Question

∑ (from n=1 to ∞) (1 / √(n + 1)) divergesb. What should n be in order that the partial sum sₙ = ∑ (from i=1 to n) (1 / √(i + 1)) satisfies sₙ \u003e 1000?

Accepted Answer

Recognize that the series $$ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}} $$ diverges, meaning its partial sums grow without bound as $$ n 	o \infty . $$Therefore, for any large number like 1000, there exists some $$ n $$ such that the partial sum $$ s_n = \sum_{i=1}^n \frac{1}{\sqrt{i+1}} $$ exceeds 1000. To estimate the value of $$ n $$ for which $$ s_n \u003e 1000 $$, use the integral test approximation. Since $$ f(x) = \frac{1}{\sqrt{x+1}}  is $$positive and decreasing for $$ x \geq 1 $$, the sum $$ s_n $$ can be approximated by the integral $$ \int_1^{n} \frac{1}{\sqrt{x+1}} \, dx . $$Set up the integral $$ I = \int_1^{n} \frac{1}{\sqrt{x+1}} \, dx . To $$solve this integral, perform the substitution $$ u = x + 1 $$, so $$ du = dx $$, and the limits change from $$ x=1  to$$ $$ u=2 $$, and $$ x=n  to$$ $$ u = n+1 . $$The integral becomes $$ \int_2^{n+1} u^{-1/2} \, du . $$Evaluate the integral $$ \int_2^{n+1} u^{-1/2} \, du = 2(\sqrt{n+1} - \sqrt{2}) . $$This gives an approximation for the partial sum $$ s_n  as$$ $$ s_n \approx 2(\sqrt{n+1} - \sqrt{2}) . $$Set the inequality $$ 2(\sqrt{n+1} - \sqrt{2}) \u003e 1000  to $$find the smallest $$ n $$ such that the partial sum exceeds 1000. Solve this inequality for $$ n  by $$isolating $$ \sqrt{n+1} $$ and then squaring both sides to find $$ n .$$

∑ (from n=1 to ∞) (1 / √(n + 1)) diverges
b. What should n be in order that the partial sum sₙ = ∑ (from i=1 to n) (1 / √(i + 1)) satisfies sₙ > 1000?

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Key Concepts

Infinite Series and Partial Sums

Divergence of a Series

Integral Test for Series Estimation

∑ (from n=1 to ∞) (1 / √(n + 1)) divergesb. What should n be in order that the partial sum sₙ = ∑ (from i=1 to n) (1 / √(i + 1)) satisfies sₙ > 1000?