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## Series Contest (SC3)

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Moderators: Random Variable, sos440

### Series Contest (SC3)

Sat Sep 28, 2013 2:37 pm

Posts: 850
Location: Jaipur, India

Series Contest 3

### Re: Series Contest (SC3)

Sat Sep 28, 2013 4:20 pm

Posts: 138
Location: North Londinium, UK
Nice!!!

Not sure I'll be able to contribute much, as you guys are light years ahead of me, but either way, I'll enjoy watching...

### Re: Series Contest (SC3)

Tue Oct 01, 2013 9:05 am

Posts: 850
Location: Jaipur, India

Here we go!

Problem 1

(i) Prove that

$\displaystyle \sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1}=-\gamma +\log 2$

$\gamma$ is the well known Euler's constant.

(ii) [BONUS]

$\displaystyle \sum_{k=2}^\infty \frac{\left\{ 2 \left( \frac{3}{2}\right)^k-3\right\}}{k}\left[ \zeta(k)-1\right]=\log(\pi)$

### Re: Series Contest (SC3)

Tue Oct 01, 2013 7:32 pm
zaidalyafey
Global Moderator

Posts: 354
Mathematica says the sum does not converge.
$\displaystyle \sum_{k\geq 0}\frac{f^{(k)} (s)}{(s+1)_k} (-s)^{k}= \int^{1}_0x^{s} f(s x) \left( \frac{x}{s}\right) \, dx$

Wanna learn what we discuss , see tutorials

### Re: Series Contest (SC3)

Wed Oct 02, 2013 2:17 am

Posts: 850
Location: Jaipur, India

zaidalyafey wrote:
Mathematica says the sum does not converge .

Mathematica does not say that. Both the series are convergent!

### Re: Series Contest (SC3)

Wed Oct 02, 2013 5:41 am
Random Variable Integration Guru

Posts: 381
If $\displaystyle \zeta(2n+1)$ goes to $\displaystyle 1$ fast enough that $\displaystyle \sum_{k=1}^{\infty} \Big( \zeta(2k+1)-1 \Big)$ converges, how could $\displaystyle \sum_{k=1}^{\infty} \frac{\zeta(2k+1)-1}{k+1}$ not converge?

$\displaystyle \sum_{k=1}^{\infty} \frac{\zeta(2k+1)-1}{k+1} = \sum_{k=2}^{\infty} \frac{\zeta(2k-1)-1}{k}$

$\displaystyle \sum_{k=2}^{\infty} \sum_{m=2}^{\infty} \frac{1}{k m^{2k-1}} = \sum_{m=2}^{\infty} \sum_{k=2}^{\infty} \frac{m}{km^{2k}}$ since all terms are positive

$\displaystyle = \sum_{m=2}^{\infty} \Big( -m \ln \Big( 1- \frac{1}{m^{2}} \Big) - \frac{1}{m} \Big) = \sum_{m=2}^{\infty} \Big(m \ln \Big( \frac{m^{2}}{m^{2}-1} \Big) - \frac{1}{m} \Big)$

$\displaystyle = \sum_{m=2}^{\infty} \Bigg( m \Big( \ln(m^{2}) - \ln(m^{2}-1) \Big) - \frac{1}{m} \Bigg)$

$\displaystyle = \lim_{M \to \infty} \sum_{m=2}^{M} \Big( 2m \ln(m) - m \ln(m+1) - m \ln(m-1) - \frac{1}{m} \Big)$

$\displaystyle = \lim_{M \to \infty} \Big( \ln 2 + (M+1) \ln(M) - M \ln(M+1) - H_{M}+1 \Big)$

$\displaystyle = \lim_{M \to \infty} \Bigg( \ln 2 - H_{M} + \ln(M) +1 - M \ln \Big( 1 + \frac{1}{M} \Big) \Bigg)$

$\displaystyle = \lim_{M \to \infty} \Bigg( \ln 2 - H_{M} + \ln(M) +1 - M \Big( \frac{1}{M} + \mathcal{O}(M^{-2}) \Big) \Bigg)$

$\displaystyle = \ln 2 - \gamma$

### Re: Series Contest (SC3)

Wed Oct 02, 2013 7:09 am
zaidalyafey
Global Moderator

Posts: 354
Ok, sorry for confusion.
$\displaystyle \sum_{k\geq 0}\frac{f^{(k)} (s)}{(s+1)_k} (-s)^{k}= \int^{1}_0x^{s} f(s x) \left( \frac{x}{s}\right) \, dx$

Wanna learn what we discuss , see tutorials

### Re: Series Contest (SC3)

Wed Oct 02, 2013 9:56 am

Posts: 850
Location: Jaipur, India

Very good RV!! The second problem can be done by using the formula:

$\sum_{k=2}^\infty \frac{\zeta(k,a) t^{k-1}}{k}=\frac{\log \Gamma(a-t)}{t}-\frac{\log\Gamma(a)}{t}+ \psi_0(a)$

where $\zeta(s,a)$ is the Hurwitz zeta function.

