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
[tex]f(x)=\dfrac4{2x+3}=\dfrac{\frac43}{1-\left(-\frac{2x}3\right)}[/tex]
Now recall that for [tex]|x|<1[/tex], we have
[tex]\dfrac1{1-x}=\displaystyle\sum_{n\ge0}x^n[/tex]
so that for this function, we get
[tex]f(x)=\dfrac43\displaystyle\sum_{n\ge0}\left(-\frac{2x}3\right)^n[/tex]
Because this is a geometric sum, this converges when [tex]\left|-\dfrac{2x}3\right|<1[/tex], or [tex]|x|<\dfrac32[/tex]. 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:
[tex]F(x)=\displaystyle\int f(x)\,\mathrm dx=\int\frac{\mathrm dx}{(1+x)^2}=-\frac1{1+x}+C[/tex]
We also have
[tex]\displaystyle-\frac1{1+x}=-\frac1{1-(-x)}=-\sum_{n\ge0}(-x)^n\implies F(x)=C-1-\sum_{n\ge1}(-x)^n[/tex]
and differentiating this gives
[tex]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[/tex]
By the ratio test, this converges when
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{(n+2)(-1)^{n+2}x^{n+1}}{(n+1)(-1)^{n+1}x^n}\right|<1[/tex]
The limit reduces to
[tex]\displaystyle|x|\lim_{n\to\infty}\frac{n+2}{n+1}=|x|[/tex]
and so the series converges absolutely for [tex]|x|<1[/tex]. Checking the endpoints is also easy in this case. The factor of [tex]n+1[/tex] is a clear sign that the series will diverge at either extreme.
