# Thread: cool series

1. ## cool series

Mathstud likes series, as well as others. I enjoy a good one too. Here is one I thought was cool because of its familiar punchline.

"For $\displaystyle n=1,2,3,.......$ let $\displaystyle c_{n}=1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{n}$

Evaluate the sum:

$\displaystyle S=\sum_{n=1}^{\infty}\frac{c_{n}}{n(n+1)}$"

Not that this is particularly difficult. I just liked it. Enjoy the holiday for those of you who celebrate the Fourth of July. Even if you're not American, take the time to enjoy anyway. Moo, you have Bastille Day coming up I think July 14th?. Do the French use it as an excuse to party as Americans do the fourth?.

If no one bites by tomorrow morning, I will post my attempt.

2. Originally Posted by galactus
Mathstud likes series, as well as others. I enjoy a good one too. Here is one I thought was cool because of its familiar punchline.

"For $\displaystyle n=1,2,3,.......$ let $\displaystyle c_{n}=1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{n}$

Evaluate the sum:

$\displaystyle S=\sum_{n=1}^{\infty}\frac{c_{n}}{n(n+1)}$"

Not that this is particularly difficult. I just liked it. Enjoy the holiday for those of you who celebrate the Fourth of July. Even if you're not American, take the time to enjoy anyway. Moo, you have Bastille Day coming up I think July 14th?. Do the French use it as an excuse to party as Americans do the fourth?.

If no one bites by tomorrow morning, I will post my attempt.
We can prove that $\displaystyle \lim_{n \to \infty} \frac{c_{n}}{n} = 0$.

$\displaystyle S=\sum_{n=1}^{\infty}\frac{c_{n}}{n(n+1)} = \sum_{n=1}^{\infty}\left(\dfrac{c_n}{n} - \frac{c_{n+1}- \frac{1}{n+1}}{n+1}\right)$$\displaystyle =\sum_{n=1}^{\infty}\left(\dfrac{c_n}{n} - \frac{c_{n+1}}{n+1}\right)+\sum_{n=1}^{\infty} \frac1{(n+1)^2}$

Clearly the first part telescopes to give:

$\displaystyle \sum_{n=1}^{\infty}\left(\dfrac{c_n}{n} - \frac{c_{n+1}}{n+1}\right) = c_1(=1) - \lim_{n \to \infty} \frac{c_{n+1}}{n+1}$

Now $\displaystyle \lim_{n \to \infty} \frac{c_{n+1}}{n+1} = \lim_{n \to \infty} \frac{c_n}{n}=0$

Thus
$\displaystyle S = 1 + \sum_{n=1}^{\infty} \frac1{(n+1)^2} = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi ^2}{6}$

Note: Proof of $\displaystyle \lim_{n \to \infty} \frac{c_n}{n}=0$

$\displaystyle \lim_{n \to \infty} (c_n - \log n) = \gamma \Rightarrow \lim_{n \to \infty} \frac{c_n}{n} = \lim_{n \to \infty} \frac{\log n}{n} = 0$

3. Well done, Iso. Notice the familiar ending?. The ever popular Basel problem solution. Here is what I done. It's pretty much your approach.

$\displaystyle \sum_{n=1}^{k}\frac{c_{n}}{n(n+1)}$

$\displaystyle =\sum_{n=1}^{k}\left(\frac{c_{n}}{n}-\frac{c_{n}}{n+1}\right)$

$\displaystyle =\sum_{n=1}^{k}\frac{c_{n}}{n}-\sum_{n=2}^{k+1}\frac{c_{n-1}}{n}$

$\displaystyle =c_{1}+\sum_{n=2}^{k}\frac{c_{n}-c_{n-1}}{n}-\frac{c_{k}}{k+1}$

$\displaystyle =1+\sum_{n=2}^{k}\frac{1}{n^{2}}-\frac{c_{k}}{k+1}$

$\displaystyle =\sum_{n=1}^{k}\frac{1}{n^{2}}-\frac{c_{k}}{k+1}$

$\displaystyle =\sum_{n=1}^{k}\frac{1}{n^{2}}-\frac{c_{k}-ln(k)}{k+1}-\frac{ln(k)}{k+1}$

Letting $\displaystyle n\to {\infty}$, and using the fact that:

$\displaystyle \lim_{k\to {\infty}}(c_{k}-ln(k))$ exists and

$\displaystyle \lim_{k\to {\infty}}\frac{ln(k)}{k+1}=0$

Finally:

$\displaystyle S=\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\boxed{\frac{ {\pi}^{2}}{6}}$

4. Originally Posted by galactus
Well done, Iso. Notice the familiar ending?. The ever popular Basel problem solution.
Of course, how else would I have computed that final sum

Actually I was trying a different approach too. I worked it out and posted it in the Definitely Integral thread. Actually your sum amounts to this integral:
$\displaystyle \int_0^1 \int_0^x \frac{\ln (1 - y)}{y(y-1)} ~ dy ~ dx$ and its easy to work this out. Check out wingless's solution.