1

$\{S_n\}= \sum_{k=1}^n\frac{1.3.5.7..(2k-1)}{3.6.9..(3k)}$

I applied the D Alembert Ratio test on:

$S_\infty=\frac{1}{3}+\frac{1.3}{3.6}+\frac{1.3.5}{3.6.9}+\frac{1.3.5.7..(2k-1)}{3.6.9..(3k)}$ as $k$ tends to $\infty$

so, $\frac{a_n}{a_{n+1}}=\frac{(3n+1).(3n+2).(3n+3)}{2n.(2n+1)}$ which will go to $\infty (>1)$ as $n$ tends $\infty$, so the series is convergent.

But how to find the actual limit of the series? Kindly help.

Anne Bauval
  • 34,650
санкет мхаске
  • 395
  • 2
    $\mathcal{S}n$ can be written as ${\mathcal{S}_n } = \sum{k=1}^n \frac{(2k)!}{6^k , (k!)^2} $ – Lucky Chouhan May 09 '23 at 05:12
  • sorry @AnneBauval I am not able to correlate the two questions – санкет мхаске May 09 '23 at 09:10
  • The correlation is direct: $1+\lim S_n=\sum_{k=0}^\infty\binom{2k}k(1/6)^k=\frac1{\sqrt{1-4/6}}.$ – Anne Bauval May 09 '23 at 09:14
  • @AnneBauval I could spot the correlation by writing first few terms but still I am not able to follow the solution there from the start.. For example I dont know how to make sense of $C_{2n}^{n}$ ...how can we select $2n$ items from $n$ items...thats how I know it – санкет мхаске May 09 '23 at 09:31
  • You don't need that combinatorial interpretation (btw, $C_{2n}^n=\binom{2n}n$ is select $n$ items from $2n$, not the converse). Simply remember that $\binom mk=\frac{m!}{k!(n-k)!}$ hence the $S_n=\sum_{k=1}^n \frac{(2k)!}{6^k , (k!)^2}$ in the first comment above becomes $S_n=\sum_{k=1}^n\binom{2k}k\frac1{6^k}.$ – Anne Bauval May 09 '23 at 09:50
  • @AnneBauval I could understand the outline of the solution mentioned in the post. But still I was wondering if there is a solution which does not use calculus, and without solving differential equations...I mean a more direct solution without first generalizing the problem.. – санкет мхаске May 10 '23 at 03:47
  • 2
    That proof is so usual that I doubt there is a simpler one. See also https://en.wikipedia.org/wiki/Binomial_series#Summation_of_the_binomial_series – Anne Bauval May 10 '23 at 06:00

0 Answers0