Problems in the convergence or divergence of number series
 W I N I O N O W
Mathematical Series
The Contest Center
Wappingers Falls, NY 12590
 W I N I O N O W

Determine whether these series converge or diverge.
We will post the names of anyone who solves them.
Solutions that do not use calculus are preferred.
All logarithms are natural logarithms to the base e=2.71828...

Series #1A
1/(log 2) + 1/(log 2)(log 3) + 1/(log 2)(log 3)(log 4) + ...

Solved by:   William Alber, Arijit Bhattacharyya, Swapnil Jain, Ritwik Chaudhuri, P.M.A. Hakeem, Mu Lu

Series #1B
2/(log 2) + 3/(log 2)(log 3) + 4/(log 2)(log 3)(log 4) + ...

Solved by:   Ritwik Chaudhuri, P.M.A. Hakeem, Mu Lu

Series #2
1/(log 2) + 1/(log 2 + log 3) + 1/(log 2 + log 3 + log 4) + ...

Solved by:   Jean Jacquelin, Swapnil Jain, Mu Lu, Ritwik Chaudhuri

Series #3A
(2/3)2 + (3/4)3 + (4/5)4 + ...

Solved by:   Steven Kennedy, William Alber, Gaurav Agrawal, Arijit Bhattacharyya, Swapnil Jain, Ritwik Chaudhuri, P.M.A. Hakeem, Mu Lu

Series #3B
(2/(2+log 2))2 + (3/(3+log 3))3 + (4/(4+log 4))4 + ...

Solved by:   Jean-Yves Degos, Mu Lu, P.M.A. Hakeem

Series #4
1/(1+log 3/2) + 1/(1+log 3/2)(1+log 4/3) + 1/(1+log 3/2)(1+log 4/3)(1+log 5/4) + ...

Solved by:   Ante Turudic, Mu Lu, P.M.A. Hakeem

Series #5
1/√2 + 1/(√2+√3) + 1/(√2+√3+√4) + ...

Solved by:   Jean Jacquelin, Mu Lu

Series #6A
1/(log 3)log 3 + 1/(log 4)log 4 + 1/(log 5)log 5 + ...

Solved by:   Jean Jacquelin, Ante Turudic, James Herterich, P.M.A. Hakeem, Mu Lu

Series #6B
1/3log log 3 + 1/4log log 4 + 1/5log log 5 + ...

Solved by:   P.M.A. Hakeem, Mu Lu

Series #6C
1/(log 3)log log 3 + 1/(log 4)log log 4 + 1/(log 5)log log 5 + ...

Solved by:   James Herterich, P.M.A. Hakeem, Mu Lu

Series #6D
1/(log log 3)log 3 + 1/(log log 4)log 4 + 1/(log log 5)log 5 + ...

Solved by:   James Herterich, Mu Lu

Series #6E
1/(log log 3)log log 3 + 1/(log log 4)log log 4 + 1/(log log 5)log log 5 + ...

Solved by:   Mu Lu

Series #7
1/(2/(log 2)) + 1/(2/(log 2)+3/(log 3)) + 1/(2/(log 2)+3/(log 3)+4/(log 4)) + ...

Solved by:   Ante Turudic, Mu Lu

Series #8
1/23/2 + 1/34/3 + 1/45/4 + ...

Solved by:   Ante Turudic, Arijit Bhattacharyya, Swapnil Jain, Mu Lu

Series #9
(log 2)/(2 log 2)2 + ((log 2)+(log 3)))/(3 log 3)2 + ((log 2)+(log 3)+(log 4))/(4 log 4)2 + ...

Solved by:   Ante Turudic, Mu Lu

Series #10A
1/R(12) + 1/R(22) + 1/R(32) + ...
Where the function R(x) returns a random integer in the range 1 to x, inclusive, for any positive integer x.

Solved by:   Mu Lu

Series #10B
1/R(1)2 + 1/R(2)2 + 1/R(3)2 + ...
Where the function R(x) returns a random integer in the range 1 to x, inclusive, for any positive integer x.

Solved by:   Mu Lu

Series #11A
1/(log 4/3) + 1/(log 4/3)(log 9/4) + 1/(log 4/3)(log 9/4)(log 16/5) + 1/(log 4/3)(log 9/4)(log 16/5)(log 25/6) + ...

Solved by:   Ante Turudic, Swapnil Jain, Mu Lu, P.M.A. Hakeem

Series #11B
1/(log 4/3) + 1/(log 5/3)(log 6/4) + 1/(log 6/3)(log 7/4)(log 8/5) + 1/(log 7/3)(log 8/4)(log 9/5)(log 10/6) + ...

Solved by:   Mu Lu

Series #12
1/(sin²1)² + 1/(sin²1+sin²2)² + 1/(sin²1+sin²2+sin²3)² + 1/(sin²1+sin²2+sin²3+sin²4)² + ...

