You could also just notice that the expansion of e^x is summation(n=0 to infinity) (x^n/n!) , just multiply both sides with x which leads to an (n+1) in the power, and taking an integral from 0 to 1 on both sides yeilds us the n+2 term in the denominator giving us the answer 1 for the required summation but notice that we need to minus the n=0 term aswell from both sides which will yield us the required answer (1/2)
@@devilcreedgamer2291 Coaching material... that's interesting. If you email me the photocopy or captured image of that material, I will take a closer look to see if I can solve it. (cornerstonesofmath@gmail.com)
Hi, if I understood it right, I guess you can do it like that: first, notice that 1/n(n+1)(n+1)! = (1/n - 1/(n+1))1/(n+1)! so the sum becomes (1/1 - 1/2) 1/2! + (1/2 - 1/3) 1/3! + (1/3 - 1/4) 1/4! + (1/4 - 1/5) 1/5! + ... Rearranging, 1/2! - (1/2! - 1/3!) 1/2 - (1/3! - 1/4!) 1/3 - (1/4! - 1/5!) 1/4 - ... meaning the sum is equal to 1/2! - Sum (1/n! - 1/(n+1)!) 1/n with n = 2 to ∞. But we can simplify the term (1/n! - 1/(n+1)!) 1/n = n/(n+1)! 1/n = 1/(n+1)! Now, it is done: 1/2 - Sum 1/(n+1)! = 1/2 - (e - 1/0! - 1/1! - 1/2!) = 1/2 - (e - 1 - 1 - 1/2) = 3 - e You just have to write it properly.
Very nice video.
This person make it so easy I don't believe this to simple thanks sir🎉🎉🎉❤.
Thank you!❤️
I didn't see a single video that explains this in Spanish. I appreciate it very much, greetings.
As a non-native English speaker, I understand your struggle. Thank you for your kind comment.
@@CornerstonesOfMath Bro, thanks to you I got the best mark. (10/10)
@@centella8 That's great!💯 Glad that I could be of help.
You could also just notice that the expansion of e^x is summation(n=0 to infinity) (x^n/n!) , just multiply both sides with x which leads to an (n+1) in the power, and taking an integral from 0 to 1 on both sides yeilds us the n+2 term in the denominator giving us the answer 1 for the required summation but notice that we need to minus the n=0 term aswell from both sides which will yield us the required answer (1/2)
You're right! Always glad to see different methods for one problem:)
Hlo sir can u help me with
Σ n=1 to ∞, 1/ n(n+1).(n+1)!
It seems like no one (including internet) has succeeded to calculate its sum algebraically. Is it from the published book or the problem set?
@@CornerstonesOfMath it is from coaching material
@@CornerstonesOfMath sir if I can get your contact i can send you the question any contact instagram or telegram or WhatsApp
@@devilcreedgamer2291 Coaching material... that's interesting. If you email me the photocopy or captured image of that material, I will take a closer look to see if I can solve it. (cornerstonesofmath@gmail.com)
Hi, if I understood it right, I guess you can do it like that:
first, notice that
1/n(n+1)(n+1)!
= (1/n - 1/(n+1))1/(n+1)!
so the sum becomes
(1/1 - 1/2) 1/2!
+ (1/2 - 1/3) 1/3!
+ (1/3 - 1/4) 1/4!
+ (1/4 - 1/5) 1/5!
+ ...
Rearranging,
1/2!
- (1/2! - 1/3!) 1/2
- (1/3! - 1/4!) 1/3
- (1/4! - 1/5!) 1/4
- ...
meaning the sum is equal to
1/2! - Sum (1/n! - 1/(n+1)!) 1/n
with n = 2 to ∞. But we can simplify the term
(1/n! - 1/(n+1)!) 1/n
= n/(n+1)! 1/n
= 1/(n+1)!
Now, it is done:
1/2 - Sum 1/(n+1)!
= 1/2 - (e - 1/0! - 1/1! - 1/2!)
= 1/2 - (e - 1 - 1 - 1/2)
= 3 - e
You just have to write it properly.