Notes: `sum_(n=1)^oo (n+2)/(n+1)` Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the...

Sunday, 22 January 2017

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{{n}+{2}}}{{{n}+{1}}}$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the...

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{{n}+{2}}}{{{n}+{1}}}$

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

For the given series $\displaystyle{a}_{{n}}=\frac{{{n}+{2}}}{{{n}+{1}}}$

Consider $\displaystyle{f{{\left({x}\right)}}}=\frac{{{x}+{2}}}{{{x}+{1}}}$

Refer to the attached graph of the function. From the graph, we observe that the function is positive , continuous and decreasing for $\displaystyle{x}\ge{1}$

We can also determine whether the function...

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{{n}+{2}}}{{{n}+{1}}}$

For the given series $\displaystyle{a}_{{n}}=\frac{{{n}+{2}}}{{{n}+{1}}}$

Consider $\displaystyle{f{{\left({x}\right)}}}=\frac{{{x}+{2}}}{{{x}+{1}}}$

Refer to the attached graph of the function. From the graph, we observe that the function is positive , continuous and decreasing for $\displaystyle{x}\ge{1}$

We can also determine whether the function is decreasing by finding its derivative f'(x).

Apply quotient rule to find f'(x),

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

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

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

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

$\displaystyle{f{'}}{\left({x}\right)}<{0}$ which implies that f(x) is decreasing for $\displaystyle{x}\ge{1}$

We can apply integral test, as the function satisfies all the conditions for the integral test.

Now let's determine whether the corresponding improper integral $\displaystyle{\int_{{1}}^{\infty}}\frac{{{x}+{2}}}{{{x}+{1}}}{\left.{d}{x}\right.}$ converges or diverges,

$\displaystyle{\int_{{1}}^{\infty}}\frac{{{x}+{2}}}{{{x}+{1}}}{\left.{d}{x}\right.}=\lim_{{{b}\to\infty}}{\int_{{1}}^{{b}}}\frac{{{x}+{2}}}{{{x}+{1}}}{\left.{d}{x}\right.}$

Let's first evaluate the indefinite integral,

$\displaystyle\int\frac{{{x}+{2}}}{{{x}+{1}}}{\left.{d}{x}\right.}=\int\frac{{{x}+{1}+{1}}}{{{x}+{1}}}{\left.{d}{x}\right.}$

$\displaystyle=\int{\left({1}+\frac{{1}}{{{x}+{1}}}\right)}{\left.{d}{x}\right.}$

Apply the sum rule,

$\displaystyle=\int{1}{\left.{d}{x}\right.}+\int\frac{{1}}{{{x}+{1}}}{\left.{d}{x}\right.}$

$\displaystyle={x}+{\ln}{\left|{x}+{1}\right|}+{C}$

$\displaystyle{\int_{{1}}^{\infty}}\frac{{{x}+{2}}}{{{x}+{1}}}{\left.{d}{x}\right.}=\lim_{{{b}\to\infty}}{{\left[{x}+{\ln}{\left|{x}+{1}\right|}\right]}_{{1}}^{{b}}}$

$\displaystyle=\lim_{{{b}\to\infty}}{\left[{b}+{\ln}{\left|{b}+{1}\right|}\right]}-{\left({1}+{\ln}{\left|{1}+{1}\right|}\right)}$

$\displaystyle=\lim_{{{b}\to\infty}}{\left[{b}+{\ln}{\left|{b}+{1}\right|}\right]}-{\left({1}+{\ln{{2}}}\right)}$

$\displaystyle=\infty-{\left({1}+{\ln{{2}}}\right)}$

$\displaystyle=\infty$

Since the integral $\displaystyle{\int_{{1}}^{\infty}}\frac{{{x}+{2}}}{{{x}+{1}}}{\left.{d}{x}\right.}$ diverges, we conclude from the integral test that the series diverges.

Notes

Sunday, 22 January 2017

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{{n}+{2}}}{{{n}+{1}}}$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the...

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Sunday, 22 January 2017

Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the...

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{{n}+{2}}}{{{n}+{1}}}$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the...

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?