Question

Does the attached series converge or diverge? Note: it is a complex series and not a real series. I've been told the limit comparison test is not a valid approach to solving this problem. sum_{n=1}^inftyfrac{1}{n-i^n}

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asked 2021-01-04
Does the attached series converge or diverge? Note: it is a complex series and not a real series. I've been told the limit comparison test is not a valid approach to solving this problem.
\(\sum_{n=1}^\infty\frac{1}{n-i^n}\)

Answers (1)

2021-01-05
The given series is
\(\sum_{n=1}^\infty\frac{1}{n-i^n}\)
Let us solve the given problem by the help of the Ratio test.
The Ratio test is,
For a series \(a_n,\text{ If }\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|=L\)
Then, the series is convergent if L
the series is divergent if L>1
And if L=1, we have no result or we can say the test fails.
Here \(a_n=\frac{1}{n-i^n}\)
So, \(\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|=\lim_{n\rightarrow\infty}|\frac{n-i^n}{(n+1)-i^{n+1}}|(\frac{\infty}{\infty})\)
\(=\lim_{n\rightarrow\infty}|\frac{1-i^n\ln i}{1-i^{n+1}\ln i}|(\frac{\infty}{\infty})\)
\(=\lim_{n\rightarrow\infty}|\frac{i^n(\ln i)^2}{-i^{n+1}(\ln i)^2}|\)
\(=\frac1i\)
\(=\frac1i(\frac{i}{i})\)
=-i,(as \(i^2=-1\))
Therefore, L=-i
So, the given series is convergent.
ANSWER: The given series is convergent.
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