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questions on radius of convergence

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Find the radius of convergence of the following power series:

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Recall that for a given power series
\\[\\displaystyle\\sum_{n=1}^\\infty a_n,\\]

one can use
\\[\\displaystyle\\lim_{n\\to\\infty}\\left| \\frac{a_n}{a_{n+1}}\\right|\\]

to find its radius of convergence.

a) The radius of convergence is

\\[\\displaystyle\\lim_{n\\to\\infty}\\left| \\frac{a_n}{a_{n+1}}\\right| = \\lim_{n\\to\\infty}\\left| \\frac{\\frac{(-1)^{n+1}}{n}}{\\frac{(-1)^{n+2}}{n+1}}\\right|
= \\lim_{n\\to\\infty}\\frac{\\frac{1}{n}}{\\frac{1}{n+1}} = 1.\\]

So the power series is convergent when $|x|<1$. Next we need to analyse the behaviour at the end points to determine the convergence.

At $x=1$ the series becomes $\\displaystyle\\sum_{n=1}^\\infty \\frac{(-1)^{n+1}}{n}$, which, by alternating series test is convergent.
At $x=-1$ the series becomes $\\displaystyle\\sum_{n=1}^\\infty \\frac{(-1)^{2n+1}}{n}$, which is again convergent by alternating series test.

Therefore the interval of convergence is $[-1,1]$ (or $-1\\leq x \\leq 1$).

b) The radius of convergence is

\\[\\displaystyle\\lim_{n\\to\\infty}\\left| \\frac{a_n}{a_{n+1}}\\right| = \\lim_{n\\to\\infty}\\left| \\frac{\\frac{1}{(2n)!}}{\\frac{1}{(2n+2)!}}\\right|
= \\lim_{n\\to\\infty} (2n+2)(2n+1) = \\infty.\\]

So the power series is convergent for all $x\\in\\mathbb R$.

c) The radius of convergence is

\\[\\displaystyle\\lim_{n\\to\\infty}\\left| \\frac{a_n}{a_{n+1}}\\right| = \\lim_{n\\to\\infty}\\left| \\frac{(-1)^{n+1}}{(-1)^{n+2}}\\right|
= 1.\\]

So the power series is convergent when $|x|<1$. Next we need to analyse the behaviour at the end points to determine the convergence.

At $x=1$ the series becomes $\\displaystyle\\sum_{n=1}^\\infty (-1)^{n+1}$, which is divergent.
At $x=-1$ the series becomes $\\displaystyle\\sum_{n=1}^\\infty (-1)^{n+1}$, which is divergent.

Therefore the interval of convergence is $(-1,1)$ (or $-1< x < 1$).

d) The radius of convergence is

\\[\\displaystyle\\lim_{n\\to\\infty}\\left| \\frac{a_n}{a_{n+1}}\\right| = \\lim_{n\\to\\infty}\\left| \\frac{\\frac{n}{5^{n-1}}}{\\frac{(n+1)}{5^{n}}}\\right|
= 5.\\]

So the power series is convergent when $|x+2|<5$; equivalently when $-7 < x < 3$. Next we need to analyse the behaviour at the end points to determine the convergence.

At $x=3$ the series becomes $\\displaystyle\\sum_{n=1}^\\infty n$, which is divergent.
At $x=-7$ the series becomes $\\displaystyle\\sum_{n=1}^\\infty (-1)^{n}n$, which is divergent.

Therefore the interval of convergence is $(-7,3)$ (or $-7< x < 3$).

e) The radius of convergence is

\\[\\displaystyle\\lim_{n\\to\\infty}\\left| \\frac{a_n}{a_{n+1}}\\right| = \\lim_{n\\to\\infty}\\left| \\frac{\\frac{2^n}{3^n}}{\\frac{2^{n+1}}{3^{n+1}}}\\right|
= \\frac{3}{2}.\\]

So the power series is convergent when $|x|<\\frac{3}{2}$. Next we need to analyse the behaviour at the end points to determine the convergence.

At $x= 3/2 $ the series becomes $\\displaystyle\\sum_{n=1}^\\infty 1$, which is divergent.
At $x=-3/2$ the series becomes $\\displaystyle\\sum_{n=1}^\\infty (-1)^{n}$, which is divergent.

Therefore the interval of convergence is $(-3/2,3/2)$ (or $-3/2< x < 3/2$).

f) The radius of convergence is

\\[\\displaystyle\\lim_{n\\to\\infty}\\left| \\frac{a_n}{a_{n+1}}\\right| = \\lim_{n\\to\\infty}\\left| \\frac{\\frac{(-1)^n}{n^2+1}}{\\frac{(-1)^{n+1}}{(n+1)^2+1}}\\right|
= \\lim_{n\\to\\infty}\\frac{(n+1)^2+1}{n^2+1} = 1.\\]

So the power series is convergent when $|x + 2| < 1$, i.e. $-3< x< -1$. Next we need to analyse the behaviour at the end points to determine the convergence.

At $x= -1 $ the series becomes $\\displaystyle\\sum_{n=1}^\\infty \\frac{(-1)^n}{n^2+1}$, which is convergent, by alternating series test.
At $x=-3$ the series becomes $\\displaystyle\\sum_{n=1}^\\infty \\frac{1}{n^2+1}$, which is convergent by comparison test (as $\\frac{1}{n^2+1} <\\frac{1}{n^2}$, and $\\displaystyle\\sum_{n=1}^\\infty\\frac{1}{n^2}$ converges).

Therefore the interval of convergence is $[-3,-1]$ (or $-3\\leq x \\leq -1$).

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The power series $\\displaystyle\\sum_{n=1}^\\infty \\frac{(-1)^{n+1}x^n}{n}$ has radius of convergence [[0]]

The interval of convergence is [[1]]

The power series $\\displaystyle\\sum_{n=1}^\\infty (-1)^{n+1}x^{2n}$ has radius of convergence [[0]]

The interval of convergence is [[1]]

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