-1
$\begingroup$

Let $$T_n = 1 \cdot 3 \cdot \dots (2n+1)$$

What will be $\sum_{n=1}^N T_n$?

If we put value to understand we get $$ (1\cdot3 )+ (1\cdot3\cdot5 )+ (1\cdot3\cdot5\cdot7)\dots + (1\cdot3\cdot5\cdot7 \dots(2n+1)) $$

I have tried to do something, but couldn't even able to write first step.

$\endgroup$
3
  • $\begingroup$ So sounds like $T_n = \dfrac{(2n+1)!}{2^n n!}$ $\endgroup$
    – gt6989b
    Sep 19, 2022 at 13:20
  • 1
    $\begingroup$ What makes you think there is a closed-form expression for your sum (OEIS A076795$ -2$)? $\endgroup$
    – Henry
    Sep 19, 2022 at 13:22
  • $\begingroup$ @Henry thanks for the OEIS link -- it also gives the exponential generating function... $\endgroup$
    – gt6989b
    Sep 19, 2022 at 13:44

1 Answer 1

Reset to default
2
$\begingroup$

Just for your curiosity.

If you enjoy special functions $$S_N=\sum_{n=1}^N T_n=\sum_{n=1}^N \dfrac{(2n+1)!}{2^n \,n!}$$ $$S_N=\frac{2^{N+1} E_{N+\frac{5}{2}}\left(-\frac{1}{2}\right) \Gamma \left(N+\frac{5}{2}\right)}{\sqrt{e \pi }}-\frac{E_{\frac{3}{2}}\left(-\frac{1}{2}\right)}{2 \sqrt{e}}-1$$ where appear the exponential integral function.

It can also write $$S_N=\sqrt{\frac{\pi }{2 e}}\left(i+ \text{erfi}\left(\frac{1}{\sqrt{2}}\right)\right)-i\frac{ (-1)^N \Gamma \left(\frac{2N+5}{2}\right) \Gamma \left(-\frac{2N+3}{2},-\frac{1}{2}\right)}{\sqrt{2 e \pi }}-2$$

$\endgroup$
1
  • 1
    $\begingroup$ It is like a bouncer for me 😅but thanks $\endgroup$
    – OpenLearner
    Sep 25, 2022 at 12:41

Not the answer you're looking for? Browse other questions tagged .