JEE Advanced 2023 Paper 2 · Mathematics · Sequences and Series · hard

Let S=Σk=1^10000(1)/(√ k). The greatest integer not exceeding S is

Numerical answer

Let

S=\displaystyle\sum_{k=1}^{10000}\frac{1}{\sqrt k}

. The greatest integer not exceeding

S

Show answer and solution

Answer: 198

For

k\ge1

\sqrt{k+1}-\sqrt k<\frac1{2\sqrt k}<\sqrt k-\sqrt{k-1}

. Summing gives

2(\sqrt{10001}-1)<S<1+2(\sqrt{10000}-1)=199

, i.e.

198.0<S<199

. Hence

\lfloor S\rfloor=198

(the actual value is

198.54\ldots

, confirmed numerically).

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