Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So for the formula $\dfrac {1}{x}$, If you were to add up all $y$ values from $x=1$ to $x=∞$, wouldn't the sum approach a number because even though you are always adding, aren't you just adding smaller and smaller numbers? Wouldn't this mean that it approached a certain number?

share|improve this question
2  
The integral or the sum? –  Qiaochu Yuan Sep 21 '10 at 4:44
1  
To elaborate on Qiaochu's comment: while the series you describe in your actual posting is obviously related to the function in the topic, they are actually quite different things. You should probably revise the topic. –  Niel de Beaudrap Sep 21 '10 at 14:51

5 Answers 5

As others have explained, the series diverges. But the divergence is very slow, indeed. See below.

A recent related idea for a first year calculus exercise:

An intelligent robot named Marvin travelled back in time to the moment of the Big Bang 13.7 billion years ago. He started calculating the partial sums of the harmonic series $$ \frac11+\frac12+\frac13+\frac14+\frac15+\cdots $$ He added one term to the partial sum per second. Using the estimates leading to the so called integral test answer the following question: As of today, has Marvin's sum reached the value 42?

share|improve this answer

Let's try a series that diverges more quickly. The sequence $b_n = \sqrt{n}$ clearly grows to infinity, but slowly. The sequence

$$a_n = b_{n+1} - b_n = \sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}}$$

goes to zero. Nevertheless, it is clear that the series $a_1 + a_2 + a_3 + ... $ diverges.

share|improve this answer

Your logic fails much more obviously for the series
$1.1 + 1.01 + 1.001 + 1.0001 + \cdots$

Though each term is smaller than the last, it is clearly divergent (think of $1 + 1 + 1 + \cdots$)

share|improve this answer
11  
That's a fine point, but I think this demonstrates imprecise language more than incorrect logic. I believe the OP meant decreasing to 0. –  Jonas Meyer Sep 21 '10 at 16:06
3  
I think the divergence of this series is pretty clear: $1 + 1/2 + 1/2 + 1/3 + 1/3 + 1/3 + 1/4 + 1/4 + 1/4 + 1/4 + 1/5 + \cdots$ –  Tanner Swett Mar 21 '12 at 16:37

Yes it is true that the numbers you are adding are getting smaller and smaller. The key is that they do not get small quick enough. There are many proofs that can be found easily online (search for proof that the harmonic series diverges) that show that you can add up enough terms of the harmonic series to make its sum as large as you wish. The key is that the rate of growth of the (partial sums of the) harmonic series is logarithmic. Even though $\ln(x)$ grows extremely slowly, it can still be made larger than any fixed value given sufficiently large $x$.

share|improve this answer

Ignoring the 1, if you group together the first term, then the next two terms, then the next four terms, and so on, you get:

$$(1/2) + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + \cdots$$

which is greater than

$$(1/2) + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + \cdots$$

where now each group is exactly equal to 1/2. This shows that the sum of the first $2^n$ terms is at least $1 + 1/2 \cdot n$, and so the sum of all the terms is unbounded.

If you know a little bit of calculus, $\int dx/x=\log x$, so $1+1/2+1/3+\cdots+1/n\ge\int_1^ndx/x=\log n-\log1$.

See http://math.stackexchange.com/questions/5035 which discusses precisely how fast the series diverges.

share|improve this answer
    
This is the Oresme proof, and see math.stackexchange.com/questions/250/840#840 as well for how you might attempt a physical demonstration. –  J. M. Sep 21 '10 at 5:08
2  
I just realized that this is actually a proof that $1/2\le\log 2\le1.$ Heh. –  Charles Dec 17 '12 at 18:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.