MathCastle
ChallengeSuperchallenge

Challenge · Superchallenge problem

Let

S=1−12+13−14+⋯+12025−12026.S = 1 - \frac12 + \frac13 - \frac14 + \cdots + \frac{1}{2025} - \frac{1}{2026}.

It turns out that SS is EXACTLY equal to a sum of consecutive unit fractions ending at 12026\dfrac{1}{2026}:

S=1k+1k+1+⋯+12026.S = \frac{1}{k} + \frac{1}{k+1} + \cdots + \frac{1}{2026}.

Find kk.

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