The following problem comes from a 1961 exam set collected by Ed Barbeau of the University of Toronto. The discontinued exams (by 2003) were for 5th year Ontario high school students seeking entrance and scholarships for the second year at a university.
“If sn denotes the sum of the first n natural numbers, find the sum of the infinite series
Unfortunately, the “Grade XIII” exam problem sets were not provided with answers, so I have no confirmation for my result. There may be a cunning way to manipulate the series to get a solution, but I could not see it off-hand. So I employed my tried and true power series approach to get my answer. It turned out to be power series manipulations on steroids, so there must be a simpler solution that does not use calculus. I assume the exams were timed exams, so I am not sure how a harried student could come up with a quick solution. I would appreciate any insights into this.
See Serious Series
(Update 1/18/2021) Another Solution
I received another solution from Evander Tandiarrang (of Hard Geometric Problem fame) from Papua New Guinea, now in 7th grade! It looks correct to me and only relies on the geometric series without explicit calculus. Needless to say, I was nowhere near such mathematical prowess in 7th grade.
Pada Senin, 18 Januari 2021 11.06.52 GMT+9
Good morning and Happy New Year,
Every time I practice with mathematic and science problems, and now I try to solved the Serious Series problem, but may be not correct. Please give me a correction for my solution.
Clearly this mathematics is known. but not that well known.
The name is Arithmetico-Geometric Series according to Wikipedia.
I have written it out nicely:
Sorry I was unable to access your link via Firefox. I looked up Arithmetico-Geometric Series in Wikipedia and their definition said the nth numerator term Sn was part of an arithmetic progression. This meant it grew linearly with n. But in the Serious Series the Sn grows quadratically with n. So I believe it is not an example of an Arithmetico-Geometric Series. Thanks for the comment.
It’s a pity you can’t read the pdf. The link in my post above works for me in Firefox.
The first thing I did was to split the original sum( n(n+1)/(2^n) ) into two sums:
sum( n²/(2^n) ) and sum( n/(2^n) )
It is this second sum which is found on Wikipedia, and Gradshteyn & Ryzhik (6th ed) 0.231
I wrote the second sum up as a problem on the AplusClick site, and there is also a copy of my paper held on that site in the answer: https://aplusclick.org/t.htm?q=13591
I had a look at Evander’s solution. I agree with the first equation for “A”. I agree with the manipulation to give “2A”, There is then a line with a minus sign in brackets suggesting
2A – A = A
But how you get to “A equals …” something starting with 4 and then terms of the sum ( (k+2)/(2^k) ) is unclear. The new expression for A is numerically correct, but it just seems to have been plucked out of the air. Maybe there are missing steps?
It looks to me like Evander subtracts the first term of A, 1, from the first two terms of 2A, 2 + 3 = 5, to get 4; then the second term of A, 3/2, from the third term of 2A, 6/2, to get 3/2; then the third term of A, 6/4, from the fourth term of 2A, 10/4, to get 4/4; and so on. And then the rest follows from basic manipulations of the geometric series. Very imaginative.
Nice! He has: [ n(n+1) – (n-1) n ] / 2 = [ ( n+1) – (n – 1) ] n /2 = [ 2 ] n / 2 = n on top (numerator)
That has removed the squared term, leaving most of the Arithmetico-Geometric series. Incidentally, that part can also be evaluated using Gabriel’s Staircase (Swain, 1994).
Thanks very much for explaining that missing step 🙂