Putting $t=1$ and $t=\frac{3}{2}$, we obtain

\displaystyle \begin{align*} \sum_{k=2}^\infty \frac{\zeta(k,a)}{k} &= \psi_0(a)+\log \frac{\Gamma(a-1)}{\Gamma(a)} \tag{1} \\ \sum_{k=2}^\infty \frac{\zeta(k,a)}{k} \left(\frac{3}{2} \right)^{k-1} &= \psi_0(a) +\frac{2}{3}\log \frac{\Gamma(a- \frac{3}{2})}{\Gamma(a)} \tag{2} \end{align*}

Subtract (1) from (2) and put $a=2$:

$\displaystyle \sum_{k=2}^\infty \frac{\left( \frac{3}{2}\right)^{k-1}-1}{k}\left[ \zeta(k)-1\right] = \frac{1}{3}\log \pi \tag{3}$

Multiplying by 3 and simplifying gives us the required answer.

### Re: Series Contest (SC3)

Wed Oct 02, 2013 10:10 am

Posts: 850
Location: Jaipur, India

Problem 2

This problem is based upon the "Bino-Harmonic" Series.

(i) Prove that $\displaystyle \sum_{n=1}^\infty \binom{2n}{n} \frac{(-1)^{n-1}}{4^n}H_n = \sqrt{2} \log \left(\frac{2\sqrt{2}}{1+\sqrt{2}} \right)$

(ii)[BONUS] Prove that $\displaystyle \sum_{n=1}^\infty \binom{2n}{n} \frac{H_n}{4^n n}=\frac{\pi^2}{3}$

### Re: Series Contest (SC3)

Thu Oct 03, 2013 1:33 am
Random Variable Integration Guru

Posts: 381
Here's a quick dervation of $\displaystyle \sum_{k=2}^\infty \frac{\zeta(k,a) t^{k-1}}{k}=\frac{\log \Gamma(a-t)}{t}-\frac{\log\Gamma(a)}{t}+ \psi_0(a)$

For $\displaystyle \text{Re} (s) > 1$ and $\displaystyle \text{Re}(a) > 0$, the Hurwitz zeta function has the integral representation $\displaystyle \zeta(s,a) = \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{x^{s-1} e^{-ax}}{1-e^{-x}} \ dx$.

There is also the much cooler Hermite integral representation which is valid for all complex values of $\displaystyle s$. But that's not needed here.

$\displaystyle \sum_{k=2}^{\infty} \zeta(k,a) t^{k-1} = \sum_{k=1}^{\infty} \zeta(k+1,a) t^{k} = \sum_{k=1}^{\infty} \Big( \frac{1}{k!} \int_{0}^{\infty} \frac{x^{k} e^{-ax}}{1-e^{-x}} \ dx \Big) \ t^{k}$

$\displaystyle = \int_{0}^{\infty} \frac{e^{-ax}}{1-e^{-x}} \sum_{k=1}^{\infty} \frac{(tx)^{k}}{k!} \ dx = \int_{0}^{\infty} \frac{e^{-ax} (e^{tx}-1)}{1-e^{-x}} \ dx$

$\displaystyle = \int_{0}^{\infty} \Big( \frac{e^{-x}-e^{-ax}}{1-e^{-x}} - \frac{e^{-x}-e^{-(a-t)x}}{1-e^{-x}} \Big) \ dx$

$\displaystyle = \int_{0}^{1} \Big( \frac{1-u^{a-1}}{1-u} - \frac{1-u^{(a-t)-1}}{1-u} \Big) \ du$

$\displaystyle = - \gamma + \int_{0}^{1} \frac{1-u^{a-1}}{1-u} \ du + \gamma - \int_{0}^{1} \frac{1-u^{(a-t)-1}}{1-u} \ du$

$\displaystyle = \psi(a) - \psi(a-t)$

Now integrate both sides with respect to $\displaystyle t$.

$\displaystyle \sum_{k=2}^{\infty} \frac{\zeta(k,a)}{k} t^{k} = \psi(a) t + \log \Gamma(a-t) + C$

The right side of the equation needs to be zero when $\displaystyle t$ is zero. So $\displaystyle C$ is $\displaystyle - \log \Gamma (a)$.

Dividing both sides by $\displaystyle t$ we have $\displaystyle \sum_{k=2}^{\infty} \frac{\zeta(k,a)}{k} t^{k-1} = \frac{\log \Gamma (a-t)}{t} - \frac{\log \Gamma (a)}{t} + \psi(a)$

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