Linked Questions

7
votes
1answer
250 views

Potentially Useful Question [duplicate]

I have been solving problems using a "Potentially Helpful Formulas" sheet from my esteemed math professor. i want to solve for: $\sum_{n=1}^{\infty} \dfrac{1}{n^4} =$ ? On my formula sheet i have: ...
2
votes
1answer
469 views

$\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}$ [duplicate]

How we can do this sum? $$ \sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90} $$ I know that we could possibly use a Fourier series decomposition however I don't know what function to start with. I ...
1
vote
1answer
151 views

Show that $\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$ [duplicate]

Show that $\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$ I have to show this using results I have found earlier. I started with $$0 = \frac{512}{10}+\sum_{n=1}^{\infty} 2048(\pi^2n^2-6)\frac{(...
1
vote
0answers
166 views

The sum of the reciprocals of fourth powers [duplicate]

This problem is an extension of the well known basel problem and involves finding the sum of 1 + 1/16 + 1/81 ... = 1/1^4 + 1/2^4 + 1/3^4 ... 1/n^4 where n tends to infinity Euler managed to prove ...
0
votes
1answer
127 views

Calculate $\sum_{n=1}^\infty \frac{1}{n^4}$. [duplicate]

Calculate $\sum_{n=1}^\infty \frac{1}{n^4}$. Remark: I know that $\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}$, but not how to prove that, I totally stalled.
3
votes
0answers
119 views

Compute $\sum_{n=1}^{\infty}\frac{1}{n^4}$ using Parseval identity [duplicate]

Starting with Fourier cosine series for $f(x)=x^2$ in the interval $(0,l),$ use Parseval identity to compute $\sum_{n=1}^{\infty}\frac{1}{n^4}$. Attempt: I have that the coefficient $A_n$ for cosine ...
2
votes
0answers
100 views

Extending the Basel problem for 1/n^4 [duplicate]

Is there any way to prove that: $$ \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90} $$ Using Euler's original proof of equating coefficients?
2
votes
0answers
90 views

$\zeta(4)=\sum_ {k=1}^{\infty}{\frac{1}{k^4}}$ [duplicate]

How to Find $$\zeta(4)=\sum_ {k=1}^{\infty}{\frac{1}{k^4}}$$ the most basic way possible? I know it's $\pi^4/90$ but to arrive at this figure? Curious, because I need it to solve the integral $\int_{...
0
votes
1answer
48 views

Sum of the following infinite series [duplicate]

My question may seem to be too silly as I am not quite in touch with these things since long, for which I apologise in advance. How should I proceed to calculate the following? $$\sum_{n=0}^\infty\...
2
votes
0answers
61 views

How come pi is in this question? [duplicate]

I was doing my homework assignment and I did this question correctly. However, I'm interested in knowing the reason behind the logic that how $1/1^4 + 1/2^4 + 1/3^4$ .... to infinity evaluates to $pi^...
1
vote
0answers
32 views

Infinite Sum Convergence [duplicate]

What value does the infinite sum $\sum_{k=1}^{\infty} \frac{1}{k^4} $ converge to? I know that $\sum_{k=1}^{\infty} \frac{1}{k^2} $ converges to $\frac{\pi^2}{6}$ but I simply do not have a clue as to ...
673
votes
43answers
96k views

Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ (Basel problem)

As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem) $$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}.$$ However, Euler was Euler ...
43
votes
5answers
4k views

Computing $\zeta(6)=\sum\limits_{k=1}^\infty \frac1{k^6}$ with Fourier series.

Let $ f$ be a function such that $ f\in C_{2\pi}^{0}(\mathbb{R},\mathbb{R}) $ (f is $2\pi$-periodic) such that $ \forall x \in [0,\pi]$: $$f(x)=x(\pi-x)$$ Computing the Fourier series of $f$ and ...
26
votes
3answers
1k views

interesting square of log sin integral

I ran across this challenging log sin integral and am wondering what may be a good approach. $$ \int_{0}^{\frac{\pi}{2}}x^{2}\ln^{2}(2\cos(x))dx=\frac{11{{\pi}^{5}}}{1440} $$ This looks like it ...
15
votes
5answers
655 views

The other ways to calculate $\int_0^1\frac{\ln(1-x^2)}{x}dx$

Prove that $$\int_0^1\frac{\ln(1-x^2)}{x}dx=-\frac{\pi^2}{12}$$ without using series expansion. An easy way to calculate the above integral is using series expansion. Here is an example \begin{...

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