# Thread: Find a value of a series

1. ## Find a value of a series

Can we find the value of the below series?

$\sum_{n=1}^{\infty}2n\int_{0}^{1}(1-t^2)^{n}t^{2n-1}dt=?$

2. The way I rewrote the integral had a fatal algebraic mistake.
Thanks Mr. F

What I wrote would give

$
\int_{0}^{1}(1-t^2)^nt^{2n-1}dt \ne =\int_{0}^{1}\int_{t^2}^{1}ns^{n-1}t^{2n-1}dsdt=\int_{0}^{1}(1^n-(t^2)^n))t^{2n-1}dt
$

3. Originally Posted by TheEmptySet
I don't think the sum converges

lets focus on the integral for a bit

$\int_{0}^{1}(1-t^2)^nt^{2n-1}dt=\int_{0}^{1}\int_{t^2}^{1}ns^{n-1}t^{2n-1}dsdt$

Perhaps I'm missing something but this doesn't seem correct:

INT{t^2, 1}(n*s^(n-1))ds = s^n|{t2,1} = 1^n - (t^2)^n = 1 - t^(2n), which doesn't equal (1 - t^2)^n

In fact, (1 - t^2)^n = 1 + INT{0,n}[(1-t^2)^x*ln(1-t^2)dx], which is a pretty disgusting expression, imo.

Tonio

Now switching the order of integration gives

$=\int_{0}^{1}\int_{0}^{\sqrt{s}}ns^{n-1}t^{2n-1}dtds=\frac{1}{2}\int_{0}^1 s^{n-1}t^{2n} \bigg|_{0}^{\sqrt{s}}ds$

$=\frac{1}{2} \int_{0}^{1}s^{2n-1}ds=\frac{1}{4n}$

So the series becomes

$\sum_{n=1}^{\infty}2n \cdot \frac{1}{4n}=\frac{1}{2}\sum_{n=1}^{\infty}1 = \infty$
......

4. Originally Posted by tonio
......
The integral seems to be way nastier than it looked at first, but taking a peek into my integral tables ("Tables of integrals, series and products", Gradshteyn and Ryzhik) its value is (1/2)B(n,n), where B is the betta function, and since B(x,y) = G(x)G(y)/G(x+y), with G = the gamma function, and since the gamma function is the factorial function, we get:

B(n,n) = G(n)^2/G(2n) = (n!)^2/(2n)! ==> the integral is (n!)^2/2(2n)! ==> the exercise gives the infinite series of
n(n!)^2/(2n)! and using the quotient test we get a_(n+1)/a_n --> 1/4 ==> the series converges.

About its value I'll pass for now.

Tonio

5. Originally Posted by Xingyuan
Can we find the value of the below series?

$\sum_{n=1}^{\infty}2n\int_{0}^{1}(1-t^2)^{n}t^{2n-1}dt=?$
The integral can be written $\frac{1}{2} \int_0^1 (t^2)^{n-1} (1 - t^2)^n d(t^2)$ which is recognised as a Beta function: Beta function - Wikipedia, the free encyclopedia

$\frac{1}{2} \int_0^1 (t^2)^{n-1} (1 - t^2)^n d(t^2) = \frac{1}{2} B(n, n+1)$.

So the job is to calculate $\frac{1}{2} \sum_{n=1}^{+\infty} (n B(n, n+1)) = \frac{1}{2} \sum_{n=1}^{+\infty} \frac{(n!)^2}{(2n)!}$.

6. $n\beta (n,n+1)=n\cdot \frac{\Gamma (n)\Gamma (n+1)}{\Gamma (2n+1)}=n\int_{0}^{1}{t^{n-1}(1-t)^{n}\,dt},$ now swap integral and series, compute the series and finally compute the remaining integral.

in fact, we could've done this immediately given the problem, just swap series and integral and then compute the series. (Find a formula for $\sum_{n=1}^\infty n a^n$ and then solve the remaining integral.)

7. Originally Posted by Xingyuan
Can we find the value of the below series?

$\sum_{n=1}^{\infty}2n\int_{0}^{1}(1-t^2)^{n}t^{2n-1}dt=?$
I agree with Krizalid:

$2n(1-t^2)^{n}t^{2n-1}=\frac{2n}{t}(1-t^2)^n(t^2)^n$

Summing it up gives:

$\frac{2}{t}\sum_{n=1}^{\infty}n(t^2-t^4)^n$

$\sum_{n=1}^{\infty} na^n=\frac{a}{(1-a)^2}$ for $a<1$.

So the above sum is $\frac{2}{t}\cdot\frac{t^2-t^4}{(1-t^2+t^4)^2}$

Simplifying an integrating yields:

$2\int_0^1\frac{-t^3+t}{(t^4-t^2+1)^2}\,dt$

That doesn't look like much fun; I evaluated it on my TI-89 and got $\frac{2\pi\sqrt{3}+9}{27}$.