# Determine whether the following series converges if “a” is real

Determine whether the following series converges if “a” is real:
$$\displaystyle{\sum_{{{k}={1}}}^{\infty}}{\left({1}–\frac{{a}}{{k}}\right)}^{{6}}{k}$$

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2k1enyvp
$$\displaystyle{\sum_{{{k}={1}}}^{\infty}}{\left({1}–\frac{{a}}{{k}}\right)}^{{6}}{k}$$
if $$\displaystyle\lim_{{{n}\to\infty}}{a}_{{n}}\ne{0},{t}{h}{e}{n}{\sum_{{{k}={1}}}^{\infty}}{a}_{{k}}$$ diverges.
Here $$\displaystyle{a}_{{k}}={\left({1}–\frac{{a}}{{k}}\right)}^{{\frac{{k}}{{6}}}}$$
So $$\displaystyle\lim_{{{k}\to\infty}}{a}_{{k}}=$$
$$\displaystyle=\lim_{{{k}\to\infty}}{\left({1}–\frac{{a}}{{k}}\right)}^{{\frac{{k}}{{6}}}}=$$
$$\displaystyle=\lim_{{{e}^{{{k}\to\infty}}}}{\left(\frac{{k}}{{6}}\right)}\times{\left(-\frac{{a}}{{k}}\right)}=$$
$$\displaystyle=\lim_{{{e}^{{{k}\to\infty}}}}-\frac{{a}}{{6}}=$$
$$\displaystyle\lim_{{{k}\to\infty}}{a}_{{k}}={e}^{{-\frac{{a}}{{6}}}}$$
a is real number, so $$\displaystyle{e}^{{-\frac{{a}}{{6}}}}\ne{0}$$ for real number a.
$$\displaystyle\lim_{{{k}\to\infty}}{a}_{{k}}\ne{0}$$
Thus, by divergent test, given series is divergent.