Sir, your solutions amaze me and inspire me that even toughest problems have easiest solutions, be it your maths problem or your life. We just need to think outside the box to solve it.
You solution is elegant but the 1/5! term should be 1/4!. You can also solve this by proving the partial sum of n terms is [(n+1)!-1]/(n+1)! (so there is a pattern to the partial sums). Easy to prove by mathematical induction . Then the answer is easily shown to be 1 if you take the limit as n approaches infinity.
I did nearly the same - but with a little bit of dry and error :-) : after calculating the first 3 elements the pattern was clear, then prove by induction. with ((n+1)!-1)/(n+1)! = (n+1)!/(n+1)! - 1/(n+1)! = 1 - 1/(n+1)! the result was clear
What I love about your videos vs. Other similar programs is your clarity of thought and demonstration of process. I am not a simpleton ie I have 3 graduate degrees and an intense job that I love. Btw, not much math with any of my work. I appreciate everyone's comments but, really, your systematic approach to math and your demystification of the process would make more people fans of math. We need more math teachers like you because the way you explain things demonstrates how brilliant you are because you know your material well. We don't need bells, whistles, fluff....that is why many people get turned off to math when it actually is a lot of fun. You provide a great service and thank you.
One of the typical methods for finding the sum of an infinite series is to find the function F(n) which meets S=Σ{F(n+1)-F(n)} and this problem is solved by using this method.
Hello, Would you please explain what happened at the time, 4:39? Is it not supposed to be +1÷4! instead of +1÷5!? Again, thank you for your videos; they are enjoyable with your voice and explanations. Sincerely,
I added the first 5 fractions together to see a pattern starting with 1/2! = 1/2 and 1/2! + 1/3! = 5/6 + small fraction. 1/3! + 1/4! = 23/24 + another small fraction. 1/4! + 1/5! = 119/120 = another small fraction. After doing that exercise, I reached the conclusion that the number is always very, very close to 1, so I guessed the answer was actually 1. Yay! Is that the brute force method?
I solved differently by noticing the gşven expression is derivative of (e^x - 1) / x with respect to x and evaluated at x=1 after derivative is taken. I also used well-known Taylor expansion of e^x = 1 + (x/1!) + (x^2/2!) + (x^3/3!) + ... to prove expression given matches the derivative term I have given above for x=1.. Same result, I felt my alternative is quicker.
For people who like mathematics and wanna pursue something related to math( charted accountant, bank work.. Etc..) Congrats cuz I am also aiming for iit jee and want tough sums like these
You have to imply Taylor series here. I also thought about derivative or something like that. e^x=1+x+x^2/2!+x^3/3!+… e^x/x=1/x+1+x/2!+x^2/3!+… Take derivative from both sides. (1-x)e^x/x^2=-1/x^2+1/2!+2x/3!+… put x=1 0=-1+1/2!+2/3!+3/4!+… Thus,the sum=1 I’m not good at English. Just trying to explain
@@Steven-ov4no Oh right, so that's why it was so familiar to me! It's on the syllabus of my Maths class but I haven't delved in it as of yet. It's high time that I learn it now! Thanks a lot!!
Alternative way would be to evaluate first few partial sums. One would see that they form a sequence n/(n+1). That can be easily proved by the induction method. And taking its limit to infinity yields one.
but we can se that if we split it there also will be 1/n! on the end (which is equal to 0 when we have lim -> infinity but it should be written i suppose ;) )
There is no "final fraction", as the elimination of terms goes on indefinetly. Where would you end an infinite series? Thus, the only term staying ins the 2/2! or 1.
approximately 1 (n!-1)/n! = n!(1-1/n!)/n! = 1-1/n!. The sum of the two first term will = 3!-1/3! or 5/6, the third 4!-1/4! or 23/24 and so on The sum will be close to numerical value on 1-1/n!, so as n becomes larger n factorial becomes larger hence 1/n! becomes smaller hence value becomes close to 1
As much as I love math, that literally makes no sense to me. Is the main difference the explanation points ? Because other than that I'm confused. Even while thinking outside the box
Your solution is attractive but my solution is that : i proved that 1/2! + 2/3! + 3/4! +... + n/(n+1)! = ((n+1)!-1)/(n+1)!. if we define k = (n+1)! we have 1/2! + 2/3! + 3/4! +... + n/(n+1)! = (k-1 )/ k. Then we define sum of statement such as s(n) =1/2! + 2/3! + 3/4! +... + n/(n+1)!. so by limit n to infinity s(n) equivalents limit k to infinity (k-1)/k that equivalent 1.
