Find explicit formula from partial sum formula.

Find explicit formula from partial sum formula.

[przemysawkwiatkowski]

If we had 4 instead of 3 in the Sn formula...? The resulting a_n would be exactly the same! Two different sums of one sequence?

Of course the difference is hidden in the 1st term. You forgot to mention it... :-(

[doctor]

There is a step missing: the difference between the nth and n-1 th partial sums only produces a_n when n is 2 or bigger. So, for a_1, we need to compute S_1.

[grumpy]

Very insightful. Thank you for the video.

[cleiven]

I was just trying to solve a problem that required this knowledge when you uploaded this video. How convenient! Great explanation!

[willbishop]

I agree with others pointing out the mistake. S_1 should equal a_1, but according to your result, a_1 = -0.5 while S_1 = 2.5.

[nmgggg]

These new thumbnails got me thinking these are 3blue1brown vids in my feed xD Great stuff all the same!!

[maximmorshin]

I have a question:
Why S1 (which is equal to a1) by the given equation is 3-(1/2)=2.5
But by calculating through an we get a1=(1-2)/2=-0.5?

[TrimutiusToo]

Logic works for all but a1 which you need to specify separately as equal to 2.5 because s1 is equal to a1. Your formula would give -0.5 for a1 which is wrong value...

[mekkinoureddine]

Maths are so beautifull with you
thanks for this video

[shpilbass]

OMG you released this just in time, I have my high school math final tomorrow and it includes series

[someperson]

The value of the nth partial sum S_n given isn't possible for any sequence a_n. You correctly showed that if S_n = 3-n/2^n, then a_n = (n-2)/2^n. However, this shows that S_1 = a_1 = -1/2. But the value of S_1given is 5/2.
The problem you probably meant to solve is: Given S_n = c - n/2^n, where c is a constant, find c and find the limit as n approaches infinity of S_n. The same calculation done in the video shows that a_n= (n-2)/2^n. Therefore, c - 1/2 = S_1 = a_1 = -1/2. So, c = 0 (not 3) and S_n = -n/2^n. We show by mathematical induction on n that
(1) if a_n = (n-2)/2^n, then S_n = -n/2^n.
Proof of (1). (1) clearly holds if n = 1. Suppose (1) held for some n = k >=1. Then,
S_(k+1) = S_k + a_(k+1) = -k/2^k + ((k+1)-2)/2^(k+1) = -(k+1)/2^(k+1),
showing that (1) holds for n = k+1.
Finally, the limit as n approaches infinity of S_n = 0.

[jakoblenke]

Heey I‘ve got a question:
When I have a graph and lay a string along the line in a given interval, how long is my string depending on the function and the interval...?

[tanmoydutta]

I wish to let you know that till date, I didn't know how to use the concepts of limits and differential and integral calculus in normal mathematical topics...thanks to blackpenredpen, now that I do

[abdalrhmanhammad]

The divergence test Does not tell you if the series convergence 😅

[funktionbro]

As if you know what I have to do in my Maths test, exactly this is what I have to do this week.

[alexdemoura]

Now I am pretty sure we are watching a "Metaseries" of videos, a series about series... pretty good I may say.

I like it, despite the fact your reasoning was too fast for me, between "Be careful, tho" and "That's the deal". No problem I can reduce YouTube play speed to 0.5x and try again.

[euriskoo]

You have a distance of 20 meter to the equal parts of each section is equal to x so that the first x---->t then the second x then 1 / 4t … and so on, Calculate the total distance in terms of X, and the time took?

[guscox]

Petition to combine blue pen and black pen, and use burgundy pen for a video

[dalek]

The answer should be You've got to make sure a1=s1.

[pierreneau]

You can calculate Sn -Sn-1 only if n > 1 because a0 is not defined so an = (n-2)/2^n if n>1 : should be indicated in the calculation.