From 05c690e1e65a729004b044b77bb7b0047b85ebf7 Mon Sep 17 00:00:00 2001 From: Muhammed Mustafa Date: Tue, 7 Jun 2022 10:54:18 +0200 Subject: [PATCH] fix(curriculum): external sum of the series link in CIP (#46375) --- .../10-coding-interview-prep/rosetta-code/sum-of-a-series.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) 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 3e7e0b4e4f3..21640c82bf7 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**th term of a [series](), 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: $S_n = \displaystyle\sum_{k=1}^n \frac{1}{k^2}$. +Compute the **n**th term of a series, i.e. the sum of the **n** first terms of the corresponding 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--