New Notes Blog: `sum_(n=1)^oo 1/n^(1/2)` Use the Integral Test to determine the convergence or divergence of the p-series.

Saturday, 23 May 2015

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{n}}^{{\frac{{1}}{{2}}}}}$ Use the Integral Test to determine the convergence or divergence of the p-series.

The Integral test is applicable if f is positive and decreasing function on the infinite interval $\displaystyle{\left[{k},\infty\right)}$ where $\displaystyle{k}\geq{1}$ and $\displaystyle{a}_{{n}}={f{{\left({x}\right)}}}$ . Then the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{a}_{{n}}$ converges if and only if the improper integral $\displaystyle{\int_{{1}}^{\infty}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ converges. If the integral diverges then the series also diverges.

For the given series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{n}}^{{\frac{{1}}{{2}}}}}$ , then $\displaystyle{a}_{{n}}=\frac{{1}}{{{n}}^{{\frac{{1}}{{2}}}}}$ then applying $\displaystyle{a}_{{n}}={f{{\left({x}\right)}}}$ , we consider:

$\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{x}}^{{\frac{{1}}{{2}}}}}$ .

As shown on the graph of $\displaystyle{f{{\left({x}\right)}}}$ , the function is positive on the interval $\displaystyle{\left[{1},\infty\right)}$ . As x at the denominator side gets larger, the function value decreases.

Therefore, we may determine the convergence of the improper integral as:

$\displaystyle{\int_{{1}}^{\infty}}\frac{{1}}{{{x}}^{{\frac{{1}}{{2}}}}}=\lim_{{{t}-\gt\infty}}{\int_{{1}}^{{t}}}\frac{{1}}{{{x}}^{{\frac{{1}}{{2}}}}}{\left.{d}{x}\right.}$

Apply Law of exponent: $\displaystyle\frac{{1}}{{{x}}^{{m}}}={{x}}^{{-{m}}}$ .

$\displaystyle\lim_{{{t}-\gt\infty}}{\int_{{1}}^{{t}}}\frac{{1}}{{{x}}^{{\frac{{1}}{{2}}}}}{\left.{d}{x}\right.}=\lim_{{{t}-\gt\infty}}{\int_{{1}}^{{t}}}{{x}}^{{-\frac{{1}}{{2}}}}{\left.{d}{x}\right.}$

Apply Power rule for integration: $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}$ .

$\displaystyle\lim_{{{t}-\gt\infty}}{\int_{{1}}^{{t}}}{{x}}^{{-\frac{{1}}{{2}}}}{\left.{d}{x}\right.}=\lim_{{{t}-\gt\infty}}{\left[\frac{{{x}}^{{-\frac{{1}}{{2}}+{1}}}}{{-\frac{{1}}{{2}}+{1}}}\right]}{{\mid}_{{1}}^{{t}}}$

$\displaystyle=\lim_{{{t}-\gt\infty}}{\left[\frac{{{x}}^{{\frac{{1}}{{2}}}}}{{\frac{{1}}{{2}}}}\right]}{{\mid}_{{1}}^{{t}}}$

$\displaystyle=\lim_{{{t}-\gt\infty}}{\left[{{x}}^{{\frac{{1}}{{2}}}}\cdot{\left(\frac{{2}}{{1}}\right)}\right]}{{\mid}_{{1}}^{{t}}}$

$\displaystyle=\lim_{{{t}-\gt\infty}}{\left[{2}{{x}}^{{\frac{{1}}{{2}}}}\right]}{{\mid}_{{1}}^{{t}}}$

or $\displaystyle\lim_{{{t}-\gt\infty}}{\left[{2}\sqrt{{{x}}}\right]}{{\mid}_{{1}}^{{t}}}$

Apply definite integral formula: $\displaystyle{F}{\left({x}\right)}{{\mid}_{{a}}^{{b}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$ .

$\displaystyle\lim_{{{t}-\gt\infty}}{\left[{2}\sqrt{{{x}}}\right]}{{\mid}_{{1}}^{{t}}}=\lim_{{{t}-\gt\infty}}{\left[{2}\sqrt{{{t}}}-{2}\sqrt{{{1}}}\right]}$

$\displaystyle=\lim_{{{t}-\gt\infty}}{\left[{2}\sqrt{{{t}}}-{2}\cdot{1}\right]}$

$\displaystyle=\lim_{{{t}-\gt\infty}}{\left[{2}\sqrt{{{t}}}-{2}\right]}$

$\displaystyle=\infty$

Note: $\displaystyle\lim_{{{t}-\gt\infty}}{\left(-{2}\right)}=-{2}$ and $\displaystyle\lim_{{{t}-\gt\infty}}{2}\sqrt{{{t}}}=\infty$ then $\displaystyle\infty-{2}\approx\infty$ .

The $\displaystyle\lim_{{{t}-\gt\infty}}{\left[{2}\sqrt{{{x}}}\right]}{{\mid}_{{1}}^{{t}}}=\infty$ implies that the integral diverges.

Conclusion: The integral $\displaystyle{\int_{{1}}^{\infty}}\frac{{1}}{{{x}}^{{\frac{{1}}{{2}}}}}$ diverges, therefore the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{n}}^{{\frac{{1}}{{2}}}}}$ must also diverge.

New Notes Blog

Saturday, 23 May 2015

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{n}}^{{\frac{{1}}{{2}}}}}$ Use the Integral Test to determine the convergence or divergence of the p-series.

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Saturday, 23 May 2015

Use the Integral Test to determine the convergence or divergence of the p-series.

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{n}}^{{\frac{{1}}{{2}}}}}$ Use the Integral Test to determine the convergence or divergence of the p-series.

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?