It's Hard ?

Calculus Level 4

Evaluate

${\displaystyle \sum_{k=0}^{\infty} (-1)^k \left(\frac{1}{4k+1} + \frac{1}{4k+3}\right)} .$

The answer is 1.11.

2 solutions

Jack D'Aurizio
May 30, 2014

We have: $\displaystyle \begin{aligned}\sum_{k=0}^{+\infty}(-1)^k\left(\frac{1}{4k+1}+\frac{1}{4k+3}\right)=\int_{0}^{1}\sum_{k=0}^{+\infty}(-1)^k(x^{4k}+x^{4k+2})\,dx &=& \int_{0}^{1}\frac{1+x^2}{1+x^4}\,dx\\ &=&\frac{1}{2}\int_{0}^{+\infty}\frac{1+x^2}{1+x^4}\,dx \\&=&\frac{1}{4}\int_{\mathbb{R}}\frac{1+x^2}{1+x^4}\,dx\end{aligned}$ and by calculating residues we have that the starting series equals $\frac{\pi}{\sqrt{8}}=1.11072\ldots$ .

You don't need to use residues for this problem. :)

Rewrite the definite integral in the following manner:

$\displaystyle \int_0^{\infty} \frac{1+x^2}{1+x^4}\,dx=\int_0^{\infty} \frac{1+1/x^2}{(x-1/x)^2+2}\,dx$

Use the substitution $x-1/x=t$ and we are done.

Pranav Arora - 7 years ago

@Pranav Arora , how can infinite sums be transformed into integrals?I have heard of methods like Borel Summation, Euler's method,The Euler-Maclaurin formula etc. but none of them correspond to the integral you used.

Bogdan Simeonov - 7 years ago

Hello Bogdan! :)

I don't think I understood your question. Are you talking about the definite integral I posted or do you have some doubts regarding Jack's solution?

Pranav Arora - 7 years ago

@Pranav Arora – The integral Jack used, which then you rewrited.

Bogdan Simeonov - 7 years ago

@Bogdan Simeonov – Jack used the following:

$\displaystyle \int_0^1 x^{4k}\,dx=\frac{1}{4k+1}$

$\displaystyle \int_0^1 x^{4k+2}\,dx=\frac{1}{4k+3}$

He then used the fact that:

$\displaystyle \sum_{k=0}^{\infty} \left(x^{4k}+x^{4k+2}\right)= \sum_{k=0}^{\infty} (1+x^2)\,x^{4k}=\frac{1+x^2}{1+x^4}$

Does this help? :)

Pranav Arora - 7 years ago

@Pranav Arora – Yes, thank you :D

Bogdan Simeonov - 7 years ago

How do you solve that in detail? It seems I haven't encounter this kind of problem before. Thanks :)

John Shadrach Abalos - 7 years ago

I use partial fractions $\int\frac{1+x^2}{1+x^4}\,dx=\frac12\int\left[\frac{1}{x^2-\sqrt{2}x+1}+\frac{1}{x^2+\sqrt{2}x+1}\right]\ dt$ and then I use completing the square method. How about this one: $\sum_{n=0}^\infty \frac{(-1)^n}{4^{4n+1}(4n+1)}$ BTW, Jack D'Aurizio is also Math SE user.

Anastasiya Romanova - 7 years ago

Gabriel Ribeiro Freitas
Jun 21, 2014

I wrote a C code to solve it (: Here is a piece of the code:

float a, k;

for(k=0;k<10000;k++)

    a+=pow(-1, k)*((1/(4*k+1))+(1/(4*k+3)));

cout << a;

It doesnt converge very fast, so i had to use a big number of iterations to see where it was going.

It's Hard ?

The answer is 1.11.

2 solutions

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