Eat a function

Calculus Level 3

Fill in the blank:

The series $1 + \dfrac1{2^2} + \dfrac1{3^2} + \dfrac1{4^2} + \cdots$ is two times the value of the alternating series $1 - \dfrac1{2^2} + \dfrac1{3^2} - \dfrac1{4^2} + \cdots$ .

The series $1 + \dfrac1{2^3} + \dfrac1{3^3} + \dfrac1{4^3} + \cdots$ is $\text{\_\_\_\_\_\_\_\_}$ times the value of the alternating series $1 - \dfrac1{2^3} + \dfrac1{3^3} - \dfrac1{4^3} + \cdots$ .

4 None of these choices 5 3

1 solution

Steven Perkins
Apr 19, 2017

It appears that the straight summation series is 4/3 times the alternating series.

Not anywhere close to any of the answers given.

How do you know that it should be 4/3?

Pi Han Goh - 4 years, 1 month ago

There's probably a formula for these that I don't know.

But I'm lazy, so I wrote a quick Python program to calculate the sums to 1000 terms.

Steven Perkins - 4 years, 1 month ago

Hmmm, in that case, why don't you show us the Python code to illustrate how you got the answer? Because right now, I don't find this solution helpful to anyone else...

And no, there's no formula involved. On the other hand, you might find "Dirichlet eta function" important.

Pi Han Goh - 4 years, 1 month ago

@Pi Han Goh – You are correct of course. It wasn't too helpful as it was. Here is the simple code:

sa=0
for i in range(1,1001):
  sa=sa+1.0/(i*i*i)*(-1)**(i-1)

s=0
for i in range(1,1001):
  s=s+1.0/(i*i*i)    

print s,sa,s/sa 
1.20205640366 0.90154267687 1.33333278002

Steven Perkins - 4 years, 1 month ago

Eat a function

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