I had to write a computer program for this way back in 1976 using Fortran I wrote the program then I have to transfer the information to IBM 360 Punch cards The program ran on a mainframe computer Me Kept the running total and print out everything 1000 times we had a 100 answers
Very nice! It is difficult to calculate the value. We know the Maclaurin series of e^x e^x = 1 + x + x^2/2! + x^3/3! + … . This is not logical. But formally we have e^x - 1 = x + x^2/2! + x^3/3! + …, ( e^x - 1 )/x = 1 + x/2! + x^2/3! + x^3/4! + … . Taking derivative of above we have ( e^x・x - ( e^x - 1 )・1 )/x^2 = 1/2! + 2x/3! + 3x^2/4! + … ( ( x - 1 )・e^x + 1 )/x^2 = 1/2! + 2x/3! + 3x^2/4! + … Applying x = 1 we have 1/2! + 2/3! + 3/4! + … = 1.
Unfortunately there is a mistake in the solution. Of course your method is right but you forgot that from the last component there will be 1/n! left finally. It is easy to observe when you finish on 4, that you have crossed out 1/2!s and 1/3!s but you will be left with 1/4!. So the final correct solution is 1 - 1/n! But not 1 only.
Yes, but the sum is supposed to be infinite, with n--> infty so what we are looking for is the limit. But yes, if you only compute the sum of the first n numbers there is a residual 1/n!
Very nice sir (madam?). And I am very sorry if I offended you in any manner if I got your personal pronoun wrong. You should include your personal pronouns in the description of your video so as to not have this problem in the future. Thank you person.
Excuse me, I think your conclusion that all the fractions other than the first fraction of 2/2! are cancelled out is false; and therefore your answer is also false. Could you check please?
Your explanation is very wrong in this case. You can't reduce telescopic the terms of an infinite series. Let's take this example: 1-1+1-1+1-1.... From your explanation is 0. But is not, is a divergent series, becouse there are two partial sums, one is 0 and other is 1. In fact, here we must consider the partial sum which, in this case, after we reduce the terms telescopic, it finished with the negative term. So, not even in the partial sum the terms are not reduced all beginning with the second term. The last one is the minus one and remain. So, the partial sum is 2/2! - 1/(n+1)!. And becouse the limit of 1/(n+1)! Is zero, and becouse an infinite series is convergent if and only if his partial sum is convergent, it result the series is limit (2/2!-1/(n+1)!) which is 2/2!.
Sir, your solutions amaze me and inspire me that even toughest problems have easiest solutions, be it your maths problem or your life. We just need to think outside the box to solve it.
Wow, thanks
Glad you enjoyed it!
Thank you for your amazing feedback! Cheers!
You are awesome, Jindal. Keep it up 😀
Love and prayers from the USA!
@Bhavya Jindal yep you r right
@@PreMath What is the value of infint series n/(1*3*5*...*(2n+1))
You solution is elegant but the 1/5! term should be 1/4!. You can also solve this by proving the partial sum of n terms is [(n+1)!-1]/(n+1)! (so there is a pattern to the partial sums). Easy to prove by mathematical induction . Then the answer is easily shown to be 1 if you take the limit as n approaches infinity.
I did nearly the same - but with a little bit of dry and error :-) : after calculating the first 3 elements the pattern was clear, then prove by induction.
with ((n+1)!-1)/(n+1)! = (n+1)!/(n+1)! - 1/(n+1)! = 1 - 1/(n+1)! the result was clear
S
What I love about your videos vs. Other similar programs is your clarity of thought and demonstration of process. I am not a simpleton ie I have 3 graduate degrees and an intense job that I love. Btw, not much math with any of my work. I appreciate everyone's comments but, really, your systematic approach to math and your demystification of the process would make more people fans of math. We need more math teachers like you because the way you explain things demonstrates how brilliant you are because you know your material well. We don't need bells, whistles, fluff....that is why many people get turned off to math when it actually is a lot of fun. You provide a great service and thank you.
