A Nice Infinite Sum

🤩 Hello everyone, I'm very excited to bring you SyberMath Shorts!
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡
/ @sybermath
/ @aplusbi
⭐ Join this channel to get access to perks:→ bit.ly/3cBgfR1
My merch → teespring.com/...
Follow me → / sybermath
Subscribe → www.youtube.co...
⭐ Suggest → forms.gle/A5bG...
If you need to post a picture of your solution or idea:
in...
#algebra #counting #geometry #numbertheory #calculus
via @RUclips @Apple @Desmos @NotabilityApp @googledocs @canva
PLAYLISTS 🎵 :
▶ Trigonometry: • Trigonometry
▶ Algebra: • Algebra
▶ Complex Numbers: • Complex Numbers
▶ Calculus: • Calculus
▶ Geometry: • Geometry
▶ Sequences And Series: • Sequences And Series

[@Blaqjaqshellaq]

The solution can also be presented as ln(1-r)^-r.

[@snowmanking120]

no, it can't😀

[@JerrySu-k3n]

thats a really interesting approach

[@grrgrrgrr0202]

You could've gotten rid of the c much more cleanly by integrating over [0, r] instead of taking the indefinite integral.

[@chaosredefined3834]

You had r + r^2 / 2 + r^3 / 3 + ... = -ln(1-r) + c.
Set r = 0. We have 0 = 0 + c. So... c = 0. No limits needed.

[@hazalouldi7130]

good

[@spdas5942]

You made the thing complex by dividing both sides with r. Before division, putting r=0 leads rhs = 0 and lhs = C => C=0. 🎉

[@yurenchu]

Why take the limit of the green expression, when you could just plug in r = 0 into the pink expression before (at 3:14), and determine C from there?

[@shmuelzehavi4940]

That's a very good question !

[@shmuelzehavi4940]

The domain of r may be restricted to: -1 < r < 1.

[@tetsuyaikeda4319]

congrats! your voice has been cured completely.

[@Dae-Ying-Kim12345]

* 安安安 ! *

[@RashmiRay-c1y]

ln(1-r) = -r-r^2/2-r^3/3..... So, the sum is S=-1/r ln(1-r).

[@yurenchu]

_Answer_ : -(1/r)*ln(1-r)
_Calculation_ :
Let f(r) = 1 + r/2 + (r^2)/3 + (r^3)/4 + ...
Let g(r) = r*f(r) = r + (r^2)/2 + (r^3)/3 + (r^4)/4 + ...
==>
g'(r) = 1 + r + r^2 + r^3 + ...
= 1/(1 - r)
==>
g(r) = ∫ 1/(1-t) dt , from t = 0 to t = r
= [ -ln(1-r) ] - [ -ln(1-0) ]
= - ln(1-r) + ln(1)
= - ln(1-r)
g(r) = r*f(r) ==>
-ln(1-r) = r*f(r)
f(r) = -(1/r)*ln(1-r)
Check:
f(0) = 1 + 0/2 + (0^2)/3 + (0^3)/4 + ... = 1
lim_{r--> 0} f(r) =
= lim_{r--> 0} [ -(1/r)*ln(1-r) ]
... substitute ln(1-r) = x , r = 1 - (e^x) ...
= lim_{x--> 0} [ -(1/(1-e^x))*x ]
= lim_{x--> 0} [ (1/(e^x - 1))*x ]
= lim_{x--> 0} [ x/(e^x - 1) ]
= lim_{x--> 0} [ (x-0)/(e^x - 1) ]
= 1/h'(0) , where h(x) = e^x
... Note: h'(x) = [e^x]' = e^x ...
= 1/(e^0)
= 1/1
= 1
==> checks out!