't this be done without using $log(1+x)$ expansion? I was trying $lim_ntoinftysum_r=1^nfraclog(1+frac rn)r=lim_xto0sum_r=1^nfraclog(1+x)xcdotfrac1n$ where $x=frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here?
– tatan
Sep 2 at 8:13
@tatan You can't take limits whilst the term is being multiplied by another in a summation.
– TheSimpliFire
Sep 2 at 8:15
@TheSimpliFire Okay thanks. But why can you please explain?
– tatan
Sep 2 at 8:15
 |Â
show 4 more comments
up vote
3
down vote
accepted
HINT:
Write beginalignfrac1rlogleft(1+frac rnright)&=frac1rleft[frac rn-frac12left(frac rnright)^2+frac13left(frac rnright)^3-cdotsright]\&=frac1n-frac r2n^2+fracr^23n^3-cdotsendalign so beginalignsum_r=1^nfraclogleft(1+fracrnright)r&=1-fracsum r2n^2+fracsum r^23n^3-cdots\&=1-frac12int_0^1r,dr+frac13int_0^1r^2,dr-cdotsendalign as each term is essentially a Riemann sum.
answered Sep 2 at 8:02
TheSimpliFire
10.8k62054
1
Or treat each term $$frac(-1)^k(k+1),n^k+1,sum_r=1^n,r^k$$ as the Riemann sum for $$frac(-1)^k(k+1),int_0^1,x^k,textdx,.$$
– Batominovski
Sep 2 at 8:09
@Batominovski Nice. I'll add this in.
– TheSimpliFire
Sep 2 at 8:10
Can't this be done without using $log(1+x)$ expansion? I was trying $lim_ntoinftysum_r=1^nfraclog(1+frac rn)r=lim_xto0sum_r=1^nfraclog(1+x)xcdotfrac1n$ where $x=frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here?
– tatan
Sep 2 at 8:13
@tatan You can't take limits whilst the term is being multiplied by another in a summation.
– TheSimpliFire
Sep 2 at 8:15
@TheSimpliFire Okay thanks. But why can you please explain?
– tatan
Sep 2 at 8:15
 |Â
show 4 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
HINT:
Write beginalignfrac1rlogleft(1+frac rnright)&=frac1rleft[frac rn-frac12left(frac rnright)^2+frac13left(frac rnright)^3-cdotsright]\&=frac1n-frac r2n^2+fracr^23n^3-cdotsendalign so beginalignsum_r=1^nfraclogleft(1+fracrnright)r&=1-fracsum r2n^2+fracsum r^23n^3-cdots\&=1-frac12int_0^1r,dr+frac13int_0^1r^2,dr-cdotsendalign as each term is essentially a Riemann sum.
answered Sep 2 at 8:02
TheSimpliFire
10.8k62054
HINT:
Write beginalignfrac1rlogleft(1+frac rnright)&=frac1rleft[frac rn-frac12left(frac rnright)^2+frac13left(frac rnright)^3-cdotsright]\&=frac1n-frac r2n^2+fracr^23n^3-cdotsendalign so beginalignsum_r=1^nfraclogleft(1+fracrnright)r&=1-fracsum r2n^2+fracsum r^23n^3-cdots\&=1-frac12int_0^1r,dr+frac13int_0^1r^2,dr-cdotsendalign as each term is essentially a Riemann sum.
answered Sep 2 at 8:02
TheSimpliFire
10.8k62054
edited Sep 2 at 8:11
answered Sep 2 at 8:02
TheSimpliFire
10.8k62054
answered Sep 2 at 8:02
TheSimpliFire
10.8k62054
answered Sep 2 at 8:02
TheSimpliFire
10.8k62054
10.8k62054
1
Or treat each term $$frac(-1)^k(k+1),n^k+1,sum_r=1^n,r^k$$ as the Riemann sum for $$frac(-1)^k(k+1),int_0^1,x^k,textdx,.$$
– Batominovski
Sep 2 at 8:09
@Batominovski Nice. I'll add this in.
– TheSimpliFire
Sep 2 at 8:10
Can't this be done without using $log(1+x)$ expansion? I was trying $lim_ntoinftysum_r=1^nfraclog(1+frac rn)r=lim_xto0sum_r=1^nfraclog(1+x)xcdotfrac1n$ where $x=frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here?
– tatan
Sep 2 at 8:13
@tatan You can't take limits whilst the term is being multiplied by another in a summation.
– TheSimpliFire
Sep 2 at 8:15
@TheSimpliFire Okay thanks. But why can you please explain?
– tatan
Sep 2 at 8:15
 |Â
show 4 more comments
1
Or treat each term $$frac(-1)^k(k+1),n^k+1,sum_r=1^n,r^k$$ as the Riemann sum for $$frac(-1)^k(k+1),int_0^1,x^k,textdx,.$$
– Batominovski
Sep 2 at 8:09
@Batominovski Nice. I'll add this in.