Love your patience in explaining each bit slowly.you are awesome sir❤️❤️
This is old classic question! Thanks!
Bro. Your hint to the solution of the problem is Amazing. Wellcome.
How amazing!!! Thank you.
Solução perfeita. Parabéns professor!
One of the typical methods for finding the sum of an infinite series is to find the function F(n) which meets
S=Σ{F(n+1)-F(n)}
and this problem is solved by using this method.
What an incredible method. 😮😮😮😮
Way of teaching is very excellent
Hello,
Would you please explain what happened at the time, 4:39? Is it not supposed to be +1÷4! instead of +1÷5!?
Again, thank you for your videos; they are enjoyable with your voice and explanations.
Sincerely,
The eye is blurry - in the solution, it's as if it were 1/4 :D
I have known many type of math and solve.I am grateful to you.
So good explain , I try to know more .
Excellent solution for the telescopic sum.
Amazing sir. Very good analysis &good example 🙏🙏🙏🇪🇬
What a beautiful solution! Thank you sir
Thanks sir 😊😊
Another great video👍
Thank you so much for sharing😊😊
thankx sir.
Very nicely solved.
very nice solution
thank u so much i love u'r channel and u'r videos too 😁 its the best math channel ever keep rocking 🙃
Thank you sir for this amazing solution
Very nice thank you
you explained the solution to this problem very well, your step-by-step tutorial is amazing
I added the first 5 fractions together to see a pattern starting with 1/2! = 1/2 and 1/2! + 1/3! = 5/6 + small fraction. 1/3! + 1/4! = 23/24 + another small fraction. 1/4! + 1/5! = 119/120 = another small fraction. After doing that exercise, I reached the conclusion that the number is always very, very close to 1, so I guessed the answer was actually 1. Yay! Is that the brute force method?
Nice solution
I solved differently by noticing the gşven expression is derivative of
(e^x - 1) / x with respect to x and evaluated at x=1 after derivative is taken. I also used well-known Taylor expansion of e^x = 1 + (x/1!) + (x^2/2!) + (x^3/3!) + ... to prove expression given matches the derivative term I have given above for x=1..
Same result, I felt my alternative is quicker.
Yes this is the generating function approach. Definitely to be preferred
OncE again the question of level is also GooD!!
Really. math is amazing!!!
Great
Thank you
fantastic
For people who like mathematics and wanna pursue something related to math( charted accountant, bank work.. Etc..) Congrats cuz I am also aiming for iit jee and want tough sums like these
Is there no way to use some limit or derivative here? My first intuition was 1 regarding the solution but I couldn't find the way to get there.
You have to imply Taylor series here.
I also thought about derivative or something like that.
e^x=1+x+x^2/2!+x^3/3!+…
e^x/x=1/x+1+x/2!+x^2/3!+…
Take derivative from both sides.
(1-x)e^x/x^2=-1/x^2+1/2!+2x/3!+…
put x=1
0=-1+1/2!+2/3!+3/4!+…
Thus,the sum=1
I’m not good at English.
Just trying to explain
@@Steven-ov4no Oh right, so that's why it was so familiar to me! It's on the syllabus of my Maths class but I haven't delved in it as of yet. It's high time that I learn it now! Thanks a lot!!
Thanks for video. Good luck sir!!!!
You're welcome!
So nice of you
Thank you for your feedback! Cheers!
You are awesome. Keep it up 😀
Sir, super problem! But I have failed to solve it. Please give this types of problems. I love your problems.
Yeah!! Did it myself 🙌❤
Yes, there's an error @5:18 minute, since the denominators are not equal. But the result seems to me correct
That's quite unexpected and genious. :)
Is the answer the exact value? or just a value that is very close
hah! You took an infinite sum and made it 'sum' what bigger, allowing cancelling, pretty cool!
Great video!
Glad you enjoyed it!
Thank you for your feedback! Cheers!
You are awesome, Tobias. Keep it up 😀
What is the value of infint series n/(1*3*5*...*(2n+1))
Simply outstanding explanation
Glad you enjoyed it!
Thank you for your feedback! Cheers!
You are awesome, Ramani. Keep it up 😀
Alternative way would be to evaluate first few partial sums. One would see that they form a sequence n/(n+1). That can be easily proved by the induction method. And taking its limit to infinity yields one.
