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Question 50 of 65

Q.Evaluate: lim(n→∞) [1/√(n²-1²) + 1/√(n²-2²) + ... + 1/√(n²-(n-1)²)].

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2023Subjective· 5mImportance★★★★★
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Rewrite the sum so each term has a factor 1/n1/n and depends on r/nr/n — this converts the limit of a sum into a definite integral (the definition of the Riemann sum).

Step 1 — factor out nn from inside the square root.

1n2−r2=1n1−(r/n)2.\frac{1}{\sqrt{n^2-r^2}} = \frac{1}{n\sqrt{1-(r/n)^2}}.

Step 2 — write the sum as a Riemann sum.

Sn=∑r=1n−11n2−r2=1n∑r=1n−111−(r/n)2.S_n=\sum_{r=1}^{n-1}\frac{1}{\sqrt{n^2-r^2}} = \frac1n\sum_{r=1}^{n-1}\frac{1}{\sqrt{1-(r/n)^2}}.

This has the form 1n∑g(r/n)\dfrac1n\displaystyle\sum g(r/n) with g(t)=11−t2g(t)=\dfrac{1}{\sqrt{1-t^2}}, which as n→∞n\to\infty converges to ∫01g(t) dt\displaystyle\int_0^1 g(t)\,dt.

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