– TheSimpliFire
Sep 2 at 8:10
Can't this be done without using $log(1+x)$ expansion? I was trying $lim_ntoinftysum_r=1^nfraclog(1+frac rn)r=lim_xto0sum_r=1^nfraclog(1+x)xcdotfrac1n$ where $x=frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here?
– tatan
Sep 2 at 8:13
@tatan You can't take limits whilst the term is being multiplied by another in a summation.
– TheSimpliFire
Sep 2 at 8:15
@TheSimpliFire Okay thanks. But why can you please explain?
– tatan
Sep 2 at 8:15
1
1
Or treat each term $$frac(-1)^k(k+1),n^k+1,sum_r=1^n,r^k$$ as the Riemann sum for $$frac(-1)^k(k+1),int_0^1,x^k,textdx,.$$
– Batominovski
Sep 2 at 8:09
Or treat each term $$frac(-1)^k(k+1),n^k+1,sum_r=1^n,r^k$$ as the Riemann sum for $$frac(-1)^k(k+1),int_0^1,x^k,textdx,.$$
– Batominovski
Sep 2 at 8:09
@Batominovski Nice. I'll add this in.
– TheSimpliFire
Sep 2 at 8:10
@Batominovski Nice. I'll add this in.
– TheSimpliFire
Sep 2 at 8:10
Can't this be done without using $log(1+x)$ expansion? I was trying $lim_ntoinftysum_r=1^nfraclog(1+frac rn)r=lim_xto0sum_r=1^nfraclog(1+x)xcdotfrac1n$ where $x=frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here?
– tatan
Sep 2 at 8:13
Can't this be done without using $log(1+x)$ expansion? I was trying $lim_ntoinftysum_r=1^nfraclog(1+frac rn)r=lim_xto0sum_r=1^nfraclog(1+x)xcdotfrac1n$ where $x=frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here?
– tatan
Sep 2 at 8:13
@tatan You can't take limits whilst the term is being multiplied by another in a summation.
– TheSimpliFire
Sep 2 at 8:15
@tatan You can't take limits whilst the term is being multiplied by another in a summation.
– TheSimpliFire
Sep 2 at 8:15
@TheSimpliFire Okay thanks. But why can you please explain?
– tatan
Sep 2 at 8:15
@TheSimpliFire Okay thanks. But why can you please explain?
– tatan
Sep 2 at 8:15
 |Â
show 4 more comments
up vote
1
down vote
HINT
Following the previous suggestion by TheSimpliFire
$$sum_r=1^nfraclogleft(1+fracrnright)r=1-fracsum r2n^2+fracsum r^23n^3-cdotsto1-frac12cdot 2+frac13cdot 3ldots=$$$$=sum_k=0^infty frac(-1)^k(k+1)^2=sum_k=1^infty frac(-1)^k+1k^2=fracpi^212$$
indeed from the well known results
- Different methods to compute $sumlimits_k=1^infty frac1k^2$ (Basel problem)
- Sum of Reciprocals of Squares of Odd Integers
we have
$$sum_k=1^infty frac(-1)^k+1k^2=sum_k=1^infty frac1(2k-1)^2-sum_k=1^infty frac1(2k)^2=sum_k=1^infty frac1(2k-1)^2-frac14sum_k=1^infty frac1k^2=fracpi^28-fracpi^224=fracpi^212$$
answered Sep 2 at 8:13
gimusi
72.2k73888
Can't this be done without using $log(1+x)$ expansion? I was trying $lim_ntoinftysum_r=1^nfraclog(1+frac rn)r=lim_xto0sum_r=1^nfraclog(1+x)xcdotfrac1n$ where $x=frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here?
– tatan
Sep 2 at 8:15
@tatan No we can't since we need all the terms of the expansion and take 1 corresponds to take only one terms which leads to a wrong result.
– gimusi
Sep 2 at 8:31
add a comment |Â
up vote
1
down vote
HINT
Following the previous suggestion by TheSimpliFire
$$sum_r=1^nfraclogleft(1+fracrnright)r=1-fracsum r2n^2+fracsum r^23n^3-cdotsto1-frac12cdot 2+frac13cdot 3ldots=$$$$=sum_k=0^infty frac(-1)^k(k+1)^2=sum_k=1^infty frac(-1)^k+1k^2=fracpi^212$$
indeed from the well known results
- Different methods to compute $sumlimits_k=1^infty frac1k^2$ (Basel problem)
- Sum of Reciprocals of Squares of Odd Integers
we have
$$sum_k=1^infty frac(-1)^k+1k^2=sum_k=1^infty frac1(2k-1)^2-sum_k=1^infty frac1(2k)^2=sum_k=1^infty frac1(2k-1)^2-frac14sum_k=1^infty frac1k^2=fracpi^28-fracpi^224=fracpi^212$$
answered Sep 2 at 8:13
gimusi
72.2k73888
Can't this be done without using $log(1+x)$ expansion? I was trying $lim_ntoinftysum_r=1^nfraclog(1+frac rn)r=lim_xto0sum_r=1^nfraclog(1+x)xcdotfrac1n$ where $x=frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here?
