Determine whether each series converges or diverges. If it converges, find its sum.

Solution (a) Except for the first three terms, the series $\sum\limits_{k\,=\,4}^{\infty }\dfrac{1}{k}=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{ 1}{6}+\cdots$ is identical to the harmonic series, which diverges. So, it follows that $\sum\limits_{k\,=\,4}^{\infty }\dfrac{1}{k}$ also diverges.

(b) $\sum\limits_{k\,=\,1}^{\infty }\dfrac{2}{k}\,{=}\,\sum\limits_{k\,=\,1}^{\infty }\,\left(2\cdot \dfrac{1}{k}\right)$. Since the harmonic series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}$ diverges, the series $\sum\limits_{k\,=\,1}^{\infty }\,\left( 2\cdot \dfrac{1}{k}\right)=\sum\limits_{k\,=\,1}^{\infty }\dfrac{2}{k}$ diverges.

(c) Since the series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{ 2^{k-1}}$ and the series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{3^{k-1}}$ are both convergent geometric series, the series defined by $\sum\limits_{k\,=\,1}^{\infty }\,\left( \dfrac{1}{2^{k-1}}+\dfrac{1}{3^{k-1}} \right)$ is also convergent. The sum is $$\sum\limits_{k\,=\,1}^{\infty }\,\left( \dfrac{1}{2^{k-1}}+\dfrac{1}{3^{k-1}} \right) =\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{2^{k-1}} +\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{3^{k-1}}=\frac{1}{1-\dfrac{1}{2}}+ \frac{1}{1-\dfrac{1}{3}}=2+\frac{3}{2}=\dfrac{7}{2}$$