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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$\sum\frac{1}{2^n} = 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$$

$$\sum \frac{1}{n} = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$$

Why is the first one convergent and the second divergent? First has $$\lim_ {n\to \infty}\ \frac{1}{2^n}$$ and the second $$\lim_{n\to \infty}\ \frac{1}{n}$$ Both limis are $0$. Right? Now, why does the first series converge and the second diverge? The first one looks like "the half of the other".

share|cite|improve this question
The limit of the terms you're adding is 0. The sum of these terms is another matter... – David Mitra Nov 18 '11 at 17:58
Are you confusing $2^n$ with $2n$? Because half of the second series is $\sum_{n} \frac{1}{2n}$, not $\sum_n \frac{1}{2^n}$ – Thomas Andrews Nov 18 '11 at 18:03
Write down about $10$ more terms of each, and see whether you still think that one looks like half of the other. – André Nicolas Nov 18 '11 at 18:40
Does one of them look like half the other after you've written down about 15 terms of each instead of just four of each? Try it. You'll see what I mean. – Michael Hardy Nov 18 '11 at 19:17
up vote 10 down vote accepted

The fact that the terms go to zero is a necessary condition for convergence, but it is not sufficient. These two series show that it is not enough for the terms to get small in order for the series to converge, you need the partial sums to converge.

Remember that the series is really the limit of the partial sums: $$\sum_{i=1}^{\infty}a_i = \lim_{n\to\infty}\left(\sum_{i=1}^n a_i\right).$$ In order for the sequence on the right to converge, the terms must get closer to one another; that's why we need (require) $a_n\to 0$ as $n\to\infty$. But this is not sufficient in order to get a sequence that converges.

The first series, $$\sum_{i=0}^{\infty}\frac{1}{2^i}$$ converges because the sequence of partial sums converges. The $n$th partial sum is equal to $$\sum_{i=0}^{n}\frac{1}{2^i} = \frac{1 - \frac{1}{2^{n+1}}}{\frac{1}{2}} = 2 - \frac{1}{2^n},$$ so the sequence converges to $2$ as $n\to\infty$. See also this answer.

The second series, $$\sum_{i=1}^{\infty}\frac{1}{i}$$ does not converge because the sequence of partial sums does not converge; even though the terms differ from the preceding one by very little, they eventually get very far away from any particular term. For instance, $$\begin{align*} \sum_{i=1}^2\frac{1}{i} &= 1+\frac{1}{2} =\frac {3}{2}\\ \sum_{i=1}^4\frac{1}{i} &= 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\ &\gt 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4} = 2\\ \sum_{i=1}^8 \frac{1}{i} &= 1+ \frac{1}{2}+\frac{1}{3}+\frac{1}{4} + \frac{1}{5}+\frac{1}{6} + \frac{1}{7}+\frac{1}{8}\\ &\gt \left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) + \left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)\\ &\gt 2 + \frac{1}{2}\\ \sum_{i=1}^{16}\frac{1}{i} &= \sum_{i=1}^{8}\frac{1}{i} + \sum_{i=9}^{16}\frac{1}{i}\\ &\gt \frac{5}{2} + \sum_{i=9}^{16}\frac{1}{16} = \frac{5}{2}+\frac{8}{16}\\ &= 3\\ &\vdots \end{align*}$$ And so on. The sum up to $\frac{1}{2^n}$ will be greater than $\frac{n}{2}+1$, which gets infinitely large.

Note that the first series is not "half the other"; half the second series would be $$\frac{1}{2}+\frac{1}{4}+\frac{1}{6} + \frac {1}{8} + \frac{1}{10} + \frac{1}{12} + \frac{1}{14} + \frac{1}{16}+\cdots$$ you keep losing more and more terms for larger $n$ (first you are missing $\frac{1}{6}$, then $\frac{1}{10}$ and $\frac{1}{12}$ and $\frac{1}{14}$, etc.). If you did in fact have the series of "half the second", that is, if you had $$\sum_{i=1}^{\infty}\frac{1}{2i}\quad\text{instead of}\quad \sum_{i=1}^{\infty}\frac{1}{2^i}$$ then the series would not converge, precisely because it is half a divergent series.

share|cite|improve this answer
Thank you a lot for your detailed answer. – Andrew Nov 18 '11 at 18:31
Oresme's argument is also here. – J. M. Nov 18 '11 at 18:48

Your Answer


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.