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integral test
- Thread starter cazza90
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D
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cazza90 said:use the integral test to determine convergence of this series (sum of n=1 to infinity): 1/((n^2) - 4n + 5)
Factorize the denominator - use partial fraction - then integrate.
Please show your work, indicating exactly where you are stuck - so that we may know where to begin to help you.
BigGlenntheHeavy
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\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2-4n+5} \ (does \ it \ converge?), \ let \ f(n) \ = \ \frac{1}{n^2-4n+5}, \ then\)
\(\displaystyle f(x) \ = \ \frac{1}{x^2-4x+5} \ \implies \ \lim_{b\to\infty}\int_{1}^{b}\frac{dx}{x^2-4x+5} \ = \ \lim_{b\to\infty}\int_{1}^{b}\frac{dx}{(x-2)^2+1}.\)
\(\displaystyle Since \ f(x) \ is \ decreasing \ for \ all \ x \ > \ 2, \ integral \ test \ is \ appropriate.\)
\(\displaystyle Hence, \ let \ u \ = \ x-2, \ du \ = \ dx.\)
\(\displaystyle \lim_{b\to\infty}\int_{-1}^{b}\frac{du}{u^2+1}, \ again \ let \ u \ = \ tan(\theta), \ then \ du \ = \ sec^2(\theta)d\theta.\)
\(\displaystyle Ergo, \ we \ have \ \int_{-\pi/4}^{\pi/2}\frac{sec^2(\theta)d\theta}{sec^2(\theta)} \ = \ \theta\bigg]_{-\pi/4}^{\pi/2}= \ \frac{3\pi}{4}, \ hence \ converges.\)
\(\displaystyle Note: \ Partial \ fractions \ won't \ get \ it \ (in \ real \ number \ land).\)
\(\displaystyle f(x) \ = \ \frac{1}{x^2-4x+5} \ \implies \ \lim_{b\to\infty}\int_{1}^{b}\frac{dx}{x^2-4x+5} \ = \ \lim_{b\to\infty}\int_{1}^{b}\frac{dx}{(x-2)^2+1}.\)
\(\displaystyle Since \ f(x) \ is \ decreasing \ for \ all \ x \ > \ 2, \ integral \ test \ is \ appropriate.\)
\(\displaystyle Hence, \ let \ u \ = \ x-2, \ du \ = \ dx.\)
\(\displaystyle \lim_{b\to\infty}\int_{-1}^{b}\frac{du}{u^2+1}, \ again \ let \ u \ = \ tan(\theta), \ then \ du \ = \ sec^2(\theta)d\theta.\)
\(\displaystyle Ergo, \ we \ have \ \int_{-\pi/4}^{\pi/2}\frac{sec^2(\theta)d\theta}{sec^2(\theta)} \ = \ \theta\bigg]_{-\pi/4}^{\pi/2}= \ \frac{3\pi}{4}, \ hence \ converges.\)
\(\displaystyle Note: \ Partial \ fractions \ won't \ get \ it \ (in \ real \ number \ land).\)
D
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Note: Partial fraction will work because:
\(\displaystyle \int\frac{du}{u^2+1} = \frac{1}{2i}ln\left (\frac{u-i}{u+i}\right ) \ + C_1 \ = \ tan^{-1}(u) \ + \ C_2\)
\(\displaystyle \int\frac{du}{u^2+1} = \frac{1}{2i}ln\left (\frac{u-i}{u+i}\right ) \ + C_1 \ = \ tan^{-1}(u) \ + \ C_2\)
BigGlenntheHeavy
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\(\displaystyle \lim_{b\to\infty}\int_{-1}^{b}\frac{du}{u^2+1} \ = \ \lim_{b\to\infty}\bigg[arctan(u)\bigg]_{-1}^{b} \ = \ \frac{3\pi}{4}.\)
\(\displaystyle \lim_{b\to\infty}\int_{-1}^{b}\frac{du}{u^2+1} \ = \ \lim_{b\to\infty}\int_{-1}^{b}\frac{du}{(u+i)(u-i)} \ What \ is \ the \ point?\)
\(\displaystyle Why \ interject \ imaginary \ numbers?\)
\(\displaystyle \lim_{b\to\infty}\int_{-1}^{b}\frac{du}{u^2+1} \ = \ \lim_{b\to\infty}\int_{-1}^{b}\frac{du}{(u+i)(u-i)} \ What \ is \ the \ point?\)
\(\displaystyle Why \ interject \ imaginary \ numbers?\)
D
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BigGlenntheHeavy said:\(\displaystyle \ What \ is \ the \ point?\) .... Partial fraction will work because:
\(\displaystyle Why \ interject \ imaginary \ numbers?\) .... because I like imagery .... imagine that....
BigGlenntheHeavy
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\(\displaystyle cazza90, \ disregard \ Subhotosh \ Khan \ analysis \ on \ this \ subject, \ as \ he \ obviously \ isn't \ familiar\)
\(\displaystyle with \ the \ integral \ test.\)
\(\displaystyle Additional \ note: \ When \ dealing \ with \ series, \ we \ want \ to \ know \ whether \ said \ series\)
\(\displaystyle diverges \ or \ converges \ and \ if \ it \ converges \ to \ what, \ if \ possible, \ which \ at \ times \ it \ is \ not.\)
\(\displaystyle Ergo. \ the \ integral \ test \ tells \ us \ whether \ a \ series \ diverges \ (\infty) \ or \ converges.\)
\(\displaystyle However, \ if \ it \ converges, \ the \ integral \ test \ doesn't \ tell \ us \ to \ what.\)
\(\displaystyle Don't \ mistake \ the \ value \ of \ the \ integral \ for \ the \ value \ of \ the \ converging \ series,\)
\(\displaystyle for \ example \ \sum_{n=1}^{\infty}\frac{1}{n^2} \ = \ \frac{\pi^2}{6} \ (see \ galactus), \ but \ \int_{1}^{\infty}\frac{1}{x^2}dx \ = \ 1.\)
\(\displaystyle All \ the \ integral \ test \ does \ is \ tell \ you \ whether \ said \ series \ converges \ or \ not, \ but \ doesn't \ tell \ you\)
\(\displaystyle what \ it \ converges \ to \ (assuming \ it \ converges).\)
\(\displaystyle with \ the \ integral \ test.\)
\(\displaystyle Additional \ note: \ When \ dealing \ with \ series, \ we \ want \ to \ know \ whether \ said \ series\)
\(\displaystyle diverges \ or \ converges \ and \ if \ it \ converges \ to \ what, \ if \ possible, \ which \ at \ times \ it \ is \ not.\)
\(\displaystyle Ergo. \ the \ integral \ test \ tells \ us \ whether \ a \ series \ diverges \ (\infty) \ or \ converges.\)
\(\displaystyle However, \ if \ it \ converges, \ the \ integral \ test \ doesn't \ tell \ us \ to \ what.\)
\(\displaystyle Don't \ mistake \ the \ value \ of \ the \ integral \ for \ the \ value \ of \ the \ converging \ series,\)
\(\displaystyle for \ example \ \sum_{n=1}^{\infty}\frac{1}{n^2} \ = \ \frac{\pi^2}{6} \ (see \ galactus), \ but \ \int_{1}^{\infty}\frac{1}{x^2}dx \ = \ 1.\)
\(\displaystyle All \ the \ integral \ test \ does \ is \ tell \ you \ whether \ said \ series \ converges \ or \ not, \ but \ doesn't \ tell \ you\)
\(\displaystyle what \ it \ converges \ to \ (assuming \ it \ converges).\)
D
Deleted member 4993
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BigGlenntheHeavy said:\(\displaystyle cazza90, \ disregard \ Subhotosh \ Khan \ analysis \ on \ this \ subject, \ as \ he \ obviously \ isn't \ familiar\)
\(\displaystyle with \ the \ integral \ test.\)
Mr. Heavy,
It will be useful if you stop "judging" work done by others with your limited knowledge.