Once again, I make sure that the most difficult tasks have the simplest answer! Thank you so much, sir!
You're welcome!
Thank you for your feedback! Cheers!
You are awesome, Anatoliy. Keep it up 😀
Love and prayers from the USA! Peace!
Great video.
Glad you enjoyed it!
Thank you for your feedback! Cheers!
You are awesome. Keep it up 😀
Term n=n/(n+1)! = (n+1) - 1 divided by (n+1)! That gives term n = 1/n! - 1/ (n+1)! That gives sum of n terms = 1- (1/n+1!). Then sum to infinity =1.
Sir try calculus and limits sums plz. ❤️
Wow nice question thanks
but we can se that if we split it there also will be 1/n! on the end (which is equal to 0 when we have lim -> infinity but it should be written i suppose ;) )
I used a series method by using s(1), s(2)... and then found out the formula for s(n). Then just put a limit on a formula and the answer is clear
This question is in my school's maths book
I know the final fraction will eventually be so small it's negligible but still feels wrong saying it equals one...
There is no "final fraction", as the elimination of terms goes on indefinetly. Where would you end an infinite series? Thus, the only term staying ins the 2/2! or 1.
V nice vquick way of solving
So nice of you, Niru
Thank you for your feedback! Cheers!
You are the best. Keep it up 😀
Thank you, it's very amazing for me.
You're welcome!
Glad you enjoyed it!
Thank you for your feedback! Cheers!
You are awesome, Panya. Keep it up 😀
You can consider e^x = 1 + x + x^2/2! +...
Consider d/dx((e^x - 1)/x)
Then evaluate the derivative at x = 1.
approximately 1
(n!-1)/n! = n!(1-1/n!)/n! = 1-1/n!. The sum of the two first term will = 3!-1/3! or 5/6, the third 4!-1/4! or 23/24 and so on The sum will be close to numerical value on 1-1/n!, so as n becomes larger n factorial becomes larger hence 1/n! becomes smaller hence value becomes close to 1
This is a simple jee main level question but your solution is nice
[Σ(n=1->infinity)]n/(1+n)!
Sir can you prove the following equation in the next video?
(1-x)^-½=1+1/2x+3/8x²+5/16x³ .....................infinite
please give the solution.
Super sir
how does -1/4! +1/5! cancel out??
Very nice problem
Super, Mahalakshmi
Thank you for your feedback! Cheers!
You are the best. Keep it up 😀
4:59
It isn't 1/5! ,but 1/4!
Ans : 1
Mostly Hard Factorial question, though you made it Easy 😃😀
Glad you think so!
Thank you for your feedback! Cheers!
You are awesome, Atharv. Keep it up 😀
sorry, but I don't understand how simplify (- 1/4!) in black by (1/5!) in red in the fourth line
solved very easily
As much as I love math, that literally makes no sense to me. Is the main difference the explanation points ? Because other than that I'm confused. Even while thinking outside the box
Your solution is attractive but my solution is that : i proved that 1/2! + 2/3! + 3/4! +... + n/(n+1)! = ((n+1)!-1)/(n+1)!. if we define k = (n+1)! we have 1/2! + 2/3! + 3/4! +... + n/(n+1)! = (k-1 )/ k. Then we define sum of statement such as s(n) =1/2! + 2/3! + 3/4! +... + n/(n+1)!. so by limit n to infinity s(n) equivalents limit k to infinity (k-1)/k that equivalent 1.
Very smart solution,
But the correct answer is 1-(1/(n+1)!)
I got the same as you but if it is an infinite series the lim1-(1/(n+1)! As n tends to infinity=1
I think the solution is 1 - ( 1/n!), because the last term is (n-1)/n! . That is to say, n/n! -1/n!. The part n/n! is eliminated but -1/n!, doesnt.
Mr premath doesn't wrote it but we are adding this sume to infinity so we have lim -> infinity. From this we have that 1/n! = 0 but it isn't written)
@@maxstrozyk7805 You are right. I didnt think about limits.
