A Simple Sum of the Series Containing Factorials

In this video we evaluate the sum of series which contains factorial in the denominator. The key here is to change the individual terms so that they are expressed in the form of differences, so that they can be cancelled out when carrying out the summation.
#Sequences #Series
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[@centella8]

I didn't see a single video that explains this in Spanish. I appreciate it very much, greetings.

[@CornerstonesOfMath]

As a non-native English speaker, I understand your struggle. Thank you for your kind comment.

[@centella8]

​@@CornerstonesOfMath Bro, thanks to you I got the best mark. (10/10)

[@CornerstonesOfMath]

@@centella8 That's great!💯 Glad that I could be of help.

[@lakshay3745]

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)

[@CornerstonesOfMath]

You're right! Always glad to see different methods for one problem:)

[@devilcreedgamer2291]

Hlo sir can u help me with
Σ n=1 to ∞, 1/ n(n+1).(n+1)!

[@CornerstonesOfMath]

It seems like no one (including internet) has succeeded to calculate its sum algebraically. Is it from the published book or the problem set?

[@devilcreedgamer2291]

@@CornerstonesOfMath it is from coaching material

[@devilcreedgamer2291]

@@CornerstonesOfMath sir if I can get your contact i can send you the question any contact instagram or telegram or WhatsApp

[@CornerstonesOfMath]

@@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. ([e-mail])

[@samueldeandrade8535]

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.