So for the formula $\dfrac {1}{n}$, If you were to add up all $y$ values from $n=1$ to $n=∞$, wouldn't the sum

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

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?

  • 5
    $\begingroup$ The integral or the sum? $\endgroup$ Sep 21, 2010 at 4:44
  • 1
    $\begingroup$ 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. $\endgroup$ Sep 21, 2010 at 14:51
  • 1
    $\begingroup$ $\frac1x$ converges to $0$ as $x\to\infty$. The OP probably won't see this comment anyway, as they have not logged in recently. The posted answers are correct, and another way to illustrate that the above reasoning (posted in the question) is not, is to consider $\frac{1+x}x$ instead. Again we are "adding smaller and smaller numbers", but each of them is bigger than $1$. $\endgroup$
    – Mirko
    Feb 18, 2016 at 2:30
  • $\begingroup$ Can anyone give a geometric demonstration of this series? $\endgroup$
    – I.R
    Mar 20, 2017 at 16:28

5 Answers 5


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 Is there any formula for the series $1 + \frac12 + \frac13 + \cdots + \frac 1 n = ?$ which discusses precisely how fast the series diverges.


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?


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 $x$.

  • 1
    $\begingroup$ the rate of growth is logarithmic? Isn't ln the integral of 1/x rather than the rate, so ln literally approximates the value of the sum at some point rather than its rate of growth. The rate of growth is the derivative. But your point still makes sense $\endgroup$ Apr 4, 2021 at 22:15

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 $1 + 1 + 1 + \cdots$)

  • 19
    $\begingroup$ That's a fine point, but I think this demonstrates imprecise language more than incorrect logic. I believe the OP meant decreasing to 0. $\endgroup$ Sep 21, 2010 at 16:06
  • 9
    $\begingroup$ I think the divergence of this series is pretty clear: $1 + 1/2 + 1/2 + 1/3 + 1/3 + 1/3 + 1/4 + 1/4 + 1/4 + 1/4 + 1/5 + \cdots$ $\endgroup$ Mar 21, 2012 at 16:37
  • $\begingroup$ the op clearly wasn't referring to this $\endgroup$ Apr 4, 2021 at 22:10

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .