I need help finding the kth partial sum of the question below:

Consider the series $\sum_{n=1}^{\infty}a_n$ where $a_n=\ln\left(\frac{6n-1}{6n+5}\right)$. Find a formula, in closed form, for the $k^{th}$ partial sum $s_k$.

I think this is a telescoping series but I'm sorta lost on how to answer the question. Any help would be appreciated

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1 Answer 1


$$\begin{aligned}a_n \;=\;& \ln\left(\frac{6n-1}{6n+5}\right) \;=\; \ln(6n-1) \;-\; \ln(6n+5) \\[5mm] =\;& \ln(6n-1) \;-\; \ln(6(n+1)-1)\end{aligned}$$

Which means that $\sum a_n$ is a telescoping sum, giving us:

$$\sum^n_{k=1}a_k \;=\; \ln(5) \;-\; \ln(6(n+1)-1)$$

  • $\begingroup$ When $n=2$, your final expression reads as $a_1+a_2=a_1-a_2$. Looks weird to me. $\endgroup$ Mar 10, 2021 at 16:02
  • $\begingroup$ Thank you for pointing that out, I changed it $\endgroup$
    – SimonK
    Mar 10, 2021 at 16:09
  • $\begingroup$ Now I agree, +1 :) $\endgroup$ Mar 10, 2021 at 16:13
  • $\begingroup$ Sounded right in my head; obviously wasn't. I guess thinking of it as "all but first and last cancel" can lead to some confusion $\endgroup$
    – SimonK
    Mar 10, 2021 at 16:17

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