# Math Question: (1/1)+(1/2)+(1/3)+(1/4)... **URL:** \ **Category:** Factual Questions\ **Created:** [March 10, 2003, 6:51pm UTC](https://boards.straightdope.com/t/math-question-1-1-1-2-1-3-1-4/160368 "2003-03-10T18:51:34Z")\ **Posts on this page:** 8\ **Page:** 2
**Author:** ![Northern\_Piper](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/northern_piper/32/5304_2.png) [@Northern\_Piper](https://boards.straightdope.com/u/Northern_Piper)\ **Post date:** [March 11, 2003, 3:32am UTC](https://boards.straightdope.com/t/math-question-1-1-1-2-1-3-1-4/160368/21 "2003-03-11T03:32:50Z")
People doing math. :eek: In their spare time. :eek: :eek: For fun. :eek: :eek: :eek: [**Piper** slowly backs out of the room, not making any sudden moves.] ---
**Author:** ![Orbifold](https://avatars.discourse-cdn.com/v4/letter/o/779978/32.png) [@Orbifold](https://boards.straightdope.com/u/Orbifold)\ **Post date:** [March 11, 2003, 5:04am UTC](https://boards.straightdope.com/t/math-question-1-1-1-2-1-3-1-4/160368/22 "2003-03-11T05:04:50Z")
[hijack] **Piper** , I’ve been known to convert numbers into base 8 in my head just because I have nothing else to do, and in my spare moments I occasionally try to visualize five- and six-dimensional objects. And everyone else in this thread _beat me to the answer_. Be very afraid. [/hijack] **I, Brian** : perhaps you’re thinking of 1-(1/3)+(1/5)-(1/7)+(1/9)-…, which converges to pi/4. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [March 11, 2003, 6:39am UTC](https://boards.straightdope.com/t/math-question-1-1-1-2-1-3-1-4/160368/23 "2003-03-11T06:39:44Z")
> [@](#): > > \*Originally posted by jinx \* > \*\*Karl Gauss, I believe you, but I must ask: I ain’t a mathematician, but it sure looks like it should converge, or you should be able to take the limit. I mean, each additional term is adding just a smaller and smaller piece to the total. You’d think it’d reach a limit of some value as n grows larger and larger in the expression “1/n”. > > I could express the OP’s question as a summation notation of 1/n as n ranges from 1 to infinity, couldn’t I? Then, couldn’t I find the limit as n -\> infinity? > > Please feel free to explain in greater detail where my thinking may be wrong. But, please be gentle! - Jinx \*\* Here is another version of **Karl Gauss’** proof from _Mathematics for Scientists and Engineers_, Sokolnikof and Redheffer. The series 1/n as n goes from 1 to n can be arranged as follows: 1+1/2+(1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+…+1/16)+… compare this term by term with: 1+1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)+(1/16+…+1/16)+… Every term in the top sequence is equal to or greater than the corresponding term in the bottom sequence. The bottom sequence can also be written as: 1+1/2+1/2+1/2+1/2+… which obviously increases without limit, _i.e._ diverges. The first sequence must diverge also since all the terms in it are at least as big as the terms in this sequence, ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [March 11, 2003, 6:53am UTC](https://boards.straightdope.com/t/math-question-1-1-1-2-1-3-1-4/160368/24 "2003-03-11T06:53:51Z")
> [@](#): > > \*Originally posted by I, Brian \* > \*\*Hmmm…isn’t this another way of writing out pi? Sorry - I forget the exact different factorials involved. \*\* There are several, including: Pi/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - … See [here](http://mathforum.org/library/drmath/view/58286.html) for more. ---
**Author:** ![Qwertyasdfg](https://avatars.discourse-cdn.com/v4/letter/q/ecc23a/32.png) [@Qwertyasdfg](https://boards.straightdope.com/u/Qwertyasdfg)\ **Post date:** [March 11, 2003, 10:19pm UTC](https://boards.straightdope.com/t/math-question-1-1-1-2-1-3-1-4/160368/25 "2003-03-11T22:19:03Z")
> [@](#): > > \*Originally posted by Northern Piper \* > \*\*People doing math. :eek: > > In their spare time. :eek: :eek: > > For fun. :eek: :eek: :eek: > > [**Piper** slowly backs out of the room, not making any sudden moves.] \*\* You have 3000+ posts, and you’re shocked to see a math thread? Anyway, can anyone confirm or debunk my method (integrals)? The fact that I thought of it makes me think its wrong, so I’m really curious as to how smart I should feel. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [March 11, 2003, 10:51pm UTC](https://boards.straightdope.com/t/math-question-1-1-1-2-1-3-1-4/160368/26 "2003-03-11T22:51:45Z")
> [@](#): > > \*Originally posted by Qwertyasdfg \* > \*\*Anyway, can anyone confirm or debunk my method (integrals)? The fact that I thought of it makes me think its wrong, so I’m really curious as to how smart I should feel. \*\* In this case, your method is pretty close. But if f varies a lot between integers, the integral and the sum can be pretty far off. Consider f(x) = x + |40,000 \* sin(2[symbol]p[/symbol]x)|. The sum of f(x) over [1, 10] is 55. The integral of f(x) over the same interval is (50[symbol]p[/symbol] + 800000)/[symbol]p[/symbol], which is roughly 255000. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [March 11, 2003, 10:56pm UTC](https://boards.straightdope.com/t/math-question-1-1-1-2-1-3-1-4/160368/27 "2003-03-11T22:56:32Z")
I should add that if you were to accidentally mistype f without the absolute value and take the integral of that, you’d get 50. It’s cause there’s a lot of area under the curve in that case. ---
**Author:** ![ftg](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ftg/32/2801_2.png) [@ftg](https://boards.straightdope.com/u/ftg)\ **Post date:** [March 12, 2003, 12:26am UTC](https://boards.straightdope.com/t/math-question-1-1-1-2-1-3-1-4/160368/28 "2003-03-12T00:26:57Z")
(With apologies to **Northern Piper**.) **Qwertyasdfg** , here’s how to do summation approximation using integrals. First the function must by strictly increasing or decreasing. If not, then you have to work on it piecewise, assuming that is practical, and no asymptotes in the immediate area. 1/n fits the criteria if we are careful to stay away from n=0. Think of building a bar chart using 1/n. Make a bar, width 1, of height 1/1 between 1 and 2. Then another of height 1/2 between 2 and 3, then height 1/3, etc. Finish with one height 1/n between n and n+1. Note that the sum is equal to the areas of these bars. Plot the curve 1/n over this and note that the area under the 1/n curve is a little less than the sum. So a lower bound for the sum is the integral from 1 to n+1 (!) of 1/n. (Yeah, I shouldn’t use n twice like this.) For an upper bound, plot 1/(n-1) instead. Oops, n=1 looks bad. But the rest of the sum, from 2 to n, is less than the integral from 2 to n+1 of 1/(n-1). Add in a +1 for the missing part. If you do the math right, you’ll see that the upper and lower bound are ln n + lower order stuff. (You gotta know how to approximate ln (n-1). That’s another “lecture.”) So the derivative of any nicely behaved, slow growing function, can be plugged into a summation that diverges. E.g., the derivative of ln ln ln ln n in a sum will also diverge, and much more slowly too. (“More slowly”, “slower”? Ah forget it, I’m going to watch “The Simpsons.”) [Previous page](https://boards.straightdope.com/t/math-question-1-1-1-2-1-3-1-4/160368.md?page=1)