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.

I don't understand why ${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n} }$ is divergent, but ${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n^2} }$ is convergent and its limit is equal to ${ \displaystyle\frac{\pi^2}{6} }$. In both cases, ${n^{th}}$ term tends to zero, so what makes these series different?

${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + }$ ...

${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n^2} =1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + }$ ...

share|improve this question
1  
The $n^{th}$ term $\to 0$ is only a necessary condition but not a sufficient one to decide the convergence of the series. –  user17762 Nov 24 '12 at 0:27
    
One way to see why one converges and the other diverges is to use the integral test. –  Rankeya Nov 24 '12 at 0:27
3  
In one series, the $n$th term tends to zero much faster than in the other series. –  littleO Nov 24 '12 at 1:11
    
The concept that the terms tend towards zero, yet one diverges and the other converges, causes quite a confusion. Think of it as, given any positive number $n$, the first series eventually, eventually increases beyond $n$. However, the second series, given however big a number, does not get past its limit value. –  Sawarnik Feb 24 at 20:45

4 Answers 4

up vote 46 down vote accepted

The following example may be easier to grasp. Consider the series $$1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\cdots. $$ So we have $2$ copies of $\frac{1}{2}$, $4$ copies of $\frac{1}{4}$, $8$ copies of $\frac{1}{8}$, $16$ copies of $\frac{1}{16}$, and so on like that forever.

The $2^k$ copies of $\frac{1}{2^k}$ add up to $1$, so partial sums get arbitrarily large, and therefore our series diverges.

Now look at the series of the squares of the above numbers. We get $$1+\frac{1}{4}+\frac{1}{4}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}+\cdots $$ The first entry is $1$. The next $2$ entries add up to $\frac{1}{2}$. The next $4$ entries add up to $\frac{1}{4}$. The next $8$ entries add up to $\frac{1}{8}$. And so on forever. So the full sum is $$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$$ This geometric series has sum $2$.

share|improve this answer
1  
Thank you again! I like your explanations so much, they are really clear and helpful. –  Edward Ruchevits Nov 24 '12 at 1:04
5  
Cauchy condenstion test. –  FrenzY DT. Nov 24 '12 at 1:21
    
Small MathJax/LaTeX tip: If you write $$ <stuff> $$. you get the full stop on a new line after the equation. I took the liberty of editing it out of your answer. –  kahen Nov 24 '12 at 12:41
    
@kahen I don't get your comment. I can look at the previous LaTeX, and he didn't do a period after double dollar signs. He did the period inside the double dollar signs, which is correct. What am I missing? –  Graphth Nov 28 '12 at 2:46

We can see that the harmonic series, $\displaystyle\sum_{k=1}^\infty\frac1k$, diverges using the classical observation: $$ \frac11+\frac12+\underbrace{\frac13+\frac14}_{\mbox{$2$ terms}}+\underbrace{\frac15+\frac16+\frac17+\frac18}_{\mbox{$4$ terms}}+\dots+\underbrace{\frac1{2^n+1}+\dots+\frac1{2^{n+1}}}_{\mbox{$2^n$ terms}}+\dots $$ where each grouping of terms totals at least $\frac12$.


The series $\displaystyle\sum_{k=1}^\infty\frac1{k^2}$ converges to a value $\le2$ by comparison: $$ \frac1{1^2}+\underbrace{\frac1{2^2}}_{\Large\lt\frac1{1\cdot2}}+\underbrace{\frac1{3^2}}_{\Large\lt\frac1{2\cdot3}}+\underbrace{\frac1{4^2}}_{\Large\lt\frac1{3\cdot4}}+\dots+\underbrace{\frac1{n^2}}_{\Large\lt\frac1{(n-1)n}}+\dots $$ and $$ \begin{align} &\frac1{1\cdot2}+\frac1{2\cdot3}+\frac1{3\cdot4}+\dots+\frac1{(n-1)n}+\dots\\ &=\left(\frac11-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\dots+\left(\frac1{(n-1)}-\frac1n\right)+\dots\\ &=1 \end{align} $$ This last series is called a "telescoping sum" since the last part of each term is cancelled by the first part of the next term, leaving only the first part of the first term and the last part of the last term. Since the last part of the last term vanishes, this series converges to the first part of the first term.

