Thoughts on Infinity (Part 3b)

The five most important numbers in mathematics are 0, 1, e, \pi, and i. In sixth place (a distant sixth place) is probably \gamma, the Euler-Mascheroni constant. See Mathworld or Wikipedia for more details. (For example, it’s astounding that we still don’t know if \gamma is irrational or not.)

In yesterday’s post, we’ve seen the curious phenomenon that the commutative and associative laws do not apply to a conditionally convergent series or infinite product. In tomorrow’s post, I’ll present another classic example of this phenomenon due to Cauchy. However, to be ready for this fact, I’ll need to see how \gamma arises from a certain conditionally convergent series.

Separately define the even and odd terms of the sequence \{a_n\} by

a_{2n} = \displaystyle \int_n^{n+1} \frac{dx}{x}

and

a_{2n-1} = \displaystyle \frac{1}{n}.

It’s pretty straightforward to show that this sequence is decreasing. The function f(x) = \displaystyle \frac{1}{x} is clearly decreasing for x > 0, and so the maximum value of f(x) on the interval [n,n+1] must occur at the left endpoint, while the minimum value must occur at the right endpoint. Since the length of this interval is 1, we have

\displaystyle \frac{1}{n+1} \cdot 1 < \displaystyle \int_n^{n+1} \frac{dx}{x} < \displaystyle \frac{1}{n} \cdot 1,

or

a_{2n+1} < a_{2n} < a_{2n-1}.

Since the subsequence \{a_{2n-1}\} clearly decreases to 0, this shows the full sequence \{a_n\} is a decreasing sequence with limit 0.

By the alternating series test, this implies that the series

\displaystyle \sum_{n=1}^\infty (-1)^{n-1} a_n

converges. This limit is called the

Since this series converges, that means that the limit of the partial sums converges to \gamma:

\displaystyle \lim_{M \to \infty} \sum_{n=1}^M (-1)^{n-1} a_n = \gamma.

Let’s take the upper limit to be an odd number M, where M = 2N-1 and N is an integer. Then by separating the even and odd terms, we obtain

\displaystyle \sum_{n=1}^{2N-1} (-1)^{n-1} a_n = \displaystyle \sum_{n=1}^{N} (-1)^{2n-1-1} a_{2n-1} + \sum_{n=1}^{N-1} (-1)^{2n-1} a_{2n}

= \displaystyle \sum_{n=1}^N a_{2n-1} - \sum_{n=1}^{N-1} a_{2n}

= \displaystyle \sum_{n=1}^N \frac{1}{n} - \sum_{n=1}^{N-1} \int_n^{n+1} \frac{dx}{x}

= \displaystyle \sum_{n=1}^N \frac{1}{n} - \int_1^N \frac{dx}{x}.

Therefore,

\displaystyle \lim_{N \to \infty} \left( \sum_{n=1}^N \frac{1}{n} - \int_1^N \frac{dx}{x} \right) = \gamma.

With this interpretation, the sum can be viewed as the sum of the N rectangles in the above picture, while the integral is the area under the hyperbola. Therefore, the limit \gamma can be viewed as the limit of the blue part of the above picture.

In other words, it’s an amazing fact that while both

\displaystyle \sum_{n=1}^\infty \frac{1}{n}

and

\displaystyle \int_1^\infty \frac{dx}{x}

diverge, somehow the difference

\displaystyle \lim_{N \to \infty} \left(\sum_{n=1}^N \frac{1}{n} - \int_1^N \frac{dx}{x} \right)

converges… and this limit is defined to be the number \gamma.

Leave a comment

24 Comments

  1. The difference function 1/x – 1/(x+1) encloses the blue bits, and its integral is INT(1/(x(x+1)_, which is less than INT(1/(x^2), which inturn is finite when taken from 0 to infinity. Hence convergence of the sum of blue bits.
    Tell us more about gamma.
    Are you going to write about the weirdness of the finite volume,infinite surface area of the rotation of 1/x around the x axis ? This is one of the more bizarre results of mathematics.

    Reply
    • Thanks for the suggestions. I hadn’t planned to say anything about the surface area/volume of that solid of revolution, since all I think I can say is, “This is a very weird result.” But let me think about it.

      Reply
  2. Two right parentheses missing !!!

    Reply
  1. Thoughts on Infinity: Index | Mean Green Math
  2. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 1 | Mean Green Math
  3. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 2 | Mean Green Math
  4. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 3 | Mean Green Math
  5. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 4 | Mean Green Math
  6. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 5 | Mean Green Math
  7. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 6 | Mean Green Math
  8. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 7 | Mean Green Math
  9. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 8 | Mean Green Math
  10. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 9 | Mean Green Math
  11. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 10 | Mean Green Math
  12. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 11 | Mean Green Math
  13. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 12 | Mean Green Math
  14. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 13 | Mean Green Math
  15. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 14 | Mean Green Math
  16. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 15 | Mean Green Math
  17. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 16 | Mean Green Math
  18. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 17 | Mean Green Math
  19. What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 18 | Mean Green Math
  20. What I Learned by Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Index | Mean Green Math

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