Closed form for $sum_mgeq 2, ngeq 2frac1m^n-1$
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Is there a nice closed form for $displaystyle sum_substackmgeq 2\ ngeq 2frac1m^n-1$ ?
It's easy to prove that $displaystyle sum_substackmgeq 2\ ngeq 2frac1m^n = 1$ by summing first over $n$ (geometric sum) and then over $m$ (telescopic series).
It can be proven that $ sum_q in Q frac1q-1=1 quad textwhere Q=m^n, ngeq 2, mgeq 2$. A natural extension is to compute $displaystyle sum_mgeq 2, ngeq 2frac1m^n-1$.
Here's what I've noticed:
$$beginalign
sum_substackmgeq 2\ ngeq 2frac1m^n-1&=sum_substackmgeq 2\ ngeq 2frac1/m^n1-1/m^n=sum_substackmgeq 2\ ngeq 2sum_kgeq 1frac1m^nk\
&=sum_mgeq 2sum_kgeq 1frac1m^k(m^k-1) = sum_mgeq 2sum_kgeq 1left(frac 1m^k-1-frac 1m^kright)
endalign$$
but that looks like a dead-end.
According to Mathematica, an approximation of the limit is $1.130396449$.
sequences-and-series summation closed-form
asked Aug 26 at 9:11
Gabriel Romon
17.2k53182
add a comment |Â
up vote
2
down vote
favorite
Is there a nice closed form for $displaystyle sum_substackmgeq 2\ ngeq 2frac1m^n-1$ ?
It's easy to prove that $displaystyle sum_substackmgeq 2\ ngeq 2frac1m^n = 1$ by summing first over $n$ (geometric sum) and then over $m$ (telescopic series).
It can be proven that $ sum_q in Q frac1q-1=1 quad textwhere Q=m^n, ngeq 2, mgeq 2$. A natural extension is to compute $displaystyle sum_mgeq 2, ngeq 2frac1m^n-1$.
Here's what I've noticed:
$$beginalign
sum_substackmgeq 2\ ngeq 2frac1m^n-1&=sum_substackmgeq 2\ ngeq 2frac1/m^n1-1/m^n=sum_substackmgeq 2\ ngeq 2sum_kgeq 1frac1m^nk\
&=sum_mgeq 2sum_kgeq 1frac1m^k(m^k-1) = sum_mgeq 2sum_kgeq 1left(frac 1m^k-1-frac 1m^kright)
endalign$$
but that looks like a dead-end.
According to Mathematica, an approximation of the limit is $1.130396449$.
sequences-and-series summation closed-form
asked Aug 26 at 9:11
Gabriel Romon
17.2k53182
1
Why did you delete math.stackexchange.com/questions/2894258/… ?
– Lord Shark the Unknown
Aug 26 at 9:18
@LordSharktheUnknown the equality I had written is wrong, since $sum _n=2^10 sum _m=2^10 frac1m^n-1$ evaluates to something greater than $1.02$ What's true is $sum_q in Q frac1q-1=1 quad textwhere Q=m^n, ngeq 2, mgeq 2$.
– Gabriel Romon
Aug 26 at 9:28
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is there a nice closed form for $displaystyle sum_substackmgeq 2\ ngeq 2frac1m^n-1$ ?
It's easy to prove that $displaystyle sum_substackmgeq 2\ ngeq 2frac1m^n = 1$ by summing first over $n$ (geometric sum) and then over $m$ (telescopic series).
It can be proven that $ sum_q in Q frac1q-1=1 quad textwhere Q=m^n, ngeq 2, mgeq 2$. A natural extension is to compute $displaystyle sum_mgeq 2, ngeq 2frac1m^n-1$.
Here's what I've noticed:
$$beginalign
sum_substackmgeq 2\ ngeq 2frac1m^n-1&=sum_substackmgeq 2\ ngeq 2frac1/m^n1-1/m^n=sum_substackmgeq 2\ ngeq 2sum_kgeq 1frac1m^nk\
&=sum_mgeq 2sum_kgeq 1frac1m^k(m^k-1) = sum_mgeq 2sum_kgeq 1left(frac 1m^k-1-frac 1m^kright)
endalign$$
but that looks like a dead-end.
