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Surd telescoping

Authors
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    Name
    Vu Hung
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Problem Statement

Let

ak=1k+1+k,Sn=∑k=1nak.a_k=\frac1{\sqrt{k+1}+\sqrt{k}},\qquad S_n=\sum_{k=1}^n a_k.
  • Show ak=k+1−ka_k=\sqrt{k+1}-\sqrt{k}.
  • Find SnS_n.
  • Evaluate S99S_{99}.
  • Find the smallest integer NN such that SN>10S_N>10.

Hints

Rationalize with the conjugate and then telescope.

Solutions

ak=k+1−k(k+1)−k=k+1−k.a_k=\frac{\sqrt{k+1}-\sqrt{k}}{(k+1)-k}=\sqrt{k+1}-\sqrt{k}.

Hence

Sn=(2−1)+⋯+(n+1−n)=n+1−1.S_n=(\sqrt2-\sqrt1)+\cdots+(\sqrt{n+1}-\sqrt n)=\sqrt{n+1}-1.

So S99=10−1=9S_{99}=10-1=9. For SN>10S_N>10:

N+1−1>10  ⟹  N>120,\sqrt{N+1}-1>10 \implies N>120,

smallest such integer is N=121N=121.

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