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$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{n^2+n+4372}{(2n+1)^7(n+1)}\right]=\frac{61}{184320}\pi^7\tag1$$

Step 1:

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{4372}{(2n+1)^7(n+1)}\right]+\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{n^2+n}{(2n+1)^7(n+1)}\right]\tag2$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{4372}{(2n+1)^7(n+1)}\right]+\sum_{n=2}^{\infty}(-1)^n\left[\frac{1}{(2n+1)^7}\right]\tag3$$

Recall $$\sum_{n=0}^{\infty}(-1)^n\frac{1}{(2n+1)^7}=\frac{61}{184320}\pi^7$$

It looks like that this sum $$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{4372}{(2n+1)^7(n+1)}\right]$$ is a rational number

I am not able to show that sum $(1)=\frac{61}{184320}\pi^7$

Any help.Thank you!

Dragon
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  • Does this come from trying to evaluate the Riemann zeta function at some specific values (ie $s=7$)? Is that where your recall fact originates? – quarague Feb 04 '19 at 11:41
  • You can try Telescoping series. – user90369 Feb 04 '19 at 11:46
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2 Answers2

2

It comes from telescoping series:

$$ \frac{(-1)^n}n\left(\frac{1}{(2 n+1)^7 (n+1)}+\frac{1}{(2 n-1)^7 (n-1)}\right) = (-1)^n\left(\frac{1}{n-1} + \frac{1}{n}\right) + (-1)^n\left(\frac{1}{n}+\frac{1}{n+1}\right) - 4(-1)^n\left(\frac{1}{(2 n-1)^7}+\frac{1}{(2 n+1)^7}\right) - 4(-1)^n\left(\frac{1}{(2 n-1)^5}+\frac{1}{(2 n+1)^5}\right) - 4(-1)^n\left(\frac{1}{(2 n-1)^3}+\frac{1}{(2 n+1)^3}\right) - 4(-1)^n\left(\frac{1}{2 n-1}+\frac{1}{2 n+1}\right) $$

Each parenthesis has two terms. The second term of $n=n_0$ is equal to the first term of $n=n_0+1$ with the opposite sign (coming from $(-1)^n$). So only the first term of $n=2$ survives.

Edit. P.S. By the way, it's not necessary to calculate partial fractions, if you just want to show the rationality. We can see that the function is odd ($n\to -n$), so there are no fractions with even exponent and all $(2n\pm1)^{-k}$ parentheses have the same coefficient. We only need to calculate the coefficients of $n^{-1}$, $(n+1)^{-1}$ and $(n-1)^{-1}$. We can easily do it with multiplying by corresponing monomial and letting $n=-1,0,1$.

Vasily Mitch
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2

Let $a_n$ be $$ a_n= \frac{4372}{(2n-1)^7\cdot n(n-1)}\ . $$ Then we sum $(-1)^n(a_n+a_{n+1})$.

dan_fulea
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