Another way is to observe that the sum appears to be the Taylor series of the derivative of e^x / x evaluated at x = 1. Long story short: (x - 1) * e^x / x² = -1 / x² + sum [ (n - 1) * x^(n -2) / n! ] with n = 2 ... ∞ and with x = 1: 0 = -1 + sum sum = 1 🤠
Notice that you find the partial sums are S(n) = 1 - 1/(n!) So, as the sum is the limit of the series of the partials sums when n goes to infinity, it's easy to see that S = 1.
You can also solve it with calculus. f(x) = sum from 1 to infinity of nx^(n-1) / (n+1)! f(1) results in the original sum in your question. You can integrate f(x) to get an antiderivative: F(x) = sum from 1 to infinity of x^n / (n+1)! which is 1/x * the sum .. of x^(n+1) / (n+1)! using the power series formula for the exponential function we can write this as: F(x) = 1/x * (e^x - x -1) Differentiate to get back f: f(x) = 1/x^2 * (x*e^x - e^x + 1) and indeed f(1) = 1
I solved it differently: the partial sums have the form S_n = ((n+1)!-1)/(n+1)!; prove that with induction and then take the limit as n approaches infinity.
My first approach was, that if you add (e - 2), you get the Taylor series for (e -1): sum = (e - 1) - (e - 2) = 1 (I used (n+1) / (n+1)! = 1 / n!) The first expression was missing only the 0th term (e^1 - 1), the other both the 0th and 1st term (e^1 - 2).
5:32 the job is almost done here. You've hinted that Un, the sum of the first n terms of the series is equal to Un=1-1/(n+1)! You've calculated it for U1, U2, U3, U4 so it's just a matter of proving it for U(n+1) once it's valid for n. U(n+1)=Un+(n+1)/(n+2)!=1-1/(n+1)!+(n+1)/(n+2)!=1+(n+1-n-2)/(n+2)!=1-1/(n+2)! which is your formula for U(n+1). By recurrence in just 2 lines you have demonstrated that Un=1-1/(n+1)!. The infinite series is the limit of Un when n tends towards infinity, which is 1-0=1.
An alternating series is conditionally convergent if the series obtained by taking absolute values is divergent. If the series that results from taking absolute values converges, the alternating series is absolutely convergent.
There is an easy way to make this long story short : n/(n+1)! = (n+1)/(n+1)! - 1/(n+1)! = 1/n! - 1/(n+1)! . If you now write down the infinite sums you have the difference 0 ,except for the first term .Hence the original sum = 1.
I'd just increase every numerator by 1, which is equivalent to adding e to the sum. Of course, n/n! = 1/(n - 1)!, so the expression is S + e = 1 + e, so S = 1.
wolframalpha actually didn't do anything wrong there. it is you who just assumes after 3 terms that you already know the infinite pattern of the series. that's a classic problem with induction, you just assume something to be right for any n by just checking the first couple of cases. wolframalpha just makes an educated guess that you gave it a rational expression as terms for the series. only with 4 terms it gives up that approach in favor of the gamma function. just like the usual problems like "continue the pattern" have infinitely many solutions since there are infinitely many sequences with the same finite number of starting terms
Sum (n-1)/n! from 1 to infinity =Sum (n-1)/n! from 1 to infinity + e - e =Sum (n-1)/n! from 1 to infinity+Sum 1/n! from 0 to infinity - e =Sum (n-1)/n! from 1 to infinity+Sum 1/n! from 1 to infinity + 1 - e =Sum n/n! from 1 to infinity + 1 - e =Sum 1/(n-1)! from 1 to infinity + 1 - e =Sum 1/n! from 0 to infinity + 1 - e =e+1-e =1
Based on this result, an interesting numbering system can be created. Let's call it "base factorial." Here's how it will work: First recall how our base 10 system works. To the left of the decimal point, the rightmost digit has a weighting factor of 10⁰ = 1. The digit to the left of the rightmost digit has a weighting factor of 10¹, and the weighting factors for each successive digit continue 10², 10³, 10⁴, etc. To the right of the decimal point, the digits have weighting factors of 10⁻¹, 10⁻², 10⁻³, etc. Using only the digits 0, 1, 2, etc. up to 9, this numbering system allows us to express any real number uniquely, so long as we