Question

Ah! Okay, need help solving 14, and just checking for 16 and 4.

Answer 1

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

$f(x)=\dfrac4{2x+3}=\dfrac{\frac43}{1-\left(-\frac{2x}3\right)}$

Now recall that for $|x|$, we have

$\dfrac1{1-x}=\displaystyle\sum_{n\ge0}x^n$

so that for this function, we get

$f(x)=\dfrac43\displaystyle\sum_{n\ge0}\left(-\frac{2x}3\right)^n$

Because this is a geometric sum, this converges when $\left|-\dfrac{2x}3\right|$, or $|x|$. 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:

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

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

$\displaystyle\lim_{n\to\infty}\left|\frac{(n+2)(-1)^{n+2}x^{n+1}}{(n+1)(-1)^{n+1}x^n}\right|$

The limit reduces to

$\displaystyle|x|\lim_{n\to\infty}\frac{n+2}{n+1}=|x|$

and so the series converges absolutely for $|x|$. 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.