Solutions to Golomb's Puzzle Column Number 30:

Identities

Solomon W. Golomb

1. From , we see that , for all .

   With , we get
   .

2. If , we know that , since , and .

   With n=5, , where radians = .

   For any t, . In particular, with radians = , .

   Thus, , as required.

   [ Note: Problem 1 can also be approached this way, with .]

3. For positive real numbers A and B, we have .

   Thus, .

4. , which converges because converges.

   Note that , where is the same sequence as , and is the same sequence as , except for the starting points of these sequences. Thus,

   

   since all the other factors cancel. More rigorously, the k partial product is , which goes to as k goes to infinity.

5. Let , which converges by the alternating series test.

   Then

   

   if you remember your trigonometric substitutions from freshman calculus.

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