Solutions - 0190-07

(a) We first check for absolute convergence:

$$\left|\frac{{(-1)}^n}{1+\frac{1}{n}}\right|\gg \gg \frac{1}{1+\frac{1}{n}}\stackrel{n\to \infty }{⟶}\frac{1}{1+0}=1\ne 0$$

This fails the SDT, so the series diverges!

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(b) Notice that $\cos n\pi ={(-1)}^{n-1}$. Check for absolute convergence:

$$\left|\frac{\cos n\pi }{n^3+1}\right|\gg \gg \frac{1}{n^3+1}$$

Then:

$$\frac{1}{n^3+1}<\frac{1}{n^3}$$

Since ${\displaystyle \sum \frac{1}{n^3}}$ converges ($p=3>1$), the DCT says that ${\displaystyle \sum \frac{1}{n^3+1}}$ converges. So the original series converges absolutely and we are done.