Question #122305

Find the interval of convergence of P∞

n=0

x

n

n+4 .

Expert's answer

Given series is n=0xnn+4.\sum^\infin_{n=0} \frac{x^n}{n+4}.

Let an=1n+4a_n = \frac{1}{n+4}

    an+1=1n+5\implies a_{n+1} = \frac{1}{n+5}

So, Radius of convergence is R=limnanan+1=limnn+5n+4=limn1+5/n1+4/n=1R= \lim_{n \to \infin} \frac{a_n}{a_{n+1}} = \lim_{n \to \infin} \frac{n+5}{n+4} = \lim_{n \to \infin} \frac{1+5/n}{1+4/n} =1 .

So, series is convergence on (1,1)(-1,1) is convergent and on (,1)(1,)(-\infin,-1) \cup(1,\infin) series is divergent.

Now, at x=1x=1 series becomes n=01n+4\sum^\infin_{n=0} \frac{1}{n+4} which is divergent by p-test.

At x=1x=-1 series becomes n=0(1)nn+4\sum^\infin_{n=0} \frac{(-1)^n}{n+4}, which is convergent by Leibniz's test because <1n+4><\frac{1}{n+4}> is decreasing sequence and1n+40\frac{1}{n+4} \to 0 as nn\to \infin.

So interval of convergence is [1,1)[-1,1) .


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