Using Properties of Series

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

1. \(\sum\limits_{k\,=\,4}^{\infty }\dfrac{1}{k} \)
2. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{2}{k}\)
3. \(\sum\limits_{k\,=\,1}^{\infty }\,\left( \dfrac{1}{2^{k-1}}+\dfrac{1}{3^{k-1}}\right) \)

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 \[ \sum\limits_{k\,=\,1}^{\infty }\,\left( \dfrac{1}{2^{k-1}}+\dfrac{1}{3^{k-1}} \right) =\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{2^{k-1}} +\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{3^{k-1}}=\frac{1}{1-\dfrac{1}{2}}+ \frac{1}{1-\dfrac{1}{3}}=2+\frac{3}{2}=\dfrac{7}{2} \]