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KarmaSquared
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Radius of convergence, interval of convergence
Find the radius of convergence and the interval of convergence of the following series.
a) [tex]\sum_{n=0}^\infty \frac{x^n}{(n^2)+1}[/tex]
c) [tex]\sum_{n=2}^\infty \frac{x^n}{ln(n)}[/tex]
e) [tex]\sum_{n=1}^\infty \frac{n!x^n}{n^2}[/tex]
f) [tex]\sum_{n=1}^\infty \frac{n!2^nx^n}{n^n}[/tex]
I think I use the Ratio test:
[tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = L < 1[/tex]
a) Let [tex]{a_n} = \frac{x^n}{(n^2)+1}[/tex]
So:
[tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{x^(n+1)}{((n+1)^2)+1} * \frac{n^2+1}{x^n} = \lim_{n\rightarrow\infty} \frac{x*n^2+1}{n^2+2n+2} = \lim_{n\rightarrow\infty} x[/tex]
Radius: 1, Interval of convergence (-1, 1)
c) Let [tex]{a_n} = \frac{x^n}{ln(n)}[/tex]
So:
[tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{x^(n+1)}{ln(n+1)} * \frac{ln(n)}{x^n} = \lim_{n\rightarrow\infty} \frac{x*ln(n)}{ln(n+1)} = \lim_{n\rightarrow\infty} x[/tex]
Radius: 1, Interval of convergence (-1, 1)
e) Let [tex]{a_n} = \frac{n!x^n}{n^2}[/tex]
So:
[tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{(n+1)!x^(n+1)}{(n+1)^2} * \frac{n^2}{n!x^n} = \lim_{n\rightarrow\infty} \frac{n^2x}{n+1} = \lim_{n\rightarrow\infty} x[/tex]
Radius: 1, Interval of convergence (-1, 1)
f) Let [tex]{a_n} = \frac{n!2^nx^n}{n^n}[/tex]
So:
[tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{n+1!2^(n+1)x^(n+1)}{(n+1)^(n+1)} * \frac{n^n}{n!2^nx^n} = \lim_{n\rightarrow\infty} 2x * (\frac{n}{n+1})^n = \lim_{n\rightarrow\infty} 2x[/tex]
Radius: 1/2, Interval of convergence (-0.5, 0.5)
Homework Statement
Find the radius of convergence and the interval of convergence of the following series.
a) [tex]\sum_{n=0}^\infty \frac{x^n}{(n^2)+1}[/tex]
c) [tex]\sum_{n=2}^\infty \frac{x^n}{ln(n)}[/tex]
e) [tex]\sum_{n=1}^\infty \frac{n!x^n}{n^2}[/tex]
f) [tex]\sum_{n=1}^\infty \frac{n!2^nx^n}{n^n}[/tex]
Homework Equations
I think I use the Ratio test:
[tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = L < 1[/tex]
The Attempt at a Solution
a) Let [tex]{a_n} = \frac{x^n}{(n^2)+1}[/tex]
So:
[tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{x^(n+1)}{((n+1)^2)+1} * \frac{n^2+1}{x^n} = \lim_{n\rightarrow\infty} \frac{x*n^2+1}{n^2+2n+2} = \lim_{n\rightarrow\infty} x[/tex]
Radius: 1, Interval of convergence (-1, 1)
c) Let [tex]{a_n} = \frac{x^n}{ln(n)}[/tex]
So:
[tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{x^(n+1)}{ln(n+1)} * \frac{ln(n)}{x^n} = \lim_{n\rightarrow\infty} \frac{x*ln(n)}{ln(n+1)} = \lim_{n\rightarrow\infty} x[/tex]
Radius: 1, Interval of convergence (-1, 1)
e) Let [tex]{a_n} = \frac{n!x^n}{n^2}[/tex]
So:
[tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{(n+1)!x^(n+1)}{(n+1)^2} * \frac{n^2}{n!x^n} = \lim_{n\rightarrow\infty} \frac{n^2x}{n+1} = \lim_{n\rightarrow\infty} x[/tex]
Radius: 1, Interval of convergence (-1, 1)
f) Let [tex]{a_n} = \frac{n!2^nx^n}{n^n}[/tex]
So:
[tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{n+1!2^(n+1)x^(n+1)}{(n+1)^(n+1)} * \frac{n^n}{n!2^nx^n} = \lim_{n\rightarrow\infty} 2x * (\frac{n}{n+1})^n = \lim_{n\rightarrow\infty} 2x[/tex]
Radius: 1/2, Interval of convergence (-0.5, 0.5)
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