I am glad to see that you have stopped insulting students with your imprudent remarks - now you should stop your judgment regarding work produced by others - specially when you do not understand the work.
I do not address others in a confrontational way in open forum - but I feel I need to make an exception for you.
Thanks
Subhotosh Khan
BigGlenntheHeavy
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Hey Subhotosh Khan, the question posed was as follows: "use the integral test to determine convergence of this series (sum of n=1 to infinity): 1/((n^2) - 4n + 5)."
\(\displaystyle Now, \ not \ having \ any \ recourse \ to \ go \ to \ "partial \ fractions", \ you \ resorted \ to \ some \\)
\(\displaystyle mumbo-jumbo \ claptrap \ (imaginary \ numbers) \ instead \ of \ admitting \ your \ error, \ so \ take \ a \ hike \\)
\(\displaystyle as \ you \ are \ nothing \ but \ a \ pedant.\)
\(\displaystyle Note: \ A \ little \ knowledge \ is \ dangerous; \ drink \ deep \ or \ taste \ not \ of \ the \ Pierian \ well.\)
\(\displaystyle Obviously, \ you \ must \ be \ very \ thirsty.\)
\(\displaystyle Now, \ not \ having \ any \ recourse \ to \ go \ to \ "partial \ fractions", \ you \ resorted \ to \ some \\)
\(\displaystyle mumbo-jumbo \ claptrap \ (imaginary \ numbers) \ instead \ of \ admitting \ your \ error, \ so \ take \ a \ hike \\)
\(\displaystyle as \ you \ are \ nothing \ but \ a \ pedant.\)
\(\displaystyle Note: \ A \ little \ knowledge \ is \ dangerous; \ drink \ deep \ or \ taste \ not \ of \ the \ Pierian \ well.\)
\(\displaystyle Obviously, \ you \ must \ be \ very \ thirsty.\)
D
Deleted member 4993
Guest
BigGlenntheHeavy said:Hey Subhotosh Khan, the question posed was as follows: "use the integral test to determine convergence of this series (sum of n=1 to infinity): 1/((n^2) - 4n + 5)."
\(\displaystyle Now, \ not \ having \ any \ recourse \ to \ go \ to \ "partial \ fractions", \ you \ resorted \ to \ some \\)
\(\displaystyle mumbo-jumbo \ claptrap \ (imaginary \ numbers) \ instead \ of \ admitting \ your \ error, \ so \ take \ a \ hike \\)
\(\displaystyle as \ you \ are \ nothing \ but \ a \ pedant.\)
\(\displaystyle Note: \ A \ little \ knowledge \ is \ dangerous; \ drink \ deep \ or \ taste \ not \ of \ the \ Pierian \ well.\)
\(\displaystyle Obviously, \ you \ must \ be \ very \ thirsty.\)
Mr. Heavy,
Exactly right.... the question posed was..
"the question posed was as follows: "use the integral test to determine convergence"
If your knowledge does not extend to integrations involving imaginary numbers - I cannot help it. So before calling a different method wrong - study.
You should learn that there are many ways to solve a problem - do not resort to name-calling when you fail to understand another approach. You are supposed to teach younger generation.
By the way - yes I am thirsty. I make sure I learn something new everyday. I have post-graduate degrees in engineering, have published 20+ scientific paper, have 8 patents published, fluently speak 3 different languages, father of 3 children with post graduate degrees in engineering and I do not go around spewing vile diatribe at younger generation. I can confidently claim that when I leave this earth - I'll leave it a bit improved since I came in.
How about you.....
BigGlenntheHeavy
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Please show your work, indicating exactly where you are stuck - so that we may know where to begin to help you.
\(\displaystyle One \ other \ thing: \ People \ who \ think \ they \ are \ important, aren't.\)
\(\displaystyle One \ other \ thing: \ People \ who \ think \ they \ are \ important, aren't.\)