An Infinite Sum With Factorials 

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 🤠

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.

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

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.

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.

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.

Nice job!

(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

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.

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

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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. 😄

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

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

First comment ❤

Typical client of Generating functions. 3 mins. But then no revenue from RU-vid!