Evaluating Summation Problem: Floor Function Solution

[Saitama]

Problem:
Evaluate:
$$\left[\sum_{n=1}^{\infty} \sum_{k=2}^{2014} \frac{1}{n^k}\right]$$
where $[x]$ denotes the floor function.Attempt:
I can see that the above can be written as:
$$\sum_{n=1}^{\infty} \frac{1}{n^2}+\frac{1}{n^3}+\frac{1}{n^4}+\cdots + \frac{1}{n^{2014}}$$
$$=\sum_{n=1}^{\infty} \frac{1}{n^2}\left(\frac{1-1/n^{2013}}{1-1/n}\right)$$
$$=\sum_{n=1}^{\infty} \frac{1}{n^{2014}}\left(\frac{n^{2013}-1}{n-1}\right)$$

How do I proceed after this?

I am not sure if this belongs to the Algebra section.

Any help is appreciated. Thanks!

 

[chisigma]

Problem:
Evaluate:
$$\left[\sum_{n=1}^{\infty} \sum_{k=2}^{2014} \frac{1}{n^k}\right]$$
where $[x]$ denotes the floor function.Attempt:
I can see that the above can be written as:
$$\sum_{n=1}^{\infty} \frac{1}{n^2}+\frac{1}{n^3}+\frac{1}{n^4}+\cdots + \frac{1}{n^{2014}}$$
$$=\sum_{n=1}^{\infty} \frac{1}{n^2}\left(\frac{1-1/n^{2013}}{1-1/n}\right)$$
$$=\sum_{n=1}^{\infty} \frac{1}{n^{2014}}\left(\frac{n^{2013}-1}{n-1}\right)$$

How do I proceed after this?

I am not sure if this belongs to the Algebra section.

Any help is appreciated. Thanks!

In...

http://mathhelpboards.com/challenge-questions-puzzles-28/not-too-challenge-question-562.html

... it has been demonstrated that...

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

... so that we can write...

$\displaystyle S = \sum_{n=1}^{\infty} \sum_{k=2}^{2014} \frac{1}{n^k} = \sum_{k=2}^{2014} \zeta(k) = 2013 + \sum_{k=2}^{2014} \{ \zeta(k) -1\} < 2014\ (2) $

... so that is $\displaystyle \left[ S \right] = 2013$...

Kind regards

$\chi$ $\sigma$

 

[I like Serena]

In...

http://mathhelpboards.com/challenge-questions-puzzles-28/not-too-challenge-question-562.html

... it has been demonstrated that...

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

... so that we can write...

$\displaystyle S = \sum_{n=1}^{\infty} \sum_{k=2}^{2014} \frac{1}{n^k} = \sum_{k=2}^{2014} \zeta(k) = 2013 + \sum_{k=2}^{2014} \{ \zeta(k) -1\} < 2014\ (2) $

... so that is $\displaystyle \left[ S \right] = 2013$...

Kind regards

$\chi$ $\sigma$

Nice!

And it's not a $B, \Gamma, W, $ or $A$ function! ;)

If we expand the proof for $\sum\limits_{k=2}^{\infty} \{ \zeta(k) -1\} = 1$, there is not even a zeta function.

 

[alyafey22]

We need only prove that

\(\displaystyle \sum_{k=2} \zeta(k)-1 = 1\)

which can be done by interchanging the two sums.

It is clear that $\zeta(k)>1$ for all $k\geq 2$

so it must be the case that $ \sum_{n=2}^k \{\zeta(n) \} $ is less than $1$ .

 

[Saitama]

In...

http://mathhelpboards.com/challenge-questions-puzzles-28/not-too-challenge-question-562.html

... it has been demonstrated that...

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

... so that we can write...

$\displaystyle S = \sum_{n=1}^{\infty} \sum_{k=2}^{2014} \frac{1}{n^k} = \sum_{k=2}^{2014} \zeta(k) = 2013 + \sum_{k=2}^{2014} \{ \zeta(k) -1\} < 2014\ (2) $

... so that is $\displaystyle \left[ S \right] = 2013$...

Kind regards

$\chi$ $\sigma$

Thank you chisigma! :)