0
$\begingroup$

Taken from Soo T. Tan's Calculus textbook Chapter 9.7 Exercise 27-

Define $$a_n=\frac{2\cdot 4\cdot 6\cdot\ldots\cdot 2n}{3\cdot 5\cdot7\cdot\ldots\cdot (2n+1)}$$ One needs to prove the convergence or divergence of the series $$\sum_{n=1}^{\infty} a_n$$

upon finding the radius of convergence for $\sum_{n=1}^{\infty}\frac{2\cdot 4\cdot 6\cdot\ldots\cdot 2n}{3\cdot 5\cdot7\cdot\ldots\cdot (2n+1)}\cdot x^{2n+1}$ to be $1$ and checking the endpoints. Also, please use tests and methods that are taught in introductory courses.

Answers show divergence but no without explanation.

$\endgroup$
  • $\begingroup$ Do you mean the series $\sum_{n=1}^{\infty} a_n$ or the sequence $\{a_n\}_{n=1}^{\infty}$? $\endgroup$ – lulu Aug 27 '16 at 0:47
  • 1
    $\begingroup$ Show that $a_n > \frac 1n$ when $n>2$ And the series diverges by the comparison test. $\endgroup$ – Doug M Aug 27 '16 at 0:50
  • $\begingroup$ I mean series couldn't find the sigma button, sorry. @DougM Thanks $\endgroup$ – idanp Aug 27 '16 at 0:57
  • $\begingroup$ @idanp I edited your post to include the summation symbol. If you are interested, click on the "edit" button to see the syntax I used. $\endgroup$ – lulu Aug 27 '16 at 0:59
  • $\begingroup$ Thank you. @DougM is there a quick way to show it is larger than the harmonic series without induction? $\endgroup$ – idanp Aug 27 '16 at 1:06
8
$\begingroup$

Rewrite the $n$th term by sliding each of the factors in the numerator one position to the left. This gives $$ a_n = \frac 21\frac43\frac65\cdots\frac{2n}{2n-1}\frac1{2n+1}. $$ We now see $a_n$ is a product consisting of factors bigger than one, multiplied onto the final factor $\frac1{2n+1}$. Conclude $$ a_n>\frac1{2n+1}, $$ so the series $\sum a_n$ diverges.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Great, thanks a lot $\endgroup$ – idanp Aug 27 '16 at 1:34
3
$\begingroup$

Let

$$ a = \frac{2}{3} \cdot \frac{4}{5} \cdots \frac{2n}{2n+1} , \quad b = \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n-1}{2n} $$

Then $a > b$ and $ab = \dfrac{1}{2n+1}$, so actually $a > \dfrac{1}{\sqrt{2n+1}}$ - stronger than what you needed.

In other words: You can use this to prove that even $\sum a_n^2$ diverges, which is stronger than the original question.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

By using Euler's beta function we have $$ \frac{(2n)!!}{(2n+1)!!} = \frac{4^n n!^2}{(2n+1)!} = 4^n B(n+1,n+1) $$ hence: $$\begin{eqnarray*} \sum_{n=1}^{N}\frac{(2n)!!}{(2n+1)!!} = \int_{0}^{1}\sum_{n=1}^{N}\left(4x(1-x)\right)^n\,dx&=&\int_{-1/2}^{1/2}\sum_{n=1}^{N}(1-4x^2)^n\,dx\\&=&\int_{0}^{1}\sum_{n=1}^{N}(1-x^2)^n\,dx\end{eqnarray*}$$ but the last integrand function, over the interval $(0,1)$, is bounded below by $$ \max\left(0,N-\frac{N(N+1)}{2}x^2\right) $$ hence it follows that $$ \sum_{n=1}^{N}\frac{(2n)!!}{(2n+1)!!} \geq \int_{0}^{\sqrt{\frac{2}{N+1}}}\left(N-\frac{N(N+1)}{2}x^2\right)\,dx = \frac{2\sqrt{2}\,N}{3\sqrt{N+1}}$$ and the original series is divergent. By telescoping we also have the identity $$ \sum_{n=1}^{N}\frac{(2n)!!}{(2n+1)!!} = -2+\frac{\Gamma(N+2)}{\Gamma\left(N+\frac{3}{2}\right)}\sqrt{\pi} $$ and the improved bound $$\boxed{\; \sum_{n=1}^{N}\frac{(2n)!!}{(2n+1)!!} > -2+\sqrt{\pi(N+1)}\;}$$ follows from Gautschi's inequality.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Another way to look at the problem could be to consider that $$a_n=\frac {\prod_{i=1}^{n }(2i) }{\prod_{i=1}^{n }(2i+1) }$$ Using the ratio test $$\frac{a_{n+1}}{a_n}=\frac{2 (n+1)}{2 n+3}$$ which is inconclusive.

Using Raabe's test $$n\left(\frac{a_n}{a_{n+1}} -1\right)=\frac{n}{2 n+2}$$ its limit is $\frac 12$ and then divergence.

Another way, using

$$a_n=\frac {\prod_{i=1}^{n }(2i) }{\prod_{i=1}^{n }(2i+1) }=\frac {2^n\,n!}{\frac{2^{-n} (2 n+1)!}{n!}}=\frac{4^n\, (n!)^2}{(2n+1)! }$$ and using, for large values of $n$, Stirling approximation for the factorial leads to $$\log(a_n)=\left(-\frac{1}{2} \log (n)+\log \left(\frac{\sqrt{\pi }}{2}\right)\right)+O\left(\frac{1}{n}\right)$$ Using the fact that $a_n=e^{\log(a_n)}$, this leads to $$a_n=\frac{1}{2} \sqrt{\pi } \sqrt{\frac{1}{n}}+O\left(\frac{1}{n}\right)$$ which is divergent by comparaison to the $p$-series.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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