Solved by:   Ante Turudic, Mu Lu

Series #13
1/(tan²1) + 1/(tan²1+tan²2) + 1/(tan²1+tan²2+tan²3) + 1/(tan²1+tan²2+tan²3+tan²4) + ...

Solved by:   P.M.A. Hakeem

Series #14
1/(tan21) + 1/(tan42) + 1/(tan63) + 1/(tan84) + ...

Solved by:   Ante Turudic, P.M.A. Hakeem, Mu Lu

Series #15
1/(1+tan²1)2 + 1/(1+tan²2)4 + 1/(1+tan²3)6 + 1/(1+tan²4)8 + ...

Solved by:   Amogh Keni

Series #16
1/(1+abs(tan(1)))2 + 1/(1+abs(tan(2)))4 + 1/(1+abs(tan(3)))6 + 1/(1+abs(tan(4)))8 + ...

Series #17
Let d(n) be the number of divisors of the integer n.
1/d(11) + 1/d(22) + 1/d(33) + 1/d(44) + ...

Solved by:   Mu Lu, P.M.A. Hakeem

Series #18
(sin21)/1 + (sin22)/2 + (sin23)/3 + ...

Solved by:   Mu Lu

Series #19
1/11.1 + 1/21.2 + 1/31.3 + ... + 1/91.9 + 1/101.10 + 1/111.11 + ... + 1/991.99 + 1/1001.100 + 1/1011.101 + ...

Solved by:   Mu Lu, P.M.A. Hakeem

Square Root
√(1+√(4+√(9+√(16+ ... ))))

Solved by:   Mu Lu

The Rubin Number
The Rubin number is the unique positive real value R for which the infinite product
(R + sin 1) (R + sin 2) (R + sin 3) ...
neither collapses to 0 nor explodes to infinity.
Find the Rubin number, and show that if a is any real number and b>0 is any rational number then the infinite product
(R + sin(a+b)) (R + sin(a+2b)) (R + sin(a+3b)) ...
neither collapses to 0 nor explodes to infinity.

Solved by:   Sheldon Degenhardt, Mu Lu

On the Edge
It is well-known that the series
1/1p + 1/2p + 1/3p + ...
1/3(ln 3)p + 1/4(ln 4)p + 1/5(ln 5)p + ...
1/16(ln 16)((ln(ln 16))p + 1/17(ln 17)((ln(ln 17))p + 1/18(ln 18)((ln(ln 18))p + ...
etc. all converge if p>1 and diverge if p<1. The sums are taken from the appropriate starting points to infinity.
In this problem we consider series that lie on the edge between the case where p=1 and these series diverge, and p>1 where these series converge. Specifically, we wish to consider the case where the product of the logarithms becomes arbitrarily deep, (ln x) (ln ln x) (ln ln ln x) ... However, there is a difficulty as the nesting of the logarithms gets deep. Suppose we start with a fairly large number like B=10000000001000000000. Then ln B = 20723265837, ln ln B =23.755, ln ln ln B = 3.1678, ln ln ln ln B = 1.1530, ln ln ln ln ln B = .14239, ln ln ln ln ln ln B = -1.9492, and ln ln ln ln ln ln ln B is undefined, since negative numbers do not have logarithms.
This means that to get beyond 6 levels of logarithms we would have to start at a number much larger than B. Two different approaches will be used to avoid this problem with the starting point. The first approach is to step up the depth of the logarithms very slowly, so slowly that all of the terms in each product are greater than 1. Specifically, we consider the series
1/f(1) + 1/f(2) + 1/f(3) + 1/f(4) + ...
where f(n) is defined by
prod = n
term = n
while term > e do
term = ln(term)
prod = prod * term
end
f(n) = prod
Does this first "edge series" converge or diverge?

The second approach to the starting point difficulty is to substitute the function (1+ln n) for the function (ln n). The function (1+ln n) can be nested arbitrarily deeply when n>1. As before, the series
1/1p + 1/2p + 1/3p + ...
1/1(1+ln 1)p + 1/2(1+ln 2)p + 1/3(1+ln 3)p + ...
1/1(1+ln 1)((1+ln(1+ln 1))p + 1/2(1+ln 2)((1+ln(1+ln 2))p + 1/3(1+ln 3)((1+ln(1+ln 3))p + ...
etc, all converge when p>1 and diverge when p<1. The sums can now be taken from n=1 to infinity.
Now consider the following series which sits on the border between p=1 and p>1.
1/1 + 1/2(1+ln 2) + 1/3(1+ln 3)((1+ln(1+ln 3)) + 1/4(1+ln 4)((1+ln(1+ln 4))(1+ln(1+ln(1+ln 4))) + ...
Does this second "edge series" converge or diverge?

 Send us an email to submit answers to these problems, to ask for help, or to submit new convergence problems. This site grows when you contribute puzzles. Be sure to change the \$ to an @ in our email address.