disregard numbers that end in an infinite string of 9s, e.g. 1 = .9999... Generalizing the above, we can create a base b numbering system for any integer b, where the weighting factors are bⁿ for the digits to the left of the decimal point, and b⁻ⁿ for the digits to the right of the decimal point. Using only the digits 0, 1, 2, etc. up to b-1, we can express any real number uniquely, so long as we disregard numbers that end in an infinite string of k's, such as 1 = .kkkk..., where k is the digit equal to b-1. Now getting to base factorial: The weighting factor for the rightmost digit, i.e. the units digit, is 1! The digit to the left of this digit has a weighting factor of 2!, and the weighting factors continue 3!, 4!, 5!, etc. To the right of decimal point, the weighting factors are 1/2!, 1/3!, 1/4!, etc. For the units position, the permissible digits are 0 and 1. For the position to the left of the units position, the permissible digits are 0, 1, and 2. The next position has permissible digits 0, 1, 2, and 3, and this pattern continues, adding a new permissible digit for each new position. The same pattern continues to the right of the decimal point: the first digit can be 0 or 1, the next 0, 1, or 2, then 0, 1, 2, or 3, etc. As with the base b numbering systems, this scheme allows us to represent any real number uniquely, so long as we disregard infinite series such as 1 = .12345... And that's it for base factorial. Counting to 10 would go 1, 10, 11, 20, 21, 100, 101, 110, 111, 120. Considering a few rational numbers, 1/2 = .1, 1/3 = .02, 1/5 = .0104, 1/7 = .003206. In fact, for any rational number, the decimal expansion will terminate. Now, does expressing numbers in this way have any practical value? I can suggest one: In base factorial, e = 10.11111..., the 1s continuing ad infinitum. Since the decimal expansion for e in base factorial does *not* terminate, it follows immediately that e is irrational!
We know 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... converges. Consider that each term in the given series is smaller than the corresponding term in this sum. 1/2 < 1, 1/3 < 1/2, 1/8 < 1/4, 1/30 < 1/4, 1/144 < 1/16, etc. The given terms shrink a lot more than the ones in the sum I gave, so if what I gave converges, the give one must too.
Minor terminology nitpick with the names you assigned around the two minutes mark. My understanding is that the terms of a series are the items separated by the operation (addition in this case) and that a finite number of terms are evaluated to give a partial sum or a cumulative sum. It appears that you referred to partial sums as terms. I suppose that covers the situation when a partial sum is carried along as a term as the later terms of the series are unwound, but the terminology could become confusing. Eschew obfuscation per the old adage. 😄
Another way is to observe that the sum appears to be the Taylor series of the derivative of e^x / x evaluated at x = 1.
Long story short: (x - 1) * e^x / x² = -1 / x² + sum [ (n - 1) * x^(n -2) / n! ] with n = 2 ... ∞
and with x = 1: 0 = -1 + sum
sum = 1 🤠
I only got far enough in my head to be convinced this would work, but I didn't find the index, powers, final value, etc.
This is brilliant
Notice that you find the partial sums are
S(n) = 1 - 1/(n!)
So, as the sum is the limit of the series of the partials sums when n goes to infinity, it's easy to see that S = 1.
You can also solve it with calculus.
f(x) = sum from 1 to infinity of nx^(n-1) / (n+1)!
f(1) results in the original sum in your question.
You can integrate f(x) to get an antiderivative: F(x) = sum from 1 to infinity of x^n / (n+1)!
which is 1/x * the sum .. of x^(n+1) / (n+1)!
using the power series formula for the exponential function we can write this as:
F(x) = 1/x * (e^x - x -1)
Differentiate to get back f: f(x) = 1/x^2 * (x*e^x - e^x + 1)
and indeed f(1) = 1
Wow! This is nice
I solved it differently: the partial sums have the form S_n = ((n+1)!-1)/(n+1)!; prove that with induction and then take the limit as n approaches infinity.
