diff --git a/curriculum/challenges/english/10-coding-interview-prep/rosetta-code/sum-of-a-series.md b/curriculum/challenges/english/10-coding-interview-prep/rosetta-code/sum-of-a-series.md
index 69fdff5699..3e7e0b4e4f 100644
--- a/curriculum/challenges/english/10-coding-interview-prep/rosetta-code/sum-of-a-series.md
+++ b/curriculum/challenges/english/10-coding-interview-prep/rosetta-code/sum-of-a-series.md
@@ -8,7 +8,7 @@ dashedName: sum-of-a-series
 
 # --description--
 
-Compute the **n**<sup>th</sup> term of a [series](<https://en.wikipedia.org/wiki/Series (mathematics)>), i.e. the sum of the **n** first terms of the corresponding [sequence](https://en.wikipedia.org/wiki/sequence). Informally this value, or its limit when **n** tends to infinity, is also called the *sum of the series*, thus the title of this task. For this task, use: $S*n = \\sum*{k=1}^n \\frac{1}{k^2}$ and compute $S\_{1000}$ This approximates the [zeta function](<https://en.wikipedia.org/wiki/Riemann zeta function>) for S=2, whose exact value $\\zeta(2) = {\\pi^2\\over 6}$ is the solution of the [Basel problem](<https://en.wikipedia.org/wiki/Basel problem>).
+Compute the **n**<sup>th</sup> term of a [series](<https://en.wikipedia.org/wiki/Series (mathematics)>), i.e. the sum of the **n** first terms of the corresponding [sequence](https://en.wikipedia.org/wiki/sequence). Informally this value, or its limit when **n** tends to infinity, is also called the *sum of the series*, thus the title of this task. For this task, use: $S_n = \displaystyle\sum_{k=1}^n \frac{1}{k^2}$.
 
 # --instructions--