I am given the series $$\frac{1}{6} + \frac{1\cdot 8}{6\cdot 10} + \frac{1\cdot 8\cdot 15}{6\cdot 10\cdot 14}$$

I am asked to find a formula for this series $a_n$

So I have found that the last fraction multiplied at $nth$ term is $\frac{(1+7(n-1))}{(6+4(n-1))}$

But I am stuck, unsure how to proceed after this.


1 Answer 1


Try recursion.

You know that each term (after the first term, of course) is simply the previous term multiplied by a varying constant that relates to $n$.

So try something like:

$$a_1 = \frac{1}{6} $$ and $$a_n = a_{n-1} \cdot \frac{1+7(n-1)}{6+4(n-1)}.$$

  • 2
    $\begingroup$ (+1) This is extremely useful for applying the ratio test. $\endgroup$
    – robjohn
    Mar 15, 2021 at 2:23
  • $\begingroup$ Can you elaborate a bit more? The question asks me to use ratio test, which is why I am trying to find a formula for $a_n$ . I tried to substitute $a_n = \frac{1+7n}{6+4n} \cdot \frac{1+7(n-1)}{6+4(n-1)}$ and use ratio test to cancel out the $nth$ term. But I did not succeed $\endgroup$
    – Sirou Ewei
    Mar 15, 2021 at 2:29
  • 1
    $\begingroup$ Not to hijack robjohn's comment, but think about what you need to do to apply the ratio test, and think about how the $n+1^{th}$ term relates to the $n^{th}$ term in the recursive definition $\endgroup$
    – Greg V
    Mar 15, 2021 at 2:30
  • $\begingroup$ During ratio test, all the previous terms should cancel out, leaving only $\frac{(1+7(n-1))}{(6+4(n-1))}$. But this is incorrect? $\endgroup$
    – Sirou Ewei
    Mar 15, 2021 at 3:10
  • $\begingroup$ That looks spot on, except maybe an index or two being off. What's the next step of the ratio test? $\endgroup$
    – Greg V
    Mar 15, 2021 at 3:11

You must log in to answer this question.

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