Sum of zeta function upto the year

Algebra Level 5

Find the value of:

$\large \displaystyle \bigg \lfloor \zeta(2)+\zeta(3)+\zeta(4)+ \dots + \zeta(2013) + \zeta(2014) \bigg \rfloor$

The answer is 2013.

1 solution

Oliver Bel
Apr 5, 2014

The correct answer is actually $2013$ , not $2014$ .

Note that $\zeta(n)-1>0$ for $n>1$ .

It is easy to show that $\sum_{n\mathop=2}^{\infty} \zeta(n)-1=1$

Hence $0<\sum_{n\mathop=2}^{2014}\zeta(n)-1<1$

$2013<\sum_{n\mathop=2}^{2014}\zeta(n)<2014$

$\left\lfloor{\sum_{n\mathop=2}^{2014}\zeta(n)}\right\rfloor=\boxed{2013}$

Mathematica gives the following approximation of the sum:

2013.99999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999994683953969593983962
6610134919562547022344337507901680806619304019175483072471512367036506
6479505134212306879444735405234606670997803446269221021067839170113458
1597508355531421845880661993463486420862693028581331060405861146746201
5299190257315584102000510908364777591103173271801283346003150456104381
3591339775918812709402056630526686660341199531069051132240414869813180
749120900902486699228

That was the solution I intended.

Proof for $\displaystyle \sum_{n=2}^{\infty} (\zeta(n)-1) = 1$ :

Use $\large \zeta(n) - 1 = \displaystyle \sum_{s=1}^{\infty} \frac{1}{s^n} - 1 = \sum_{s=2}^{\infty} \frac{1}{s^n}$ to get

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

= $\large \displaystyle \sum_{s=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{s^n}$

= $\large \displaystyle \sum_{s=2}^{\infty} \frac{1}{s(s-1)}$

= $\large \displaystyle \sum_{s=2}^{\infty} \bigg(\frac{1}{s-1}-\frac{1}{s}\bigg) = 1$

jatin yadav - 7 years, 2 months ago

In this case, wouldn't the answer to the problem be 2014 (the closest integer?)

Test User - 7 years, 2 months ago

The floor function and the closest integer function are two different functions.

Oliver Bel - 7 years, 2 months ago

Sum of zeta function upto the year

The answer is 2013.

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