My gut instinct told me the answer would be some number infinitesimally smaller than 1, rather than 1, but I now understand where I went wrong!
one thing
series should be odd numbers
👌👌👌👌👌👌👌👌👌👌👌👌👌👌👌👌👌👌👌👌👌👌👌
I just think if base is 1/2! + 2/3! + 3/4! + ....n/(n+1)! => 1 - 1/(n+1)! And n is number of term. Sr my english bad
1/5! Should be 1/4!
It is a typo error. He meant 1/4!, but hit the 5 on the keyboard instead of 4.
I had to write a computer program for this way back in 1976 using Fortran I wrote the program then I have to transfer the information to IBM 360 Punch cards The program ran on a mainframe computer Me
Kept the running total and print out everything 1000 times we had a 100 answers
FRACTION 分數
Very nice! It is difficult to calculate the value.
We know the Maclaurin series of e^x
e^x = 1 + x + x^2/2! + x^3/3! + … .
This is not logical. But formally we have
e^x - 1 = x + x^2/2! + x^3/3! + …,
( e^x - 1 )/x = 1 + x/2! + x^2/3! + x^3/4! + … .
Taking derivative of above we have
( e^x・x - ( e^x - 1 )・1 )/x^2 = 1/2! + 2x/3! + 3x^2/4! + …
( ( x - 1 )・e^x + 1 )/x^2 = 1/2! + 2x/3! + 3x^2/4! + …
Applying x = 1 we have 1/2! + 2/3! + 3/4! + … = 1.
Thank you for your feedback! Cheers!
You are awesome. Keep it up 😀
1⁷+2⁷+3⁷+5⁷+...+n⁷=?
give the formula with prove plz
FACTORIAL 階乘
Hello
It will be 1/1!-1/(n+1)!
=1-1/(n+1)!
not 1
The limit is 1, but the sum never reaches 1 exactly.
How does 1/4! cancel with 1/5! ? :-)
Just a typo
I tried you initial approach that resulted in "no pattern"; and yes, that didn't go anywhere.🙃
Unfortunately there is a mistake in the solution. Of course your method is right but you forgot that from the last component there will be 1/n! left finally. It is easy to observe when you finish on 4, that you have crossed out 1/2!s and 1/3!s but you will be left with 1/4!. So the final correct solution is 1 - 1/n! But not 1 only.
This problem means
lim_{ n → ∞ }( 1 - 1/n! ).
So the answer 1 is correct. You know this is infinite series.
Yes, but the sum is supposed to be infinite, with n--> infty so what we are looking for is the limit. But yes, if you only compute the sum of the first n numbers there is a residual 1/n!
Very nice sir (madam?). And I am very sorry if I offended you in any manner if I got your personal pronoun wrong. You should include your personal pronouns in the description of your video so as to not have this problem in the future. Thank you person.
Excuse me, I think your conclusion that all the fractions other than the first fraction of 2/2! are cancelled out is false; and therefore your answer is also false. Could you check please?
Instead of 1÷5! It should be 1÷4!
I think U are wrong, a little. 🤔
That is not 1/5!, but 1/4! and then there is still -1/(n+1)!
Result is = 1 - 1/(n+1)! = 1.👍
-1/4! + 1/5! = 0 ?
It's a mistake, there should be -1/4! + 1/4! which is 0.
Your explanation is very wrong in this case. You can't reduce telescopic the terms of an infinite series.
Let's take this example:
1-1+1-1+1-1....
From your explanation is 0. But is not, is a divergent series, becouse there are two partial sums, one is 0 and other is 1.
In fact, here we must consider the partial sum which, in this case, after we reduce the terms telescopic, it finished with the negative term. So, not even in the partial sum the terms are not reduced all beginning with the second term. The last one is the minus one and remain.
So, the partial sum is 2/2! - 1/(n+1)!.
And becouse the limit of 1/(n+1)! Is zero, and becouse an infinite series is convergent if and only if his partial sum is convergent, it result the series is limit (2/2!-1/(n+1)!) which is 2/2!.
100% don't get it here.
1
Find the Complex Solution of this Equation; X^4+X^3+2X^2+X+1=0.... Sir Kindly Solve this problem... 🥺
Any Body Solve This Problem?
x=±¡,-1+root 3¡/2,-1-root 3¡
(x²+1)(x²+x+1)=0
= 1. Method => (n-1)/n! = (1/(n-1)! - 1/n!) is applied to all terms. As a result the sum will be equal to 1/2! + 1/2!