– tatan
Sep 2 at 8:15
@tatan No we can't since we need all the terms of the expansion and take 1 corresponds to take only one terms which leads to a wrong result.
– gimusi
Sep 2 at 8:31
add a comment |Â
up vote
1
down vote
up vote
1
down vote
HINT
Following the previous suggestion by TheSimpliFire
$$sum_r=1^nfraclogleft(1+fracrnright)r=1-fracsum r2n^2+fracsum r^23n^3-cdotsto1-frac12cdot 2+frac13cdot 3ldots=$$$$=sum_k=0^infty frac(-1)^k(k+1)^2=sum_k=1^infty frac(-1)^k+1k^2=fracpi^212$$
indeed from the well known results
- Different methods to compute $sumlimits_k=1^infty frac1k^2$ (Basel problem)
- Sum of Reciprocals of Squares of Odd Integers
we have
$$sum_k=1^infty frac(-1)^k+1k^2=sum_k=1^infty frac1(2k-1)^2-sum_k=1^infty frac1(2k)^2=sum_k=1^infty frac1(2k-1)^2-frac14sum_k=1^infty frac1k^2=fracpi^28-fracpi^224=fracpi^212$$
answered Sep 2 at 8:13
gimusi
72.2k73888
HINT
Following the previous suggestion by TheSimpliFire
$$sum_r=1^nfraclogleft(1+fracrnright)r=1-fracsum r2n^2+fracsum r^23n^3-cdotsto1-frac12cdot 2+frac13cdot 3ldots=$$$$=sum_k=0^infty frac(-1)^k(k+1)^2=sum_k=1^infty frac(-1)^k+1k^2=fracpi^212$$
indeed from the well known results
- Different methods to compute $sumlimits_k=1^infty frac1k^2$ (Basel problem)
- Sum of Reciprocals of Squares of Odd Integers
we have
$$sum_k=1^infty frac(-1)^k+1k^2=sum_k=1^infty frac1(2k-1)^2-sum_k=1^infty frac1(2k)^2=sum_k=1^infty frac1(2k-1)^2-frac14sum_k=1^infty frac1k^2=fracpi^28-fracpi^224=fracpi^212$$
answered Sep 2 at 8:13
gimusi
72.2k73888
edited Sep 2 at 8:27
answered Sep 2 at 8:13
gimusi
72.2k73888
answered Sep 2 at 8:13
gimusi
72.2k73888
answered Sep 2 at 8:13
gimusi
72.2k73888
72.2k73888
Can't this be done without using $log(1+x)$ expansion? I was trying $lim_ntoinftysum_r=1^nfraclog(1+frac rn)r=lim_xto0sum_r=1^nfraclog(1+x)xcdotfrac1n$ where $x=frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here?
– tatan
Sep 2 at 8:15
@tatan No we can't since we need all the terms of the expansion and take 1 corresponds to take only one terms which leads to a wrong result.
– gimusi
Sep 2 at 8:31
add a comment |Â
Can't this be done without using $log(1+x)$ expansion? I was trying $lim_ntoinftysum_r=1^nfraclog(1+frac rn)r=lim_xto0sum_r=1^nfraclog(1+x)xcdotfrac1n$ where $x=frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here?
– tatan
Sep 2 at 8:15
@tatan No we can't since we need all the terms of the expansion and take 1 corresponds to take only one terms which leads to a wrong result.
– gimusi
Sep 2 at 8:31
Can't this be done without using $log(1+x)$ expansion? I was trying $lim_ntoinftysum_r=1^nfraclog(1+frac rn)r=lim_xto0sum_r=1^nfraclog(1+x)xcdotfrac1n$ where $x=frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here?
– tatan
Sep 2 at 8:15
Can't this be done without using $log(1+x)$ expansion? I was trying $lim_ntoinftysum_r=1^nfraclog(1+frac rn)r=lim_xto0sum_r=1^nfraclog(1+x)xcdotfrac1n$ where $x=frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here?
– tatan
Sep 2 at 8:15
@tatan No we can't since we need all the terms of the expansion and take 1 corresponds to take only one terms which leads to a wrong result.
– gimusi
Sep 2 at 8:31
@tatan No we can't since we need all the terms of the expansion and take 1 corresponds to take only one terms which leads to a wrong result.
– gimusi
Sep 2 at 8:31
add a comment |Â
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Assuming $log(x)$ here is a natural log ( or it can be converted to one by appropriately scaling it). You can use the taylor series approximation of log(1+x) and further for cases where x tends to 0 ( because x = r/n and n tends to infinity) log(1+x) can be approximated to just x by ignoring the higher power terms in the taylor series.
– Aditya
Sep 2 at 8:06