According to Mathematica, an approximation of the limit is $1.130396449$.
sequences-and-series summation closed-form
asked Aug 26 at 9:11
Gabriel Romon
17.2k53182
Is there a nice closed form for $displaystyle sum_substackmgeq 2\ ngeq 2frac1m^n-1$ ?
It's easy to prove that $displaystyle sum_substackmgeq 2\ ngeq 2frac1m^n = 1$ by summing first over $n$ (geometric sum) and then over $m$ (telescopic series).
It can be proven that $ sum_q in Q frac1q-1=1 quad textwhere Q=m^n, ngeq 2, mgeq 2$. A natural extension is to compute $displaystyle sum_mgeq 2, ngeq 2frac1m^n-1$.
Here's what I've noticed:
$$beginalign
sum_substackmgeq 2\ ngeq 2frac1m^n-1&=sum_substackmgeq 2\ ngeq 2frac1/m^n1-1/m^n=sum_substackmgeq 2\ ngeq 2sum_kgeq 1frac1m^nk\
&=sum_mgeq 2sum_kgeq 1frac1m^k(m^k-1) = sum_mgeq 2sum_kgeq 1left(frac 1m^k-1-frac 1m^kright)
endalign$$
but that looks like a dead-end.
According to Mathematica, an approximation of the limit is $1.130396449$.
sequences-and-series summation closed-form
asked Aug 26 at 9:11
Gabriel Romon
17.2k53182
edited Aug 26 at 11:18
asked Aug 26 at 9:11
Gabriel Romon
17.2k53182
asked Aug 26 at 9:11
Gabriel Romon
17.2k53182
asked Aug 26 at 9:11
Gabriel Romon
17.2k53182
17.2k53182
1
Why did you delete math.stackexchange.com/questions/2894258/… ?
– Lord Shark the Unknown
Aug 26 at 9:18
@LordSharktheUnknown the equality I had written is wrong, since $sum _n=2^10 sum _m=2^10 frac1m^n-1$ evaluates to something greater than $1.02$ What's true is $sum_q in Q frac1q-1=1 quad textwhere Q=m^n, ngeq 2, mgeq 2$.
– Gabriel Romon
Aug 26 at 9:28
add a comment |Â
1
Why did you delete math.stackexchange.com/questions/2894258/… ?
– Lord Shark the Unknown
Aug 26 at 9:18
@LordSharktheUnknown the equality I had written is wrong, since $sum _n=2^10 sum _m=2^10 frac1m^n-1$ evaluates to something greater than $1.02$ What's true is $sum_q in Q frac1q-1=1 quad textwhere Q=m^n, ngeq 2, mgeq 2$.
– Gabriel Romon
Aug 26 at 9:28
1
1
Why did you delete math.stackexchange.com/questions/2894258/… ?
– Lord Shark the Unknown
Aug 26 at 9:18
Why did you delete math.stackexchange.com/questions/2894258/… ?
– Lord Shark the Unknown
Aug 26 at 9:18
@LordSharktheUnknown the equality I had written is wrong, since $sum _n=2^10 sum _m=2^10 frac1m^n-1$ evaluates to something greater than $1.02$ What's true is $sum_q in Q frac1q-1=1 quad textwhere Q=m^n, ngeq 2, mgeq 2$.
– Gabriel Romon
Aug 26 at 9:28
@LordSharktheUnknown the equality I had written is wrong, since $sum _n=2^10 sum _m=2^10 frac1m^n-1$ evaluates to something greater than $1.02$ What's true is $sum_q in Q frac1q-1=1 quad textwhere Q=m^n, ngeq 2, mgeq 2$.
– Gabriel Romon
Aug 26 at 9:28
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2894847%2fclosed-form-for-sum-m-geq-2-n-geq-2-frac1mn-1%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Why did you delete math.stackexchange.com/questions/2894258/… ?
– Lord Shark the Unknown
Aug 26 at 9:18
@LordSharktheUnknown the equality I had written is wrong, since $sum _n=2^10 sum _m=2^10 frac1m^n-1$ evaluates to something greater than $1.02$ What's true is $sum_q in Q frac1q-1=1 quad textwhere Q=m^n, ngeq 2, mgeq 2$.
– Gabriel Romon
Aug 26 at 9:28