P-Series Convergence: Answers to Common Questions

[foo9008]

Homework Statement

http://math.oregonstate.edu/home/pr...stStudyGuides/SandS/SeriesTests/p-series.html

Homework Equations

The Attempt at a Solution

why when the p is between 0 and 1 , the p-series diverges? when p is between 0 and 1 , the denominator still become big , when 1/ big number , the number will become smaller than before. ,So, the series will converge , right?

 

[stevendaryl]

Homework Statement

http://math.oregonstate.edu/home/pr...stStudyGuides/SandS/SeriesTests/p-series.html

Homework Equations

The Attempt at a Solution

why when the p is between 0 and 1 , the p-series diverges? when p is between 0 and 1 , the denominator still become big , when 1/ big number , the number will become smaller than before. ,So, the series will converge , right?

No, just because the terms get smaller and smaller doesn't mean that the series converges. They give the example of the series 1 + \frac{1}{2} + \frac{1}{3} + ..., which is a p-series with p=1. That doesn't converge. To see that it doesn't, we can group the terms this way:

S = 1 + \frac{1}{2} + (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}) + ...

(In each group, you have a sum of the form \frac{1}{2^n + 1} + ... + \frac{1}{2^n + 2^n})

If S converges, then certainly it would still converge if you replaced terms by smaller terms. So in each group, replace each term by the smallest term in the group:
S' = 1 + \frac{1}{2} + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}) + ...

If S converged, then so would S'. But S' = 1 + 1/2 + 1/2 + 1/2 + ... which clearly diverges.

If you choose p < 1, then the series diverges even worse.

 

[geoffrey159]

You can see that it diverges when $0<p<1$ by comparison to the harmonic serie: you have $\lim_{n\to\infty} n\times \frac{1}{n^p} = +\infty$, and therefore $\exists N\in \mathbb{N}, \ (n > N \Rightarrow \frac{1}{n^p} > \frac{1}{n})$.

 

[Ray Vickson]

Homework Statement

http://math.oregonstate.edu/home/pr...stStudyGuides/SandS/SeriesTests/p-series.html

Homework Equations

The Attempt at a Solution

why when the p is between 0 and 1 , the p-series diverges? when p is between 0 and 1 , the denominator still become big , when 1/ big number , the number will become smaller than before. ,So, the series will converge , right?

Note that for $p>0$ the function $1/x^p$ is strictly decreasing, so
\int_n^{n+1} 1/x^p \: dx &lt; 1/n^p &lt; \int_{n-1}^n 1/x^p \: dx
Summing over $n \leq N$ we have
\int_1^{N+1} x^{-p} \, dx &lt; \sum_{n=1}^N 1/n^p &lt; 1 + \int_1^N x^{-p} \, dx
The integrals are both do-able, and for $0 < p \leq 1$ we have $\sum 1/n^p \geq \int_1^{\infty} x^{-p} \, dx = +\infty$, so the sum is divergent. If $p > 1$ the partial sums are bounded above by $1+\int_1^{\infty} x^{-p} \, dx = p/(p-1)$, so $\sum 1/n^p$ converges, and its value lies between 1 and $p/(p-1)$.

 

[foo9008]

You can see that it diverges when $0<p<1$ by comparison to the harmonic serie: you have $\lim_{n\to\infty} n\times \frac{1}{n^p} = +\infty$, and therefore $\exists N\in \mathbb{N}, \ (n > N \Rightarrow \frac{1}{n^p} > \frac{1}{n})$.

i don't under stand, can you explain further?

 

[geoffrey159]

If you take a positive terms serie, it either converges or diverges to $+\infty$. In the case of the harmonic serie, it was explained in first post why it diverges ( to $+\infty$ ). From the inequality above, it is clear that the serie of $\{1/n^p\}_n$ will also diverge to $+\infty$ (when $p < 1$)

 

[foo9008]

If you take a positive terms serie, it either converges or diverges to $+\infty$. In the case of the harmonic serie, it was explained in first post why it diverges ( to $+\infty$ ). From the inequality above, it is clear that the serie of $\{1/n^p\}_n$ will also diverge to $+\infty$ (when $p < 1$)

i don't understand why the harmonic series diverges? can you explain ?

