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Maurice Perry · Answer 1 · 2018-03-05T14:58:48+0000

4. Correct. You also could have used the limit test for divergence for the same conclusion (the summand approaches infinity).

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14. I'm guessing the instructions are the same as for 16. Rewrite as

$f(x)=\frac4{2x+3}=\frac{\frac43}{1-\left(-\frac{2x}3\right)}$

Now recall that for
|x|<1

, we have

$\frac1{1-x}=\displaystyle\sum_(n\ge0)x^n$

so that for this function, we get

$f(x)=\frac43\displaystyle\sum_(n\ge0)\left(-\frac{2x}3\right)^n$

Because this is a geometric sum, this converges when
$\left|-\frac{2x}3\right|<1$ , or
$|x|<\frac32$ . This would be the interval of convergence.

Your hunch about checking the endpoints is correct. Checking is easy in this case, because at the endpoints (-3/2 and 3/2) the series obviously diverges.

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16. This one is kind of tricky, and there's more than one way to do it. The standard method would be to take the antiderivative:

$F(x)=\displaystyle\int f(x)\,\mathrm dx=\int(\mathrm dx)/((1+x)^2)=-\frac1{1+x}+C$

We also have

$\displaystyle-\frac1{1+x}=-\frac1{1-(-x)}=-\sum_(n\ge0)(-x)^n\implies F(x)=C-1-\sum_(n\ge1)(-x)^n$

and differentiating this gives

$f(x)=-\displaystyle\sum_(n\ge1)n(-x)^(n-1)=-\sum_(n\ge0)(n+1)(-x)^n=\sum_(n\ge0)(n+1)(-1)^(n+1)x^n$

By the ratio test, this converges when

$\displaystyle\lim_(n\to\infty)\left|((n+2)(-1)^(n+2)x^(n+1))/((n+1)(-1)^(n+1)x^n)\right|<1$

The limit reduces to

$\displaystyle|x|\lim_(n\to\infty)(n+2)/(n+1)=|x|$

and so the series converges absolutely for
|x|<1

. Checking the endpoints is also easy in this case. The factor of
n+1

is a clear sign that the series will diverge at either extreme.

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