Sum 1 to inf ( n / (n+1)! = (n+1 - 1) / (n+1)! = 1/n! - 1/(n+1)!) = (1/1! + 1/2! + ...) - (1/2! + 1/3! + ...) = (e-1) - (e-2) = 1
e-(e-1)=1
My first approach was, that if you add (e - 2), you get the Taylor series for (e -1): sum = (e - 1) - (e - 2) = 1
(I used (n+1) / (n+1)! = 1 / n!)
The first expression was missing only the 0th term (e^1 - 1), the other both the 0th and 1st term (e^1 - 2).
Consider adding 1/2!+1/3!+1/4!+… = e−2:
S+e−2
= (1/2!+1/2!)+(2/3!+1/3!)+(3/4!+1/4!)+…
= (1+1)/2!+(2+1)/3!+(3+1)/4!+…
= 2/2!+3/3!+4/4!+…
= 1/1!+1/2!+1/3!+…
= e−1. Therefore S = 1.
I do the same.
that's such robust method, how'd you even come up with that? that's one beautiful way to play with all the different tools
Funny, that you can "calculate" this one by induction! Ind.Hyp.: partial sum is 1 - 1/(n+2)!, n = 0, 1, ...
f(x)=x/2!+x^2/3!+x^3/4!+x^4/5!+...=(e^x-1-x)/x
f'(x)=(e^x(x-1)+1)/x^2 ----> f'(1)=1
Leverage formula for "e^1" : e - 2 = 1/2! + 1/3! .. ... So, Sum + (e-2) = 2/2! + 3/3! = 1/1! + 2/2! + ... = e - 1. So, Sum = e - 1 -(e-2) = 1
5:32 the job is almost done here. You've hinted that Un, the sum of the first n terms of the series is equal to Un=1-1/(n+1)! You've calculated it for U1, U2, U3, U4 so it's just a matter of proving it for U(n+1) once it's valid for n. U(n+1)=Un+(n+1)/(n+2)!=1-1/(n+1)!+(n+1)/(n+2)!=1+(n+1-n-2)/(n+2)!=1-1/(n+2)! which is your formula for U(n+1). By recurrence in just 2 lines you have demonstrated that Un=1-1/(n+1)!. The infinite series is the limit of Un when n tends towards infinity, which is 1-0=1.
An alternating series is conditionally convergent if the series obtained by taking absolute values is divergent. If the series that results from taking absolute values converges, the alternating series is absolutely convergent.
Nice job!
There is an easy way to make this long story short : n/(n+1)! = (n+1)/(n+1)! - 1/(n+1)! = 1/n! - 1/(n+1)! . If you now write down
the infinite sums you have the difference 0 ,except for the first term .Hence the original sum = 1.
I'd just increase every numerator by 1, which is equivalent to adding e to the sum. Of course, n/n! = 1/(n - 1)!, so the expression is S + e = 1 + e, so S = 1.
try e^x/x= Sn, then e^x(1/x-q/x^2)=S’n which is our target sum.
(n-1)/n!=1/(n-1)!-1/n!
lim[n→∞](1/2!+2/3!+3/4!+·+(n-1)/n!)
=lim[n→∞](1-1/n!)
