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

Phoebe 2021-08-18 Answered
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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Expert Answer

2k1enyvp
Answered 2021-08-19 Author has 17105 answers
\(\displaystyle{\sum_{{{k}={1}}}^{\infty}}{\left({1}–\frac{{a}}{{k}}\right)}^{{6}}{k}\)
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.
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