Is the series $ \sum_{n=2}^{\infty} \frac{1}{n^2} $ converges?

[ognik]

Hi - just done the integral test on the Riemann zeta series, came out to $\frac{1}{p-1}$

I can clearly see it therefore converges for P > 1, is singular for p=1, but for p < 1 I can't see why it diverges? In the limit p < 1 just gets smaller?

Would also like to check about p = 1, all I need do is say $\frac{1}{0}=\infty, \therefore$ it diverges, right?
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Continuing to explore the zeta function...

Test for convergence $ \sum_{n=1}^{\infty} \frac{1}{2n(2n+1)} $

I did this using the ratio test, how can I do it using the comparison test and zeta function?

 

[Guest2]

Hi - just done the integral test on the Riemann zeta series, came out to $\frac{1}{p-1}$

I can clearly see it therefore converges for P > 1, is singular for p=1, but for p < 1 I can't see why it diverges? In the limit p < 1 just gets smaller?

Would also like to check about p = 1, all I need do is say $\frac{1}{0}=\infty, \therefore$ it diverges, right?
---------------------
Continuing to explore the zeta function...

Test for convergence $ \sum_{n=1}^{\infty} \frac{1}{2n(2n+1)} $

I did this using the ratio test, how can I do it using the comparison test and zeta function?

$\displaystyle \frac{1}{2n(2n+1)} = \frac{1}{4n^2+2n} \le \frac{1}{4n^2} \le \frac{1}{n^2}$.

 

[alyafey22]

I can clearly see it therefore converges for P > 1, is singular for p=1, but for p < 1 I can't see why it diverges? In the limit p < 1 just gets smaller?

We know that $n^p < n$ for all $ p < 1 $. Hence we have

$$ \sum \frac{1}{n^p} > \sum \frac{1}{n}$$

We already know from the integral test that the harmonic series diverges.

 

[ognik]

$\displaystyle \frac{1}{2n(2n+1)} = \frac{1}{4n^2+2n} \le \frac{1}{4n^2} \le \frac{1}{n^2}$.

Thanks. The book talks about comparing with $ \zeta(2) $ which looks the same as the p series with p=2, except I seem to remember there is something deeper with the zeta function than the p series? (Even if not relevant for this exercise :-))

 

[alyafey22]

Thanks. The book talks about comparing with $ \zeta(2) $ which looks the same as the p series with p=2, except I seem to remember there is something deeper with the zeta function than the p series? (Even if not relevant for this exercise :-))

There are many representations for the zeta function. For example we have the integral representation

$$\zeta(s) = \frac{1}{\Gamma(s)}\int^{\infty}_0 \frac{t^{s-1}}{e^{t}-1}\, dt $$

The most general definition is the zeta functional equation

$$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$$

This represents an analytic continuation of the zeta function to the whole complex plain.

 

[topsquark]

There are many representations for the zeta function. For example we have the integral representation

$$\zeta(s) = \frac{1}{\Gamma(s)}\int^{\infty}_0 \frac{t^{s-1}}{e^{t}-1}\, dt $$

The most general definition is the zeta functional equation

$$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$$

This represents an analytic continuation of the zeta function to the whole complex plain.

The recursive definition isn't calculable for all complex s. s can't be calculated between -1 and 1, and is divergent for any positive integer s.

-Dan

 

[ognik]

Thanks guys, I was aware the zeta function is valuable in other aspects of math and physics, and the different representations for it are probably useful in different situations, but in this case we are dealing with reals, so to rephrase my question - why does the book use the zeta function - $\zeta(2)$ to compare with, instead of the p series $\frac{1}{n^2}$?

If the zeta is a 'better' answer, I'd like to understand why ; or alternately what is wrong with the p series in this situation?
Thanks

 

[alyafey22]

The recursive definition isn't calculable for all complex s. s can't be calculated between -1 and 1, and is divergent for any positive integer s.

-Dan

I think you are right. We can continue the zeta function to the right half plain except at $s=1$ through the equation

$$\zeta(s)=(1-2^{1-s})^{-1} \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s} $$

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Thanks guys, I was aware the zeta function is valuable in other aspects of math and physics, and the different representations for it are probably useful in different situations, but in this case we are dealing with reals, so to rephrase my question - why does the book use the zeta function - $\zeta(2)$ to compare with, instead of the p series $\frac{1}{n^2}$?

If the zeta is a 'better' answer, I'd like to understand why ; or alternately what is wrong with the p series in this situation?
Thanks

We know that the p series is defined as

$$S_p = \sum \frac{1}{n^p }$$

But we know that for $p>1$ we can say that $S_p = \zeta(p)$ so this is just a matter of notation because the zeta function is famous.

 

[ognik]

But we know that for $p>1$ we can say that $S_p = \zeta(p)$ so this is just a matter of notation because the zeta function is famous.

Thanks for that confirmation. An afterthought - is there a different formula for the (partial) sum of the p series? I have only found methods to estimate its upper and lower bounds