=1
😮
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wolframalpha actually didn't do anything wrong there. it is you who just assumes after 3 terms that you already know the infinite pattern of the series. that's a classic problem with induction, you just assume something to be right for any n by just checking the first couple of cases. wolframalpha just makes an educated guess that you gave it a rational expression as terms for the series. only with 4 terms it gives up that approach in favor of the gamma function. just like the usual problems like "continue the pattern" have infinitely many solutions since there are infinitely many sequences with the same finite number of starting terms
❤
Sum (n-1)/n! from 1 to infinity
=Sum (n-1)/n! from 1 to infinity + e - e
=Sum (n-1)/n! from 1 to infinity+Sum 1/n! from 0 to infinity - e
=Sum (n-1)/n! from 1 to infinity+Sum 1/n! from 1 to infinity + 1 - e
=Sum n/n! from 1 to infinity + 1 - e
=Sum 1/(n-1)! from 1 to infinity + 1 - e
=Sum 1/n! from 0 to infinity + 1 - e
=e+1-e
=1
Based on this result, an interesting numbering system can be created. Let's call it "base factorial." Here's how it will work:
First recall how our base 10 system works. To the left of the decimal point, the rightmost digit has a weighting factor of 10⁰ = 1. The digit to the left of the rightmost digit has a weighting factor of 10¹, and the weighting factors for each successive digit continue 10², 10³, 10⁴, etc. To the right of the decimal point, the digits have weighting factors of 10⁻¹, 10⁻², 10⁻³, etc. Using only the digits 0, 1, 2, etc. up to 9, this numbering system allows us to express any real number uniquely, so long as we disregard numbers that end in an infinite string of 9s, e.g. 1 = .9999...
Generalizing the above, we can create a base b numbering system for any integer b, where the weighting factors are bⁿ for the digits to the left of the decimal point, and b⁻ⁿ for the digits to the right of the decimal point. Using only the digits 0, 1, 2, etc. up to b-1, we can express any real number uniquely, so long as we disregard numbers that end in an infinite string of k's, such as 1 = .kkkk..., where k is the digit equal to b-1.
Now getting to base factorial: The weighting factor for the rightmost digit, i.e. the units digit, is 1! The digit to the left of this digit has a weighting factor of 2!, and the weighting factors continue 3!, 4!, 5!, etc. To the right of decimal point, the weighting factors are 1/2!, 1/3!, 1/4!, etc. For the units position, the permissible digits are 0 and 1. For the position to the left of the units position, the permissible digits are 0, 1, and 2. The next position has permissible digits 0, 1, 2, and 3, and this pattern continues, adding a new permissible digit for each new position. The same pattern continues to the right of the decimal point: the first digit can be 0 or 1, the next 0, 1, or 2, then 0, 1, 2, or 3, etc. As with the base b numbering systems, this scheme allows us to represent any real number uniquely, so long as we disregard infinite series such as 1 = .12345...
And that's it for base factorial. Counting to 10 would go 1, 10, 11, 20, 21, 100, 101, 110, 111, 120. Considering a few rational numbers, 1/2 = .1, 1/3 = .02, 1/5 = .0104, 1/7 = .003206. In fact, for any rational number, the decimal expansion will terminate.
Now, does expressing numbers in this way have any practical value? I can suggest one: In base factorial, e = 10.11111..., the 1s continuing ad infinitum. Since the decimal expansion for e in base factorial does *not* terminate, it follows immediately that e is irrational!
1/2! + 2/3! + 3/4! + 4/5! + ... =
= (2-1)/2! + (3-1)/3! + (4-1)/4! + (5-1)/5! + ...
= (2/2! + 3/3! + 4/4! + 5/5! + ...) - (1/2! + 1/3! + 1/4! + 1/5! + ...)
= (1/1! + 1/2! + 1/3! + 1/4! + ...) - [(1/1! + 1/2! + 1/3! + 1/4! + 1/5! + ...) - 1]
= S - [S - 1]
... where S = {Σ 1/(j!) , from j=1 to j=inf} = (e^1 - 1) is finite ...
= S - S + 1
= 1
but what's the true value?converges or diverges
We know 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... converges. Consider that each term in the given series is smaller than the corresponding term in this sum. 1/2 < 1, 1/3 < 1/2, 1/8 < 1/4, 1/30 < 1/4, 1/144 < 1/16, etc. The given terms shrink a lot more than the ones in the sum I gave, so if what I gave converges, the give one must too.
Minor terminology nitpick with the names you assigned around the two minutes mark. My understanding is that the terms of a series are the items separated by the operation (addition in this case) and that a finite number of terms are evaluated to give a partial sum or a cumulative sum. It appears that you referred to partial sums as terms. I suppose that covers the situation when a partial sum is carried along as a term as the later terms of the series are unwound, but the terminology could become confusing. Eschew obfuscation per the old adage. 😄
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