share|improve this answer
    
Thank you! I've understood "telescoping sum", but could you please explain what gives "comparison" for the first series? I can't see any significant difference between ${\frac{1}{n^2}}$ and ${\frac{1}{(n-1)n}}$ –  Edward Ruchevits Nov 24 '12 at 18:15
    
@EdwardRuchevits: The comparison is that $\frac1{n^2}\lt\frac1{(n-1)n}$ and since the sum of $\frac1{(n-1)n}$ converges, the sum of $\frac1{n^2}$ converges. –  robjohn Nov 24 '12 at 18:19
    
thank you again, i think I understood. :) –  Edward Ruchevits Nov 24 '12 at 18:23

Summation is closely related to integration. For a non-increasing function $f$ we have $$\sum_{n=t+1}^\infty f(n) \leq \int_t^\infty f(x)\mathrm{d}x \leq \sum_{n=t}^\infty f(n)$$ or expressed differently $$-f(t) + \sum_{n=t}^\infty f(n) \leq \int_t^\infty f(x)\mathrm{d}x \leq \sum_{n=t}^\infty f(n)$$ or equivalently $$\int_t^\infty f(x)\mathrm{d}x \leq \sum_{n=t}^\infty f(n) \leq f(t) + \int_t^\infty f(x)\mathrm{d}x$$

Illustration of the correspondence between summation and integration. Illustration of the correspondence between integraion and summation.

So if the integral exists, the sum diverges iff the integral is $\infty$ - the sum bounds the integral and vice versa.


Let's consider $p>1$: \begin{align} \int_1^{\infty} \frac{1}{x^p} \mathrm{d}x &= \int_1^{\infty} x^{-p} \mathrm{d}x \\ &= \frac{1}{1-p}(\lim_{x\rightarrow\infty} x^{1-p} - 1^{1-p}) \\ &= \frac{1}{p-1} \end{align}

which means that the corresponding sum converges. We even get some bounds, in particular $$ \frac{1}{p-1} \leq \sum_{n=1}^\infty n^{-p} \leq \frac{1}{p-1} + 1$$

If $p = 1$ then

\begin{align} \int_1^{\infty} \frac{1}{x} \mathrm{d}x &= \lim_{x\rightarrow\infty} (\ln x) - \ln 1 \\ &= \infty \end{align}

which means that the corresponding sum diverges.

So we can conclude that the sum $\sum_{n=1}^\infty n^{-p}$ converges iff $p>1$.

share|improve this answer
    
Thank you for this approach, it's always nice when something can be proved in different ways. –  Edward Ruchevits Nov 24 '12 at 18:02

As littleO's comment says, the difference is in how fast the terms are going to zero. Terms going to zero isn't enough. The rate at which the terms decay also makes a difference. In you example, if you consider $$\sum_{n=1}^{\infty}\frac{1}{n^p}$$ where $p$ can take any (let's stay with real) value for example, for $p=1$ as you already say, the series diverges. But for $p=2$ we do have convergence. It turns out that $p=1$ is the "boundary" between convergence and divergence. As long as $p$ is bigger than one, no matter how close to one, the series will converge. And as long as $p\leq1$ the series will diverge.

Every calculus student (myself included) by the way goes through these two stages. First I couldn't believe that adding up infinite terms (no matter how small) can give us a finite number. Then I thought as long as the terms go to zero we have convergence. Its tricky stuff but a lot of fun.

share|improve this answer
    
Thank you for your answer! –  Edward Ruchevits Nov 24 '12 at 2:59

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.