 

[Ray Vickson]

i don't understand why the harmonic series diverges? can you explain ?

Read reply #4; it is all in there.

 

[foo9008]

Note that for $p>0$ the function $1/x^p$ is strictly decreasing, so
\int_n^{n+1} 1/x^p \: dx &lt; 1/n^p &lt; \int_{n-1}^n 1/x^p \: dx
Summing over $n \leq N$ we have
\int_1^{N+1} x^{-p} \, dx &lt; \sum_{n=1}^N 1/n^p &lt; 1 + \int_1^N x^{-p} \, dx
The integrals are both do-able, and for $0 < p \leq 1$ we have $\sum 1/n^p \geq \int_1^{\infty} x^{-p} \, dx = +\infty$, so the sum is divergent. If $p > 1$ the partial sums are bounded above by $1+\int_1^{\infty} x^{-p} \, dx = p/(p-1)$, so $\sum 1/n^p$ converges, and its value lies between 1 and $p/(p-1)$.

can you explain on it ? i don't understand

 

[Ray Vickson]

can you explain on it ? i don't understand

1 + \int_1^N x^{-p} \, dx &lt; 1 + \int_1^{\infty} x^{-p} \, dx,
because $x^{-p} > 0$ for all $x > 0$ and
\int_1^{\infty} x^{-p} dx = \int_1^N x^{-p} dx +<br /> \underbrace{\int_N^{\infty} x^{-p} dx}_{&gt;0}

 

[foo9008]

Note that for $p>0$ the function $1/x^p$ is strictly decreasing, so
\int_n^{n+1} 1/x^p \: dx &lt; 1/n^p &lt; \int_{n-1}^n 1/x^p \: dx
Summing over $n \leq N$ we have
\int_1^{N+1} x^{-p} \, dx &lt; \sum_{n=1}^N 1/n^p &lt; 1 + \int_1^N x^{-p} \, dx
The integrals are both do-able, and for $0 < p \leq 1$ we have $\sum 1/n^p \geq \int_1^{\infty} x^{-p} \, dx = +\infty$, so the sum is divergent. If $p > 1$ the partial sums are bounded above by $1+\int_1^{\infty} x^{-p} \, dx = p/(p-1)$, so $\sum 1/n^p$ converges, and its value lies between 1 and $p/(p-1)$.

can you explain why there is 1 appear in the second line ?

 

[foo9008]

1 + \int_1^N x^{-p} \, dx &lt; 1 + \int_1^{\infty} x^{-p} \, dx,
because $x^{-p} > 0$ for all $x > 0$ and
\int_1^{\infty} x^{-p} dx = \int_1^N x^{-p} dx +<br /> \underbrace{\int_N^{\infty} x^{-p} dx}_{&gt;0}

why it will equals to p / p-1 ?

 

[Ray Vickson]

why it will equals to p / p-1 ?

Do the integral and see what you get.

 

[foo9008]

Do the integral and see what you get.

i got this

 

[geoffrey159]

Mr Vickson answers your question "why the harmonic serie diverges?" by comparing the harmonic serie to an integral. The conclusion of it is that the partial sum $H_N = \sum_{n=1}^N \frac{1}{n}$ is equivalent as $N \to +\infty$ to $\ln N$. This means that as $N\to \infty$, $H_N / \ln N \to 1$, showing that $H_N$ can't have a finite limit.

 

[stevendaryl]

i don't understand why the harmonic series diverges? can you explain ?

Several people (including me) have already explained. Just look at the partial sums:

1. The first term is greater than \frac{1}{2}
2. The sum of the first two terms is greater than \frac{2}{2}
3. The sum of the first four terms is greater than \frac{3}{2}
4. The sum of the first eight terms is greater than \frac{4}{2}
5. In general, if you sum up the first 2^n terms, you get something greater than \frac{n+1}{2}

So the partial sums keep growing bigger and bigger, without bound. For a series to be convergent, the partial sums have to be bounded.