1
$\begingroup$

Examine convergence of the series $$1-\frac{1}{2}+\frac{1\cdot3}{2\cdot4}-\frac{1\cdot3\cdot5}{2\cdot4\cdot6}+\cdots$$

$\endgroup$
3
  • 1
    $\begingroup$ Use Stirling's approximation. $\endgroup$ Jul 13, 2020 at 5:08
  • 3
    $\begingroup$ @KaviRamaMurthy: I think things are much easier here ;) $\endgroup$
    – metamorphy
    Jul 13, 2020 at 5:13
  • $\begingroup$ What have you tried so far? Where are you stuck? $\endgroup$
    – saulspatz
    Jul 13, 2020 at 5:20

2 Answers 2

0
$\begingroup$

Let's denote $a_n=\frac{1 \cdot 3 \cdot \cdots \cdot (2n-1)}{2 \cdot 4 \cdot \cdots \cdot (2n)}$ and consider: $$\frac{a_n}{a_{n+1}}=1+\frac{1}{2n+1}$$ So by Gauss test we have absolute divergence and conditional convergence by Leibniz test and for limit 0

$\endgroup$
2
  • $\begingroup$ For the Leibniz test we need to show $a_n\to0$. I don't see how this follows from the above. $\endgroup$
    – saulspatz
    Jul 13, 2020 at 5:34
  • $\begingroup$ @saulspatz. Added to answer (it automatically appeared to me on the right column). $\endgroup$
    – zkutch
    Jul 13, 2020 at 5:43
-1
$\begingroup$

You have

$$a_n=(-1)^n\frac{\prod_{k=1}^n (2k-1) } {\prod_{k=1}^n (2k) }=\binom{-\frac{1}{2}}{n}$$ and then $$S=\sum_{n=0}^\infty a_n x^n=\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n} x^n=\frac{1}{\sqrt{1+x}}$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.